Equations and Symmetry P(X ) = 0 Where P Is a Polynomial

Equations and Symmetry P(X ) = 0 where P is a polynomial

2 I Quadratic equations: X + aX + b = 0 3 2 I Cubic equations: X + aX + bX + c = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on.

Polynomial equation: 2 I Quadratic equations: X + aX + b = 0 3 2 I Cubic equations: X + aX + bX + c = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on.

Polynomial equation:

P(X ) = 0 where P is a polynomial 3 2 I Cubic equations: X + aX + bX + c = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on.

Polynomial equation:

P(X ) = 0 where P is a polynomial

2 I Quadratic equations: X + aX + b = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on.

Polynomial equation:

P(X ) = 0 where P is a polynomial

2 I Quadratic equations: X + aX + b = 0 3 2 I Cubic equations: X + aX + bX + c = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on.

Polynomial equation:

P(X ) = 0 where P is a polynomial

2 I Quadratic equations: X + aX + b = 0 3 2 I Cubic equations: X + aX + bX + c = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 Polynomial equation:

P(X ) = 0 where P is a polynomial

2 I Quadratic equations: X + aX + b = 0 3 2 I Cubic equations: X + aX + bX + c = 0 4 3 2 I Quartic equations: X + aX + bX + cX + d = 0 5 4 3 2 I Quintic equations: X + aX + bX + cX + dX + e = 0

. . . and so on. Quadratic equations: a method for solving was known to the Babylonians (ca. 2000 BCE)

“Given the sum and product of two numbers, ﬁnd the numbers.”

If p + q = A and pq = B, then p and q are the solutions of the equation X 2 − AX + B = 0

A bit of history a method for solving was known to the Babylonians (ca. 2000 BCE)

“Given the sum and product of two numbers, ﬁnd the numbers.”

If p + q = A and pq = B, then p and q are the solutions of the equation X 2 − AX + B = 0

A bit of history

Quadratic equations: “Given the sum and product of two numbers, ﬁnd the numbers.”

If p + q = A and pq = B, then p and q are the solutions of the equation X 2 − AX + B = 0

A bit of history

Quadratic equations: a method for solving was known to the Babylonians (ca. 2000 BCE) If p + q = A and pq = B, then p and q are the solutions of the equation X 2 − AX + B = 0

A bit of history

Quadratic equations: a method for solving was known to the Babylonians (ca. 2000 BCE)

“Given the sum and product of two numbers, ﬁnd the numbers.” A bit of history

Quadratic equations: a method for solving was known to the Babylonians (ca. 2000 BCE)

“Given the sum and product of two numbers, ﬁnd the numbers.”

If p + q = A and pq = B, then p and q are the solutions of the equation X 2 − AX + B = 0 particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: , Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE) , Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100) No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200) Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): “The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell) —that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions.

Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals” Cubic equations: particular equations were considered by Diophantus (ca. 250 CE), Omar Kayy´am(ca. 1100), Leonardo of Pisa (a.k.a. “Fibonacci,” 1200)

No general method of solution known

Scipione del Ferro (ca. 1500): X 3 + bX = c

Niccol`oTartaglia (1535): X 3 + aX 2 = c

“The ﬁrst clear advance in mathematics since the time of the Greeks.” (J. Stillwell)

These are “solutions by radicals”—that is, they express the solutions in terms of the coeﬃcients using only the four algebraic operations and root extractions. Lodovico Ferrari (1540); also a “solution by radicals.”

Here’s the quartic formula:

X =

v 1 1 √ u 2 3 2 2 2 3 3 2 2 2 3 − a − 1 u a2 − 2b + 2 3 (b −3ac+12d) + 2b −9abc+27c +27a d−72bd+ −4(b −3ac+12d) +(2b −9abc+27c +27a d−72bd) 4 2 t 4 3 √ 1 54 3 2b3−9abc+27c2+27a2d−72bd+ −4(b2−3ac+12d)3+(2b3−9abc+27c2+27a2d−72bd)2 3

v 1 1 √ u 2 3 2 3 u a2 4b 2 3 (b −3ac+12d) 2b3−9abc+27c2+27a2d−72bd+ −4(b2−3ac+12d) +(2b3−9abc+27c2+27a2d−72bd) u 2 − 3 − √ 1 − 54 u 3 2 3 u 3 2b3−9abc+27c2+27a2d−72bd+ −4(b2−3ac+12d) +(2b3−9abc+27c2+27a2d−72bd) u u u u + u 3 u −a +4ab−8c u − u v u u √ 1 1 3 2 ! 3 u u 2 2 3 (b2−3ac+12d) 2b3−9abc+27c2+27a2d−72bd+ −4(b2−3ac+12d) +(2b3−9abc+27c2+27a2d−72bd) u 4u a − 2b + + u 4 3 √ 1 54 t t 3 2 3 3 2b3−9abc+27c2+27a2d−72bd+ −4(b2−3ac+12d) +(2b3−9abc+27c2+27a2d−72bd)

Quartic equations ; also a “solution by radicals.”

Here’s the quartic formula:

X =

Quartic equations

Lodovico Ferrari (1540) Here’s the quartic formula:

X =

Quartic equations

Lodovico Ferrari (1540); also a “solution by radicals.” Quartic equations

Lodovico Ferrari (1540); also a “solution by radicals.”

Here’s the quartic formula:

X =

Which brings us to. . .

Joseph–Louis Lagrange (1736–1813)

. . . the ﬁrst to recognize the role of symmetry in solving equations.

Quintic equations , but without success.

Which brings us to. . .

Joseph–Louis Lagrange (1736–1813)

. . . the ﬁrst to recognize the role of symmetry in solving equations.

Quintic equations: ﬁnding a solution by radicals became a major focus of algebra for the next two centuries Which brings us to. . .

Joseph–Louis Lagrange (1736–1813)

. . . the ﬁrst to recognize the role of symmetry in solving equations.

Quintic equations: ﬁnding a solution by radicals became a major focus of algebra for the next two centuries, but without success. Joseph–Louis Lagrange (1736–1813)

. . . the ﬁrst to recognize the role of symmetry in solving equations.

Quintic equations: ﬁnding a solution by radicals became a major focus of algebra for the next two centuries, but without success.

Which brings us to...... the ﬁrst to recognize the role of symmetry in solving equations.

Quintic equations: ﬁnding a solution by radicals became a major focus of algebra for the next two centuries, but without success.

Which brings us to. . .

Joseph–Louis Lagrange (1736–1813) Quintic equations: ﬁnding a solution by radicals became a major focus of algebra for the next two centuries, but without success.

Which brings us to. . .

Joseph–Louis Lagrange (1736–1813)

. . . the ﬁrst to recognize the role of symmetry in solving equations. Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background: X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0 , then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r) X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents: X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3 = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) constant :(−p)(−q)(−r) = −pqr = −12

First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4 First, some background:

Cubic equations

X 3 + 3X 2 − 4X − 12 = 0

If the solutions are X = p, X = q and X = r, then

X 3 + 3X 2 − 4X − 12 = (X − p)(X − q)(X − r)

Multiplying and comparing coeﬃcents:

X 2 :(−p) + (−q) + (−r) = −(p + q + r) = 3

X :(−p)(−q) + (−q)(−r) + (−r)(−p) = pq + qr + rp = −4

constant :(−p)(−q)(−r) = −pqr = −12 σ2(x, y, z) = xy + yz + zx

σ3(x, y, z) = xyz

then

2 σ1(p, q, r) = − the coeﬃcient of X = −3 σ2(p, q, r) = the coeﬃcient of X = −4 σ3(p, q, r) = − the constant term = 12

and the problem of solving our cubic can be restated as:

given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), ﬁnd the values of p, q and r.

If we deﬁne the functions

σ1(x, y, z) = x + y + z σ3(x, y, z) = xyz

then

2 σ1(p, q, r) = − the coeﬃcient of X = −3 σ2(p, q, r) = the coeﬃcient of X = −4 σ3(p, q, r) = − the constant term = 12

and the problem of solving our cubic can be restated as:

given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), ﬁnd the values of p, q and r.

If we deﬁne the functions

σ1(x, y, z) = x + y + z

σ2(x, y, z) = xy + yz + zx then

2 σ1(p, q, r) = − the coeﬃcient of X = −3 σ2(p, q, r) = the coeﬃcient of X = −4 σ3(p, q, r) = − the constant term = 12

and the problem of solving our cubic can be restated as:

given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), ﬁnd the values of p, q and r.

If we deﬁne the functions

σ1(x, y, z) = x + y + z

σ2(x, y, z) = xy + yz + zx

σ3(x, y, z) = xyz and the problem of solving our cubic can be restated as:

given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), ﬁnd the values of p, q and r.

If we deﬁne the functions

σ1(x, y, z) = x + y + z

σ2(x, y, z) = xy + yz + zx

σ3(x, y, z) = xyz then

2 σ1(p, q, r) = − the coefficient of X = −3 σ2(p, q, r) = the coefficient of X = −4 σ3(p, q, r) = − the constant term = 12 given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), find the values of p, q and r.

If we deﬁne the functions

σ1(x, y, z) = x + y + z

σ2(x, y, z) = xy + yz + zx

σ3(x, y, z) = xyz then

2 σ1(p, q, r) = − the coefficient of X = −3 σ2(p, q, r) = the coefficient of X = −4 σ3(p, q, r) = − the constant term = 12 and the problem of solving our cubic can be restated as: If we define the functions

σ1(x, y, z) = x + y + z

σ2(x, y, z) = xy + yz + zx

σ3(x, y, z) = xyz then

2 σ1(p, q, r) = − the coeﬃcient of X = −3 σ2(p, q, r) = the coeﬃcient of X = −4 σ3(p, q, r) = − the constant term = 12 and the problem of solving our cubic can be restated as:

given the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), ﬁnd the values of p, q and r. no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca

= σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca

= σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value. Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca

= σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca = ab + bc + ca

= σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca Because of this, σ1, σ2 and σ3 are called symmetric functions.

The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca

= σ2(a, b, c) The functions σ1, σ2 and σ3 have the following interesting property: no matter how you permute their inputs, you get the same value.

For example, if a, b and c are any three numbers,

σ2(a, b, c) = ab + bc + ca

Also,

σ2(c, b, a) = cb + ba + ac = ab + bc + ca

= σ2(a, b, c)

Because of this, σ1, σ2 and σ3 are called symmetric functions. There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables. σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables: σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y . . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables: σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw σ4(x, y, z, w) = xyzw

σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw σ1, σ2 and σ3 are the “basic” symmetric functions of three variables.

There are two basic symmetric functions of two variables:

σ1(x, y) = x + y

σ2(x, y) = xy

. . . and four basic symmetric functions of four variables:

σ1(x, y, z, w) = x + y + z + w

σ2(x, y, z, w) = xy + xz + xw + yz + yw + zw

σ3(x, y, z, w) = xyz + xyw + xzw + yzw

σ4(x, y, z, w) = xyzw σi (x1, x2, ··· , xn) = the sum of all possible products of i of the variables x1, x2, x3, ··· , xn.

Isaac Newton:

Any polynomial in n variables which is symmetric can be expressed as a polynomial in the basic symmetric functions of the n variables.

In general, for any postive integers n and i with i ≤ n, Isaac Newton:

Any polynomial in n variables which is symmetric can be expressed as a polynomial in the basic symmetric functions of the n variables.

In general, for any postive integers n and i with i ≤ n,

σi (x1, x2, ··· , xn) = the sum of all possible products of i of the variables x1, x2, x3, ··· , xn. Any polynomial in n variables which is symmetric can be expressed as a polynomial in the basic symmetric functions of the n variables.

In general, for any postive integers n and i with i ≤ n,

σi (x1, x2, ··· , xn) = the sum of all possible products of i of the variables x1, x2, x3, ··· , xn.

Isaac Newton: In general, for any postive integers n and i with i ≤ n,

σi (x1, x2, ··· , xn) = the sum of all possible products of i of the variables x1, x2, x3, ··· , xn.

Isaac Newton:

Any polynomial in n variables which is symmetric can be expressed as a polynomial in the basic symmetric functions of the n variables. 2 2 σ1(x, y, z) = (x + y + z) = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 2 2 = x + y + z + 2σ2(x, y, z)

2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z).

Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z)

For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z. = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 2 2 = x + y + z + 2σ2(x, y, z)

2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z).

Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z)

For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z.

2 2 σ1(x, y, z) = (x + y + z) 2 2 2 = x + y + z + 2σ2(x, y, z)

2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z).

Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z)

For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z.

2 2 σ1(x, y, z) = (x + y + z) = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z).

Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z)

For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z.

2 2 σ1(x, y, z) = (x + y + z) = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 2 2 = x + y + z + 2σ2(x, y, z) Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z)

For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z.

2 2 σ1(x, y, z) = (x + y + z) = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 2 2 = x + y + z + 2σ2(x, y, z)

2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z). For example: F (x, y, z) = x2 + y 2 + z2 is a polynomial which is symmetric in x, y and z.

2 2 σ1(x, y, z) = (x + y + z) = x2 + y 2 + z2 + 2xy + 2yz + 2zx 2 2 2 = x + y + z + 2σ2(x, y, z)

2 and so F (x, y, z) = σ1(x, y, z) − 2σ2(x, y, z).

Similarly,

3 3 3 3 x + y + z = σ1(x, y, z) − 3σ1(x, y, z)σ2(x, y, z) + 3σ3(x, y, z) 2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) = 17

and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27

. . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial = (−3)2 − 2(−4) = 17

and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27

. . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = 17

and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27

. . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27

. . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) = 17 = −27

. . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) = 17 and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) . . . though we don’t yet know the values of p, q and r!

So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) = 17 and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27 So for the roots p, q and r of our cubic polynomial

2 2 2 2 p + q + r = σ1(p, q, r) − 2σ2(p, q, r) = (−3)2 − 2(−4) = 17 and

p3 + q3 + r 3 = (−3)3 − 3(−3)(−4) + 3(12) = −27

. . . though we don’t yet know the values of p, q and r! g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1.

g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange: √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1.

g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1.

g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. , but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z) = g(x, y, z)3

Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 Back to Lagrange:

g(x, y, z) = x + ωy + ω2z √ 1 3 where ω = − 2 + 2 i. ω is a primitive cube root of 1—which is to say, ω 6= 1 but ω3 = 1. g is completely unsymmetric, but Lagrange noticed that

g(z, x, y) = z + ωx + ω2y = ω(ω2z + x + ωy) (since ω3 = 1) = ω(x + ωy + ω2z) = ωg(x, y, z)

That means that g(z, x, y)3 = ω3g(x, y, z)3 = g(x, y, z)3 In fact, g(x, y, z)3 = g(z, x, y)3 = g(y, z, x)3 and g(x, z, y)3 = g(y, x, z)3 = g(z, y, x)3

but in general, g(x, y, z)3 6= g(x, z, y)3.

That is, g 3 takes on only two values as you permute the inputs in all six possible ways.

So g 3 is at least a bit more symmetric than g. and g(x, z, y)3 = g(y, x, z)3 = g(z, y, x)3

but in general, g(x, y, z)3 6= g(x, z, y)3.

That is, g 3 takes on only two values as you permute the inputs in all six possible ways.

So g 3 is at least a bit more symmetric than g.

In fact, g(x, y, z)3 = g(z, x, y)3 = g(y, z, x)3 but in general, g(x, y, z)3 6= g(x, z, y)3.

That is, g 3 takes on only two values as you permute the inputs in all six possible ways.

So g 3 is at least a bit more symmetric than g.

In fact, g(x, y, z)3 = g(z, x, y)3 = g(y, z, x)3 and g(x, z, y)3 = g(y, x, z)3 = g(z, y, x)3 That is, g 3 takes on only two values as you permute the inputs in all six possible ways.

So g 3 is at least a bit more symmetric than g.

In fact, g(x, y, z)3 = g(z, x, y)3 = g(y, z, x)3 and g(x, z, y)3 = g(y, x, z)3 = g(z, y, x)3 but in general, g(x, y, z)3 6= g(x, z, y)3. So g 3 is at least a bit more symmetric than g.

In fact, g(x, y, z)3 = g(z, x, y)3 = g(y, z, x)3 and g(x, z, y)3 = g(y, x, z)3 = g(z, y, x)3 but in general, g(x, y, z)3 6= g(x, z, y)3.

That is, g 3 takes on only two values as you permute the inputs in all six possible ways. B(x, y, z) = g(x, y, z)3g(x, z, y)3

are symmetric functions of x, y and z, so by Newton’s theorem, they can be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

(Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y)

and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) )

It follows that the functions

A(x, y, z) = g(x, y, z)3 + g(x, z, y)3 are symmetric functions of x, y and z, so by Newton’s theorem, they can be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

(Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y)

and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) )

It follows that the functions

A(x, y, z) = g(x, y, z)3 + g(x, z, y)3 B(x, y, z) = g(x, y, z)3g(x, z, y)3 are symmetric functions of x, y and z, so by Newton’s theorem, they can be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

(Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y)

and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) )

It follows that the functions

A(x, y, z) = g(x, y, z)3 + g(x, z, y)3 B(x, y, z) = g(x, y, z)3g(x, z, y)3 , so by Newton’s theorem, they can be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

(Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y)

and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) )

It follows that the functions

A(x, y, z) = g(x, y, z)3 + g(x, z, y)3 B(x, y, z) = g(x, y, z)3g(x, z, y)3 are symmetric functions of x, y and z (Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y)

and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) )

It follows that the functions

(Note that

3 3 A(x, y, z) = σ1 g(x, y, z) , g(x, z, y) and 3 3 B(x, y, z) = σ2 g(x, y, z) , g(x, z, y) ) (and in my case, a lot of help from Maple R ) to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2

Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261

It takes a little work to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2

Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261

It takes a little work (and in my case, a lot of help from Maple R ) 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2

Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261

It takes a little work (and in my case, a lot of help from Maple R ) to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261

It takes a little work (and in my case, a lot of help from Maple R ) to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2 A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261

It takes a little work (and in my case, a lot of help from Maple R ) to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2

Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: It takes a little work (and in my case, a lot of help from Maple R ) to show

3 A = 2σ1 − 9σ1σ2 + 27σ3 6 4 2 2 3 B = σ1 − 9σ1σ2 + 27σ1σ2 − 27σ2

Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate: A(p, q, r) = g(p, q, r)3 + g(p, r, q)3 = 162 B(p, q, r) = g(p, q, r)3g(p, r, q)3 = 9261 This equation is called the “resolvent quadratic” of our cubic equation.

Solving it gives the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i

So g(p, q, r)3 and g(p, r, q)3 are the roots of the quadratic equation X 2 − 162X + 9261 = 0 Solving it gives the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i

So g(p, q, r)3 and g(p, r, q)3 are the roots of the quadratic equation X 2 − 162X + 9261 = 0

This equation is called the “resolvent quadratic” of our cubic equation. So g(p, q, r)3 and g(p, r, q)3 are the roots of the quadratic equation X 2 − 162X + 9261 = 0

This equation is called the “resolvent quadratic” of our cubic equation.

Solving it gives the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i So g(p, q, r)3 and g(p, r, q)3 are the roots of the quadratic equation X 2 − 162X + 9261 = 0

This equation is called the “resolvent quadratic” of our cubic equation.

Solving it gives the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i This gives us two equations in p, q and r: √ 2 g(p, q, r) = p + ωq + ω r = −3 + 2√3i g(p, r, q) = p + ωr + ω2q = −3 − 2 3i

Note that these equations are linear!

. . . and taking cube roots: √ g(p, q, r) = −3 + 2 3i √ g(p, r, q) = −3 − 2 3i Note that these equations are linear!

. . . and taking cube roots: √ g(p, q, r) = −3 + 2 3i √ g(p, r, q) = −3 − 2 3i

This gives us two equations in p, q and r: √ 2 g(p, q, r) = p + ωq + ω r = −3 + 2√3i g(p, r, q) = p + ωr + ω2q = −3 − 2 3i . . . and taking cube roots: √ g(p, q, r) = −3 + 2 3i √ g(p, r, q) = −3 − 2 3i

This gives us two equations in p, q and r: √ 2 g(p, q, r) = p + ωq + ω r = −3 + 2√3i g(p, r, q) = p + ωr + ω2q = −3 − 2 3i

Note that these equations are linear! So we actually have a system of three linear equations in the three unknowns p, q and r:       1 1 1 p −3 √ 2  1 ω ω   q  =  −3 + 2√3i  1 ω2 ω r −3 − 2 3i

These equations are independent. . .

But we also know

p + q + r = σ1(p, q, r) = −3       1 1 1 p −3 √ 2  1 ω ω   q  =  −3 + 2√3i  1 ω2 ω r −3 − 2 3i

These equations are independent. . .

But we also know

p + q + r = σ1(p, q, r) = −3

So we actually have a system of three linear equations in the three unknowns p, q and r: These equations are independent. . .

But we also know

p + q + r = σ1(p, q, r) = −3

So we actually have a system of three linear equations in the three unknowns p, q and r:       1 1 1 p −3 √ 2  1 ω ω   q  =  −3 + 2√3i  1 ω2 ω r −3 − 2 3i But we also know

p + q + r = σ1(p, q, r) = −3

So we actually have a system of three linear equations in the three unknowns p, q and r:       1 1 1 p −3 √ 2  1 ω ω   q  =  −3 + 2√3i  1 ω2 ω r −3 − 2 3i

These equations are independent. . .  1 1 1   −3  1 √ = 1 ω2 ω −3 + 2 3i 3    √  1 ω ω2 −3 − 2 3i

 −3  =  2  −2

So the solutions of our equation are X = −3, X = 2 and X = −2.

. . . and so we can solve them:    −1   p 1 1 1 −3 √ 2  q  =  1 ω ω   −3 + 2√3i  r 1 ω2 ω −3 − 2 3i  −3  =  2  −2

So the solutions of our equation are X = −3, X = 2 and X = −2.

. . . and so we can solve them:    −1   p 1 1 1 −3 √ 2  q  =  1 ω ω   −3 + 2√3i  r 1 ω2 ω −3 − 2 3i

 1 1 1   −3  1 √ = 1 ω2 ω −3 + 2 3i 3    √  1 ω ω2 −3 − 2 3i So the solutions of our equation are X = −3, X = 2 and X = −2.

. . . and so we can solve them:    −1   p 1 1 1 −3 √ 2  q  =  1 ω ω   −3 + 2√3i  r 1 ω2 ω −3 − 2 3i

 1 1 1   −3  1 √ = 1 ω2 ω −3 + 2 3i 3    √  1 ω ω2 −3 − 2 3i

 −3  =  2  −2 . . . and so we can solve them:    −1   p 1 1 1 −3 √ 2  q  =  1 ω ω   −3 + 2√3i  r 1 ω2 ω −3 − 2 3i

 1 1 1   −3  1 √ = 1 ω2 ω −3 + 2 3i 3    √  1 ω ω2 −3 − 2 3i

 −3  =  2  −2

So the solutions of our equation are X = −3, X = 2 and X = −2. 3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps: g(x, y, z)3 = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

I Find a power of a linear function: which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways. , and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z). I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3 I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3 I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r

Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q) Key steps:

3 I Find a power of a linear function: g(x, y, z) = (x + ωy + ω2z)3 which takes on only two values as you permute the inputs in all six ways.

I The sum (σ1) and product (σ2) of these two values are symmetric functions of x, y and z, and can therefore be expressed in terms of σ1(x, y, z), σ2(x, y, z) and σ3(x, y, z).

I Since we know the values of σ1(p, q, r), σ2(p, q, r) and σ3(p, q, r), we can calculate the values of g(p, q, r)3 + g(p, r, q)3 and g(p, q, r)3g(p, r, q)3

I Solve a quadratic equation (the “resolvent quadratic”) to ﬁnd the values of g(p, q, r)3 and g(p, r, q)3

I Take cube roots to ﬁnd the values of g(p, q, r) and g(p, r, q)

I Combined with the fact that p + q + r = σ1(p, q, r), this gives three linear equations in p, q and r we could have “discovered” the solution to quadratic equations by a similar process:

X 2 − 5X + 3 = 0

If X = p and X = q are the solutions of the equation, then

σ1(p, q) = p + q = −(−5) σ2(p, q) = pq = 3

Inspired by our success with the cubic, we look for a power of a linear function of two variables which takes on only one value as we permute the inputs in all two possible ways:

Come to think of it. . . X 2 − 5X + 3 = 0

If X = p and X = q are the solutions of the equation, then

σ1(p, q) = p + q = −(−5) σ2(p, q) = pq = 3

Inspired by our success with the cubic, we look for a power of a linear function of two variables which takes on only one value as we permute the inputs in all two possible ways:

Come to think of it. . . we could have “discovered” the solution to quadratic equations by a similar process: If X = p and X = q are the solutions of the equation, then

σ1(p, q) = p + q = −(−5) σ2(p, q) = pq = 3

Inspired by our success with the cubic, we look for a power of a linear function of two variables which takes on only one value as we permute the inputs in all two possible ways:

Come to think of it. . . we could have “discovered” the solution to quadratic equations by a similar process:

X 2 − 5X + 3 = 0 Inspired by our success with the cubic, we look for a power of a linear function of two variables which takes on only one value as we permute the inputs in all two possible ways:

Come to think of it. . . we could have “discovered” the solution to quadratic equations by a similar process:

X 2 − 5X + 3 = 0

If X = p and X = q are the solutions of the equation, then

σ1(p, q) = p + q = −(−5) σ2(p, q) = pq = 3 Come to think of it. . . we could have “discovered” the solution to quadratic equations by a similar process:

X 2 − 5X + 3 = 0

If X = p and X = q are the solutions of the equation, then

σ1(p, q) = p + q = −(−5) σ2(p, q) = pq = 3

Inspired by our success with the cubic, we look for a power of a linear function of two variables which takes on only one value as we permute the inputs in all two possible ways: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value. (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R ) = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy (Anyone recognize the discriminant of the quadratic?)

f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y) f (x, y)2 = (x − y)2 is such a function: f (x, y) = x − y is linear and f 2 takes on only one value.

Since f (x, y)2 is actually symmetric, it can be expressed in terms of σ1(x, y) and σ2(x, y) (and this time, we don’t even need Maple R )

(x − y)2 = x2 − 2xy + y 2 = x2 + 2xy + y 2 − 4xy = (x + y)2 − 4xy 2 = σ1(x, y) − 4σ2(x, y)

(Anyone recognize the discriminant of the quadratic?) = 52 − 4(3) = 13

√ Taking the square root gives p − q = 13

Combined with the fact that p + q = 5 gives two linear equations in two unknowns:

p + q = 5 √ p − q = 13 √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2

So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) √ Taking the square root gives p − q = 13

Combined with the fact that p + q = 5 gives two linear equations in two unknowns:

p + q = 5 √ p − q = 13 √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2

So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) = 52 − 4(3) = 13 Combined with the fact that p + q = 5 gives two linear equations in two unknowns:

p + q = 5 √ p − q = 13 √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2

So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) = 52 − 4(3) = 13

√ Taking the square root gives p − q = 13 p + q = 5 √ p − q = 13 √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2

So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) = 52 − 4(3) = 13

√ Taking the square root gives p − q = 13

Combined with the fact that p + q = 5 gives two linear equations in two unknowns: √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2

So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) = 52 − 4(3) = 13

√ Taking the square root gives p − q = 13

Combined with the fact that p + q = 5 gives two linear equations in two unknowns:

p + q = 5 √ p − q = 13 So for the roots p and q of our quadratic:

2 2 (p − q) = σ1(p, q) − 4σ2(p, q) = 52 − 4(3) = 13

√ Taking the square root gives p − q = 13

Combined with the fact that p + q = 5 gives two linear equations in two unknowns:

p + q = 5 √ p − q = 13 √ √ 5+ 13 5− 13 which we solve to get p = 2 and q = 2 2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we f (x, y)2 = (x − y)2 which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

I Found a power of a linear function: which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways. , which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2 I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2 I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value) I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q

So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q) So again, we

2 2 I Found a power of a linear function: f (x, y) = (x − y) which assumes only one value as you permute the inputs in all two possible ways.

I Expressed this one value in terms of σ1 and σ2, which allows us to ﬁnd the value of f (p, q)2

I (This time, there’s no resolvent equation, since there’s only one value)

I Took the square root to ﬁnd the value of f (p, q)

I Combined with the fact that p + q = σ1(p, q), this gave two linear equations in p and q X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z

and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations: Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z

and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0 But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z

and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways. Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z

and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? g(x, y, z) =1 · x + ω · y + ω2 · z

and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z , and1, ω and ω2 are the three cube roots of 1.

Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z and note that1 and −1 are the two square roots of 1 Quartic equations:

X 4 + X 3 − 2X 2 − 6X − 4 = 0

Look for a power of a linear function h(x, y, z, w) which takes on only three diﬀerent values as you permute the inputs in all 24 possible ways.

But how? Recall that

f (x, y) =1 · x + −1 · y g(x, y, z) =1 · x + ω · y + ω2 · z and note that1 and −1 are the two square roots of 1, and1, ω and ω2 are the three cube roots of 1. h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w

. . . however, this doesn’t work.

Luckily, there’s a function that does work: if

h(x, y, z, w) = (x + y) − (z + w)

then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways.

So the natural thing to try would be: . . . however, this doesn’t work.

Luckily, there’s a function that does work: if

h(x, y, z, w) = (x + y) − (z + w)

then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways.

So the natural thing to try would be:

h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w Luckily, there’s a function that does work: if

h(x, y, z, w) = (x + y) − (z + w)

then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways.

So the natural thing to try would be:

h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w

. . . however, this doesn’t work. if

h(x, y, z, w) = (x + y) − (z + w)

then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways.

So the natural thing to try would be:

h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w

. . . however, this doesn’t work.

Luckily, there’s a function that does work: then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways.

So the natural thing to try would be:

h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w

. . . however, this doesn’t work.

Luckily, there’s a function that does work: if

h(x, y, z, w) = (x + y) − (z + w) So the natural thing to try would be:

h(x, y, z, w) =1 · x + i · y + −1 · z + −i · w

. . . however, this doesn’t work.

Luckily, there’s a function that does work: if

h(x, y, z, w) = (x + y) − (z + w) then h(x, y, z, w)2 (note: squared) takes on only three diﬀerent values as you permute the inputs in all 24 ways. = [(y + x) − (z + w)]2 = [(z + w) − (x + y)]2

. . . and so on.

In all, 8 permutations of the inputs leave the value unchanged, which means there will be 24/8 = 3 diﬀerent values:

h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2

For example,

[(x + y) − (z + w)]2 = [(z + w) − (x + y)]2

. . . and so on.

In all, 8 permutations of the inputs leave the value unchanged, which means there will be 24/8 = 3 diﬀerent values:

h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2

For example,

[(x + y) − (z + w)]2 = [(y + x) − (z + w)]2 . . . and so on.

In all, 8 permutations of the inputs leave the value unchanged, which means there will be 24/8 = 3 diﬀerent values:

h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2

For example,

[(x + y) − (z + w)]2 = [(y + x) − (z + w)]2 = [(z + w) − (x + y)]2 , which means there will be 24/8 = 3 diﬀerent values:

h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2

For example,

[(x + y) − (z + w)]2 = [(y + x) − (z + w)]2 = [(z + w) − (x + y)]2

. . . and so on.

In all, 8 permutations of the inputs leave the value unchanged h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2

For example,

[(x + y) − (z + w)]2 = [(y + x) − (z + w)]2 = [(z + w) − (x + y)]2

. . . and so on.

In all, 8 permutations of the inputs leave the value unchanged, which means there will be 24/8 = 3 diﬀerent values: For example,

[(x + y) − (z + w)]2 = [(y + x) − (z + w)]2 = [(z + w) − (x + y)]2

. . . and so on.

In all, 8 permutations of the inputs leave the value unchanged, which means there will be 24/8 = 3 diﬀerent values:

h(x, y, z, w)2, h(x, z, y, w)2 and h(x, w, y, z)2 = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2

then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2

then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2

then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2

then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) then A, B and C are all symmetric functions of x, y, z and w.

If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2 If we deﬁne

2 2 2 A(x, y, z, w) = σ1 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2 + h(x, z, y, w)2 + h(x, w, y, z)2 2 2 2 B(x, y, z, w) = σ2 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2 +h(x, z, y, w)2h(x, w, y, z)2 +h(x, w, y, z)2h(x, y, z, w)2 2 2 2 C(x, y, z, w) = σ3 h(x, y, z, w) , h(x, z, y, w) , h(x, w, y, z) = h(x, y, z, w)2h(x, z, y, w)2h(x, w, y, z)2 then A, B and C are all symmetric functions of x, y, z and w. This time, the computations are truly horrendous, but with even more help from Maple R , we ﬁnd that

2 A = 3σ1 − 8σ2 4 2 2 B = 3σ1 − 16σ1σ2 + 16σ2 + 16σ1σ3 − 64σ4 6 4 2 2 3 2 C = σ1 − 8σ1σ2 + 16σ1σ2 + 16σ1σ3 − 64σ1σ2σ3 + 64σ3

(how ever did Lagrange get by without Maple R ??)

Since they’re symmetric, we can express A, B and C in terms of σ1, σ2, σ3 and σ4. , but with even more help from Maple R , we ﬁnd that

2 A = 3σ1 − 8σ2 4 2 2 B = 3σ1 − 16σ1σ2 + 16σ2 + 16σ1σ3 − 64σ4 6 4 2 2 3 2 C = σ1 − 8σ1σ2 + 16σ1σ2 + 16σ1σ3 − 64σ1σ2σ3 + 64σ3

(how ever did Lagrange get by without Maple R ??)

Since they’re symmetric, we can express A, B and C in terms of σ1, σ2, σ3 and σ4.

This time, the computations are truly horrendous , we ﬁnd that

2 A = 3σ1 − 8σ2 4 2 2 B = 3σ1 − 16σ1σ2 + 16σ2 + 16σ1σ3 − 64σ4 6 4 2 2 3 2 C = σ1 − 8σ1σ2 + 16σ1σ2 + 16σ1σ3 − 64σ1σ2σ3 + 64σ3

(how ever did Lagrange get by without Maple R ??)

Since they’re symmetric, we can express A, B and C in terms of σ1, σ2, σ3 and σ4.

This time, the computations are truly horrendous, but with even more help from Maple R (how ever did Lagrange get by without Maple R ??)

Since they’re symmetric, we can express A, B and C in terms of σ1, σ2, σ3 and σ4.

This time, the computations are truly horrendous, but with even more help from Maple R , we ﬁnd that

2 A = 3σ1 − 8σ2 4 2 2 B = 3σ1 − 16σ1σ2 + 16σ2 + 16σ1σ3 − 64σ4 6 4 2 2 3 2 C = σ1 − 8σ1σ2 + 16σ1σ2 + 16σ1σ3 − 64σ1σ2σ3 + 64σ3 Since they’re symmetric, we can express A, B and C in terms of σ1, σ2, σ3 and σ4.

This time, the computations are truly horrendous, but with even more help from Maple R , we ﬁnd that

2 A = 3σ1 − 8σ2 4 2 2 B = 3σ1 − 16σ1σ2 + 16σ2 + 16σ1σ3 − 64σ4 6 4 2 2 3 2 C = σ1 − 8σ1σ2 + 16σ1σ2 + 16σ1σ3 − 64σ1σ2σ3 + 64σ3

(how ever did Lagrange get by without Maple R ??) If we call its roots p, q, r and s, then

σ1(p, q, r, s) = −1 σ2(p, q, r, s) = −2 σ3(p, q, r, s) = 6 σ4(p, q, r, s) = −4

Using these values in the above formulas gives

A(p, q, r, s) = 19 B(p, q, r, s) = 259 C(p, q, r, s) = 1521

Recall that our quartic equation is

X 4 + X 3 − 2X 2 − 6X − 4 = 0 , then

σ1(p, q, r, s) = −1 σ2(p, q, r, s) = −2 σ3(p, q, r, s) = 6 σ4(p, q, r, s) = −4

Using these values in the above formulas gives

A(p, q, r, s) = 19 B(p, q, r, s) = 259 C(p, q, r, s) = 1521

Recall that our quartic equation is

X 4 + X 3 − 2X 2 − 6X − 4 = 0

If we call its roots p, q, r and s Using these values in the above formulas gives

A(p, q, r, s) = 19 B(p, q, r, s) = 259 C(p, q, r, s) = 1521

Recall that our quartic equation is

X 4 + X 3 − 2X 2 − 6X − 4 = 0

If we call its roots p, q, r and s, then

σ1(p, q, r, s) = −1 σ2(p, q, r, s) = −2 σ3(p, q, r, s) = 6 σ4(p, q, r, s) = −4 Recall that our quartic equation is

X 4 + X 3 − 2X 2 − 6X − 4 = 0

If we call its roots p, q, r and s, then

σ1(p, q, r, s) = −1 σ2(p, q, r, s) = −2 σ3(p, q, r, s) = 6 σ4(p, q, r, s) = −4

Using these values in the above formulas gives

A(p, q, r, s) = 19 B(p, q, r, s) = 259 C(p, q, r, s) = 1521 But we have a method for solving cubic equations. . .

. . . which means that h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 are the roots of the cubic equation

X 3 − 19X 2 + 259X − 1521 = 0 . . . which means that h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 are the roots of the cubic equation

X 3 − 19X 2 + 259X − 1521 = 0

But we have a method for solving cubic equations. . . Taking square roots gives

h(p, q, r, s) = 3 h(p, r, q, s) = 3 + 2i h(p, s, q, r) = 3 − 2i

. . . which we apply to ﬁnd

h(p, q, r, s)2 = 9 h(p, r, q, s)2 = 5 + 12i h(p, s, q, r)2 = 5 − 12i . . . which we apply to ﬁnd

h(p, q, r, s)2 = 9 h(p, r, q, s)2 = 5 + 12i h(p, s, q, r)2 = 5 − 12i

Taking square roots gives

h(p, q, r, s) = 3 h(p, r, q, s) = 3 + 2i h(p, s, q, r) = 3 − 2i and of course, one more linear equation:

σ1(p, q, r, s) = p + q + r + s = −1

So we have three linear equations in the unknowns p, q, r and s:

p + q − r − s = 3 p + r − q − s = 3 + 2i p + s − q − r = 3 − 2i So we have three linear equations in the unknowns p, q, r and s:

p + q − r − s = 3 p + r − q − s = 3 + 2i p + s − q − r = 3 − 2i and of course, one more linear equation:

σ1(p, q, r, s) = p + q + r + s = −1 This coeﬃcient matrix is also invertible, and so we can calculate:

 p   2   q   −1    =    r   −1 + i  s −1 − i

Once again, we have a linear system:

 1 1 1 1   p   −1   1 1 −1 −1   q   3    ·   =    1 −1 1 −1   r   3 + 2i  1 −1 −1 1 s 3 − 2i Once again, we have a linear system:

 1 1 1 1   p   −1   1 1 −1 −1   q   3    ·   =    1 −1 1 −1   r   3 + 2i  1 −1 −1 1 s 3 − 2i

This coeﬃcient matrix is also invertible, and so we can calculate:

 p   2   q   −1    =    r   −1 + i  s −1 − i after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations. Trouble is. . . there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways. there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs.

So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . there’s no such thing. So far, so good. . . after all, we already knew there were methods for solving cubic and quartics; the point is to try to use this to come up with one for solving quintic equations.

Based on our experience so far, we should look for a power of a linear function k(x, y, z, w, v) which takes on only four diﬀerent values as we permute the inputs in all 120 ways.

Trouble is. . . there’s no such thing.

In fact, no function of ﬁve variables (power of a linear function or not) can take on exactly four diﬀerent values as you run through the 120 permutations of its inputs. Well, if there were such a thing, then its value would have to be left unchanged by exactly 120/4 = 30 of the 120 permutations.

However, the group S5 of permutations of ﬁve things doesn’t have a subgroup of order 30.

So there’s no chance of extending Lagrange’s program to quintic equations.

How do we know this? However, the group S5 of permutations of ﬁve things doesn’t have a subgroup of order 30.

So there’s no chance of extending Lagrange’s program to quintic equations.

How do we know this?

Well, if there were such a thing, then its value would have to be left unchanged by exactly 120/4 = 30 of the 120 permutations. So there’s no chance of extending Lagrange’s program to quintic equations.

How do we know this?

Well, if there were such a thing, then its value would have to be left unchanged by exactly 120/4 = 30 of the 120 permutations.

However, the group S5 of permutations of ﬁve things doesn’t have a subgroup of order 30. How do we know this?

Well, if there were such a thing, then its value would have to be left unchanged by exactly 120/4 = 30 of the 120 permutations.

However, the group S5 of permutations of ﬁve things doesn’t have a subgroup of order 30.

So there’s no chance of extending Lagrange’s program to quintic equations. Thanks to Ruﬃni, Abel and especially Galois, we know that quintic equations can’t be “solved by radicals” at all—that is, there exists a quintic equation whose roots can’t be expressed in terms of its coeﬃcients using only the four algebraic operations and root extractions.

In other words, there’s no such thing as a “quintic formula.”

Of course, we know much more than that. —that is, there exists a quintic equation whose roots can’t be expressed in terms of its coeﬃcients using only the four algebraic operations and root extractions.

In other words, there’s no such thing as a “quintic formula.”

Of course, we know much more than that.

Thanks to Ruﬃni, Abel and especially Galois, we know that quintic equations can’t be “solved by radicals” at all In other words, there’s no such thing as a “quintic formula.”

Of course, we know much more than that.

Thanks to Ruﬃni, Abel and especially Galois, we know that quintic equations can’t be “solved by radicals” at all—that is, there exists a quintic equation whose roots can’t be expressed in terms of its coeﬃcients using only the four algebraic operations and root extractions. Of course, we know much more than that.

In other words, there’s no such thing as a “quintic formula.” In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools. One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions. let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t) (This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.)

Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function. Postscript

However, you can solve quintic equations if you allow yourself more tools.

In 1858, Charles Hermite showed they could be solved using elliptic functions.

One way to deﬁne an elliptic function is as follows: let p be a polynomial of degree 3 and let

Z x 1 F (x) = p dt 0 p(t)

Then F −1 is an elliptic function.

(This type of integral comes up in the problem of calculating the arc length of an ellipse, hence the name.) let

Z x 1 F (x) = √ dt 2 0 1 − t

Then F (x) = sin−1(x), which is to say, F −1(x) = sin(x). In this way, elliptic functions are analogous to trigonometric functions.

In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions:

Here’s an analogy: Then F (x) = sin−1(x), which is to say, F −1(x) = sin(x). In this way, elliptic functions are analogous to trigonometric functions.

In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions:

Here’s an analogy: let

Z x 1 F (x) = √ dt 2 0 1 − t , which is to say, F −1(x) = sin(x). In this way, elliptic functions are analogous to trigonometric functions.

In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions:

Here’s an analogy: let

Z x 1 F (x) = √ dt 2 0 1 − t

Then F (x) = sin−1(x) In this way, elliptic functions are analogous to trigonometric functions.

In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions:

Here’s an analogy: let

Z x 1 F (x) = √ dt 2 0 1 − t

Then F (x) = sin−1(x), which is to say, F −1(x) = sin(x). In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions:

Here’s an analogy: let

Z x 1 F (x) = √ dt 2 0 1 − t

Then F (x) = sin−1(x), which is to say, F −1(x) = sin(x). In this way, elliptic functions are analogous to trigonometric functions. Here’s an analogy: let

Z x 1 F (x) = √ dt 2 0 1 − t

Then F (x) = sin−1(x), which is to say, F −1(x) = sin(x). In this way, elliptic functions are analogous to trigonometric functions.

In 1591, Fran¸coisVi`etepublished a method of solving cubic equations using the sine and inverse sine functions: a ﬁrst make the substitution Y = X + 3 ; this will transform it into an equation of the form

Y 3 + BY + C = 0

(This is exactly analogous to “completing the square.”)

Next, make the substitution r −B Y = 2 sin θ 3

Given a cubic equation

X 3 + aX 2 + bX + c = 0 ; this will transform it into an equation of the form

Y 3 + BY + C = 0

(This is exactly analogous to “completing the square.”)

Next, make the substitution r −B Y = 2 sin θ 3

Given a cubic equation

X 3 + aX 2 + bX + c = 0

a ﬁrst make the substitution Y = X + 3 (This is exactly analogous to “completing the square.”)

Next, make the substitution r −B Y = 2 sin θ 3

Given a cubic equation

X 3 + aX 2 + bX + c = 0

a ﬁrst make the substitution Y = X + 3 ; this will transform it into an equation of the form

Y 3 + BY + C = 0 Next, make the substitution r −B Y = 2 sin θ 3

Given a cubic equation

X 3 + aX 2 + bX + c = 0

a ﬁrst make the substitution Y = X + 3 ; this will transform it into an equation of the form

Y 3 + BY + C = 0

(This is exactly analogous to “completing the square.”) Given a cubic equation

X 3 + aX 2 + bX + c = 0

a ﬁrst make the substitution Y = X + 3 ; this will transform it into an equation of the form

Y 3 + BY + C = 0

(This is exactly analogous to “completing the square.”)

Next, make the substitution r −B Y = 2 sin θ 3 But the left-hand side of this last equation is sin 3θ; so our equation is just 3C r−3 sin 3θ = − 2B B 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B , from

q −B a which we can calculate Y = 2 3 sin θ, and ﬁnally X = Y − 3 .

This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B so our equation is just 3C r−3 sin 3θ = − 2B B 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B , from

q −B a which we can calculate Y = 2 3 sin θ, and ﬁnally X = Y − 3 .

This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B

But the left-hand side of this last equation is sin 3θ; 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B , from

q −B a which we can calculate Y = 2 3 sin θ, and ﬁnally X = Y − 3 .

This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B

But the left-hand side of this last equation is sin 3θ; so our equation is just 3C r−3 sin 3θ = − 2B B , from

q −B a which we can calculate Y = 2 3 sin θ, and ﬁnally X = Y − 3 .

This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B

But the left-hand side of this last equation is sin 3θ; so our equation is just 3C r−3 sin 3θ = − 2B B 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B a , and ﬁnally X = Y − 3 .

This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B

But the left-hand side of this last equation is sin 3θ; so our equation is just 3C r−3 sin 3θ = − 2B B 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B , from

q −B which we can calculate Y = 2 3 sin θ This will transform the equation into

3C r−3 −4 sin3 θ + 3 sin θ = − 2B B

But the left-hand side of this last equation is sin 3θ; so our equation is just 3C r−3 sin 3θ = − 2B B 1 −1 3C q −3 which we can solve by setting θ = 3 sin − 2B B , from

q −B a which we can calculate Y = 2 3 sin θ, and finally X = Y − 3 . but it’s significantly more complicated than Viète’ssolution to the cubic.

Hermite’s solution of the quintic uses something like a “multiple angle formula” for elliptic functions Hermite’s solution of the quintic uses something like a “multiple angle formula” for elliptic functions but it’s signiﬁcantly more complicated than Vi`ete’ssolution to the cubic. In solving quartic equations, Lagrange used the function

h(x, y, z, w) = xy + zw

instead of our

h(x, y, z, w)2 = [(x + y) − (z + w)]2

which works just as well, but requires one to solve non-linear equations at the end.

Notes which works just as well, but requires one to solve non-linear equations at the end.

Notes

In solving quartic equations, Lagrange used the function

h(x, y, z, w) = xy + zw

instead of our

h(x, y, z, w)2 = [(x + y) − (z + w)]2 Notes

In solving quartic equations, Lagrange used the function

h(x, y, z, w) = xy + zw

instead of our

h(x, y, z, w)2 = [(x + y) − (z + w)]2

which works just as well, but requires one to solve non-linear equations at the end. at which point, I said

“. . . taking cube roots, √ g(p, q, r) = −3 + 2 3i √ g(p, r, q) = −3 − 2 3i.”

More substantially, during the solution of the cubic, we solved the “resolvent quadratic” to ﬁnd the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i More substantially, during the solution of the cubic, we solved the “resolvent quadratic” to ﬁnd the values √ g(p, q, r)3 = 81 + 30 3i √ g(p, r, q)3 = 81 − 30 3i at which point, I said

“. . . taking cube roots, √ g(p, q, r) = −3 + 2 3i √ g(p, r, q) = −3 − 2 3i.” There are nine diﬀerent ways to choose one cube root for each—not all of which lead to the correct solutions of the cubic!

So how do we know which ones to pick?

√ However, the numbers 81 ± 30 3i each have three cube roots:

√ √ √ 9 3 3 5 3 −3 ± 2 3i, 2 ± 2 i and − 2 ∓ 2 i —not all of which lead to the correct solutions of the cubic!

So how do we know which ones to pick?

√ However, the numbers 81 ± 30 3i each have three cube roots:

√ √ √ 9 3 3 5 3 −3 ± 2 3i, 2 ± 2 i and − 2 ∓ 2 i

There are nine diﬀerent ways to choose one cube root for each So how do we know which ones to pick?

√ However, the numbers 81 ± 30 3i each have three cube roots:

√ √ √ 9 3 3 5 3 −3 ± 2 3i, 2 ± 2 i and − 2 ∓ 2 i

There are nine diﬀerent ways to choose one cube root for each—not all of which lead to the correct solutions of the cubic! √ However, the numbers 81 ± 30 3i each have three cube roots:

√ √ √ 9 3 3 5 3 −3 ± 2 3i, 2 ± 2 i and − 2 ∓ 2 i

There are nine diﬀerent ways to choose one cube root for each—not all of which lead to the correct solutions of the cubic!

So how do we know which ones to pick? We saw that the function

B(x, y, z) = g(x, y, z)3g(x, z, y)3

is symmetric, but as it happens, so is

B(x, y, z) = g(x, y, z)g(x, z, y)

In fact, 2 B = σ1 − 3σ2

Well, there’s one more piece to the puzzle: , but as it happens, so is

B(x, y, z) = g(x, y, z)g(x, z, y)

In fact, 2 B = σ1 − 3σ2

Well, there’s one more piece to the puzzle:

We saw that the function

B(x, y, z) = g(x, y, z)3g(x, z, y)3 is symmetric In fact, 2 B = σ1 − 3σ2

Well, there’s one more piece to the puzzle:

We saw that the function

B(x, y, z) = g(x, y, z)3g(x, z, y)3 is symmetric, but as it happens, so is

B(x, y, z) = g(x, y, z)g(x, z, y) Well, there’s one more piece to the puzzle:

We saw that the function

B(x, y, z) = g(x, y, z)3g(x, z, y)3 is symmetric, but as it happens, so is

B(x, y, z) = g(x, y, z)g(x, z, y)

In fact, 2 B = σ1 − 3σ2 and it turns out that as long as we choose cube roots of g(p, q, r)3 and g(p, r, q)3 whose product is 21, our procedure yields the correct solutions of the cubic.

So for “our” cubic, we can calculate

g(p, q, r)g(p, r, q) = (−3)2 − 3(−4) = 21 So for “our” cubic, we can calculate

g(p, q, r)g(p, r, q) = (−3)2 − 3(−4) = 21 and it turns out that as long as we choose cube roots of g(p, q, r)3 and g(p, r, q)3 whose product is 21, our procedure yields the correct solutions of the cubic. solving the resolvent cubic equation gives us the values of h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2, each of which has two square roots.

Again, not all (eight) choices lead to the correct solutions of the quartic; but this time, it turns out that the function

h(x, y, z, w)h(x, z, y, w)h(x, w, y, z)

is also symmetric.

There is a similar issue with our solution of the quartic: , each of which has two square roots.

Again, not all (eight) choices lead to the correct solutions of the quartic; but this time, it turns out that the function

h(x, y, z, w)h(x, z, y, w)h(x, w, y, z)

is also symmetric.

There is a similar issue with our solution of the quartic: solving the resolvent cubic equation gives us the values of h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 Again, not all (eight) choices lead to the correct solutions of the quartic; but this time, it turns out that the function

h(x, y, z, w)h(x, z, y, w)h(x, w, y, z)

is also symmetric.

There is a similar issue with our solution of the quartic: solving the resolvent cubic equation gives us the values of h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2, each of which has two square roots. ; but this time, it turns out that the function

h(x, y, z, w)h(x, z, y, w)h(x, w, y, z)

is also symmetric.

Again, not all (eight) choices lead to the correct solutions of the quartic There is a similar issue with our solution of the quartic: solving the resolvent cubic equation gives us the values of h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2, each of which has two square roots.

Again, not all (eight) choices lead to the correct solutions of the quartic; but this time, it turns out that the function

h(x, y, z, w)h(x, z, y, w)h(x, w, y, z) is also symmetric. which, for our quartic, is

(−1)3 + 8(6) − 4(−1)(−2) = 39

and once again, as long as we choose square roots for h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 whose product is 39, we get the correct solutions of the quartic.

In fact,

3 h(x, y, z, w)h(x, z, y, w)h(x, w, y, z) = σ1 + 8σ3 − 4σ1σ2 and once again, as long as we choose square roots for h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 whose product is 39, we get the correct solutions of the quartic.

In fact,

3 h(x, y, z, w)h(x, z, y, w)h(x, w, y, z) = σ1 + 8σ3 − 4σ1σ2 which, for our quartic, is

(−1)3 + 8(6) − 4(−1)(−2) = 39 In fact,

3 h(x, y, z, w)h(x, z, y, w)h(x, w, y, z) = σ1 + 8σ3 − 4σ1σ2 which, for our quartic, is

(−1)3 + 8(6) − 4(−1)(−2) = 39 and once again, as long as we choose square roots for h(p, q, r, s)2, h(p, r, q, s)2 and h(p, s, q, r)2 whose product is 39, we get the correct solutions of the quartic.