Modular J-Invariant Function

Modular J-invariant function

N. Saradha

T.I.F.R.

Jan 25, 2016 Student’s Seminar TIFR, Mumbai Let L = {mω1 + nω2 : m, n ∈ Z} be a lattice spanned by the two linearly independent complex numbers ω1 and ω2. An elliptic function with respect to L is a meromorphic function f on C which satisﬁes

f (z + ω) = f (z) for all ω ∈ L and z ∈ C.

D = {sω1 + tω2 : 0 ≤ s, t < 1} is called the fundamental parallelogram and any translate of this is referred to as a fundamental domain. L is called the period lattice.

Elliptic function An elliptic function with respect to L is a meromorphic function f on C which satisﬁes

f (z + ω) = f (z) for all ω ∈ L and z ∈ C.

D = {sω1 + tω2 : 0 ≤ s, t < 1} is called the fundamental parallelogram and any translate of this is referred to as a fundamental domain. L is called the period lattice.

Elliptic function

Let L = {mω1 + nω2 : m, n ∈ Z} be a lattice spanned by the two linearly independent complex numbers ω1 and ω2. D = {sω1 + tω2 : 0 ≤ s, t < 1} is called the fundamental parallelogram and any translate of this is referred to as a fundamental domain. L is called the period lattice.

Elliptic function

Let L = {mω1 + nω2 : m, n ∈ Z} be a lattice spanned by the two linearly independent complex numbers ω1 and ω2. An elliptic function with respect to L is a meromorphic function f on C which satisﬁes

f (z + ω) = f (z) for all ω ∈ L and z ∈ C. L is called the period lattice.

Elliptic function

f (z + ω) = f (z) for all ω ∈ L and z ∈ C.

D = {sω1 + tω2 : 0 ≤ s, t < 1} is called the fundamental parallelogram and any translate of this is referred to as a fundamental domain. Elliptic function

f (z + ω) = f (z) for all ω ∈ L and z ∈ C.

D = {sω1 + tω2 : 0 ≤ s, t < 1} is called the fundamental parallelogram and any translate of this is referred to as a fundamental domain. L is called the period lattice. Here is an explicit example of an elliptic function. The Weierstrass ℘− function associated with L is deﬁned by the series 1 X 1 1 ℘(z) = + − , z2 (z − ω)2 ω2 ω∈L0

where L0 denotes the set of non-zero periods. It converges absolutely and uniformly on every compact subset of C/L. It is a meromorphic function having a double pole at every point of L and no other poles.

The Weierstrass elliptic function ℘(z) The Weierstrass ℘− function associated with L is deﬁned by the series 1 X 1 1 ℘(z) = + − , z2 (z − ω)2 ω2 ω∈L0

The Weierstrass elliptic function ℘(z)

Here is an explicit example of an elliptic function. It converges absolutely and uniformly on every compact subset of C/L. It is a meromorphic function having a double pole at every point of L and no other poles.

The Weierstrass elliptic function ℘(z)

Here is an explicit example of an elliptic function. The Weierstrass ℘− function associated with L is deﬁned by the series 1 X 1 1 ℘(z) = + − , z2 (z − ω)2 ω2 ω∈L0

where L0 denotes the set of non-zero periods. It is a meromorphic function having a double pole at every point of L and no other poles.

The Weierstrass elliptic function ℘(z)

Here is an explicit example of an elliptic function. The Weierstrass ℘− function associated with L is deﬁned by the series 1 X 1 1 ℘(z) = + − , z2 (z − ω)2 ω2 ω∈L0

where L0 denotes the set of non-zero periods. It converges absolutely and uniformly on every compact subset of C/L. The Weierstrass elliptic function ℘(z)

Here is an explicit example of an elliptic function. The Weierstrass ℘− function associated with L is deﬁned by the series 1 X 1 1 ℘(z) = + − , z2 (z − ω)2 ω2 ω∈L0

X −2k G2k = G2k (L) = ω ω∈L0

the associated Eisenstein series of weight 2k. Then ℘(z) can be expressed as

∞ 1 X ℘(z) = + (2k + 1)G z2k . z2 2k+2 k=1

Eisenstein series Then ℘(z) can be expressed as

∞ 1 X ℘(z) = + (2k + 1)G z2k . z2 2k+2 k=1

Eisenstein series

Let us denote by

X −2k G2k = G2k (L) = ω ω∈L0

the associated Eisenstein series of weight 2k. Eisenstein series

Let us denote by

X −2k G2k = G2k (L) = ω ω∈L0

the associated Eisenstein series of weight 2k. Then ℘(z) can be expressed as

∞ 1 X ℘(z) = + (2k + 1)G z2k . z2 2k+2 k=1 It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6.

Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . It turns out that this discriminant is non-zero. Such curves are called elliptic curves (E).

The differential equation for ℘(z) Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . It turns out that this discriminant is non-zero. Such curves are called elliptic curves (E).

The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . It turns out that this discriminant is non-zero. Such curves are called elliptic curves (E).

The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6.

Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. It turns out that this discriminant is non-zero. Such curves are called elliptic curves (E).

The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6.

Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . Such curves are called elliptic curves (E).

The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6.

Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . It turns out that this discriminant is non-zero. The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

0 2 3 ℘ (z) = 4℘(z) − 60G4℘(z) − 140G6.

Thus (℘(z), ℘0(z)) lies on the curve deﬁned by the equation

2 3 E : y = 4x − g2x − g3

where g2 = 60G4, g3 = 140G6. This cubic polynomial has the discriminant 3 2 g2 − 27g3 . It turns out that this discriminant is non-zero. Such curves are called elliptic curves (E). Now we start with such an equation. Given (x, y) ∈ C2, lying on the curve, can we ﬁnd z such that x = ℘(z), y = ℘0(z)? If so, we can always parametrize such curves with the Weierstarss elliptic functions. Suppose ℘(z) = x has no solution. Then 1/(℘(z) − x) is an elliptic function which is holomorphic on L. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)).

We can parametrize Given (x, y) ∈ C2, lying on the curve, can we ﬁnd z such that x = ℘(z), y = ℘0(z)? If so, we can always parametrize such curves with the Weierstarss elliptic functions. Suppose ℘(z) = x has no solution. Then 1/(℘(z) − x) is an elliptic function which is holomorphic on L. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)).

We can parametrize

Now we start with such an equation. If so, we can always parametrize such curves with the Weierstarss elliptic functions. Suppose ℘(z) = x has no solution. Then 1/(℘(z) − x) is an elliptic function which is holomorphic on L. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)).

We can parametrize

Now we start with such an equation. Given (x, y) ∈ C2, lying on the curve, can we ﬁnd z such that x = ℘(z), y = ℘0(z)? Suppose ℘(z) = x has no solution. Then 1/(℘(z) − x) is an elliptic function which is holomorphic on L. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)).

We can parametrize

Now we start with such an equation. Given (x, y) ∈ C2, lying on the curve, can we ﬁnd z such that x = ℘(z), y = ℘0(z)? If so, we can always parametrize such curves with the Weierstarss elliptic functions. Suppose ℘(z) = x has no solution. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)).

We can parametrize

Now we start with such an equation. Given (x, y) ∈ C2, lying on the curve, can we ﬁnd z such that x = ℘(z), y = ℘0(z)? If so, we can always parametrize such curves with the Weierstarss elliptic functions. Suppose ℘(z) = x has no solution. Then 1/(℘(z) − x) is an elliptic function which is holomorphic on L. By periodicity, it is entire and bounded. Hence must be a constant- a contradiction. Hence y = ±℘0(z). One can check that ℘ is an even function and hence ℘0 is an odd function. So we can adjust the sign so that (x, y) = (℘(z), ℘0(z)). Suppose we are given an elliptic curve of the form y 2 = 4x3 − Ax − B (thus the discriminant is non-zero) , there exist a unique lattice L such that g2(L) = A and g3(L) = B and hence the complex points of the curve are parametrized by the Weierstrass ℘ function of L.

Converse is also true Converse is also true

Suppose we are given an elliptic curve of the form y 2 = 4x3 − Ax − B (thus the discriminant is non-zero) , there exist a unique lattice L such that g2(L) = A and g3(L) = B and hence the complex points of the curve are parametrized by the Weierstrass ℘ function of L. Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2, g3 are algebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental. As a consequence, we have that any non-zero period of an elliptic curve deﬁned over Q is transcendental. Reason: Since ℘0 is an odd function, 0 0 0 ℘ (ω1/2) = ℘ (ω2/2) = ℘ ((ω1 + ω2)/2) = 0. Hence ℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as two division points) which shows that non-zero periods are transcendental. One can also show by standard complex analytic argument, that the above two division points are distinct. And this implies that the discriminant

3 2 ∆ := g2 − 27g3 6= 0. Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. Then

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). c d 2 Z And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d

modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d modular property of g2, g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be the lattice spanned by 1 and z. Then the corresponding g2, g3 and ∆ are functions of z. a b Suppose ∈ SL ( ). Then c d 2 Z

a b az + b · z = . c d cz + d

And az + b az + b g = (cz + d)4g (z), g = (cz + d)6g (z), 2 cz + d 2 3 cz + d 3

az + b ∆ = (cz + d)12∆(z). cz + d 1728g3 j = j(z) = 2 3 2 g2 − 27g3 is called the modular invariant of the elliptic curve E. Why modular invariant:

az + b j = j(z) cz + d

so invariant under the action of SL2(Z).

The modular j− invariant function Why modular invariant:

az + b j = j(z) cz + d

so invariant under the action of SL2(Z).

The modular j− invariant function

1728g3 j = j(z) = 2 3 2 g2 − 27g3 is called the modular invariant of the elliptic curve E. az + b j = j(z) cz + d

so invariant under the action of SL2(Z).

The modular j− invariant function

1728g3 j = j(z) = 2 3 2 g2 − 27g3 is called the modular invariant of the elliptic curve E. Why modular invariant: The modular j− invariant function

1728g3 j = j(z) = 2 3 2 g2 − 27g3 is called the modular invariant of the elliptic curve E. Why modular invariant:

az + b j = j(z) cz + d

so invariant under the action of SL2(Z). (i) Two elliptic curves over C are isomorphic if and only if they have the same j. (ii) The j function takes every complex value exactly once on the quotient space H/SL ( ). √ 2 Z (iii) j(i) = 1728; j((1 + −3)/2) = 0.

Known facts (ii) The j function takes every complex value exactly once on the quotient space H/SL ( ). √ 2 Z (iii) j(i) = 1728; j((1 + −3)/2) = 0.

Known facts

(i) Two elliptic curves over C are isomorphic if and only if they have the same j. √ (iii) j(i) = 1728; j((1 + −3)/2) = 0.

Known facts

(i) Two elliptic curves over C are isomorphic if and only if they have the same j. (ii) The j function takes every complex value exactly once on the quotient space H/SL2(Z). Known facts

(i) Two elliptic curves over C are isomorphic if and only if they have the same j. (ii) The j function takes every complex value exactly once on the quotient space H/SL ( ). √ 2 Z (iii) j(i) = 1728; j((1 + −3)/2) = 0. We shall denote from now on ω2/ω1 = τ and assume that Im(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

∞ !3 ∞ 1 X qm Y J(q) = 1 + 240 m3 (1 − qn)24 q 1 − qm m=1 n=1 1 = + 744 + 196884q + 21493760q2 + ..... q

Fourier expansion of j Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

∞ !3 ∞ 1 X qm Y J(q) = 1 + 240 m3 (1 − qn)24 q 1 − qm m=1 n=1 1 = + 744 + 196884q + 21493760q2 + ..... q

Fourier expansion of j

We shall denote from now on ω2/ω1 = τ and assume that Im(τ) > 0. J(e2πiτ ) = j(τ). Then

∞ !3 ∞ 1 X qm Y J(q) = 1 + 240 m3 (1 − qn)24 q 1 − qm m=1 n=1 1 = + 744 + 196884q + 21493760q2 + ..... q

Fourier expansion of j

We shall denote from now on ω2/ω1 = τ and assume that Im(τ) > 0. Put q = e2πiτ , Then

∞ !3 ∞ 1 X qm Y J(q) = 1 + 240 m3 (1 − qn)24 q 1 − qm m=1 n=1 1 = + 744 + 196884q + 21493760q2 + ..... q

Fourier expansion of j

We shall denote from now on ω2/ω1 = τ and assume that Im(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ). Fourier expansion of j

We shall denote from now on ω2/ω1 = τ and assume that Im(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

∞ !3 ∞ 1 X qm Y J(q) = 1 + 240 m3 (1 − qn)24 q 1 − qm m=1 n=1 1 = + 744 + 196884q + 21493760q2 + ..... q All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d.

Complex Multiplication In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d.

Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. Thus τ is an algebraic integer of degree at most two over Q. End(E) is either Z or an order in an imaginary quadratic ﬁeld k. The latter case arises if and only if τ is a quadratic number. Then the ﬁeld of endomorphisms is k = Q(τ) and we say E has complex multiplication.

Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d. End(E) is either Z or an order in an imaginary quadratic ﬁeld k. The latter case arises if and only if τ is a quadratic number. Then the ﬁeld of endomorphisms is k = Q(τ) and we say E has complex multiplication.

Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d.

Thus τ is an algebraic integer of degree at most two over Q. The latter case arises if and only if τ is a quadratic number. Then the ﬁeld of endomorphisms is k = Q(τ) and we say E has complex multiplication.

Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d.

Thus τ is an algebraic integer of degree at most two over Q. End(E) is either Z or an order in an imaginary quadratic ﬁeld k. Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (where E is the associated elliptic curve) are of the form αz + L for some α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b; ατ = cτ + d.

Thus τ is an algebraic integer of degree at most two over Q. End(E) is either Z or an order in an imaginary quadratic ﬁeld k. The latter case arises if and only if τ is a quadratic number. Then the ﬁeld of endomorphisms is k = Q(τ) and we say E has complex multiplication. In CM case j(τ) IS AN ALGEBRAIC INTEGER ITS DEGREE = h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). Suppose τ ∈ H with j(τ) algebraic. Then

τ algebraic if and only if τ is imaginary quadratic.

An interesting fact in the CM case Suppose τ ∈ H with j(τ) algebraic. Then

τ algebraic if and only if τ is imaginary quadratic.

An interesting fact in the CM case

In CM case j(τ) IS AN ALGEBRAIC INTEGER ITS DEGREE = h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). An interesting fact in the CM case

In CM case j(τ) IS AN ALGEBRAIC INTEGER ITS DEGREE = h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). Suppose τ ∈ H with j(τ) algebraic. Then

τ algebraic if and only if τ is imaginary quadratic. Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. Note: (i) eπ is transcendental. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed: For α and β algebraic numbers with α 6= 0 and β 6∈ Q and for any choice of log α 6= 0, the number αβ is transcendental. √ √ Note: (i) eπ is transcendental. So are 2 2, eπ 163. (ii) The functions ez and eβz are algebraically independent. But cannot take algebraic values at log α. Schneider (1937) using elliptic analogue of the above result showed: Suppose τ ∈ H is algebraic and not an imaginary quadratic, then the value j(τ) is transcendental. Schneider himself asked if this can be proved using modular functions. √ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ Then q = −e−π 163.

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ 1+i 163 Suppose τ = 2 . j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e . i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q √ So why is this happening? What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ What is special about eπ 163?

Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. So why is this happening? Diophantine Approximation via the j function

√ eπ 163 = 262537412640768743.9999999999992...... √ √ 1+i 163 −π 163 Suppose τ = 2 . Then q = −e .

j(τ) = −(640320)3

and 1 −12 j(τ) − − 744 < 10 q

i.e √ | − (640320)3 − 744 + eπ 163| < 10−12 √ i.e j(τ) is a good approximation to eπ 163. √ So why is this happening? What is special about eπ 163? We have alreday seen that j(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREE EQUALS h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). Let us take ( √ i d if d = 2 or 3 (mod 4) τ = √ (1 + i d)/2 if d ≡ 1(mod 4). Then ( √ e−2π d if d ≡ 2 or 3(mod 4) q = √ −e−π d if d ≡ 1(mod 4).

163 j(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREE EQUALS h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). Let us take ( √ i d if d = 2 or 3 (mod 4) τ = √ (1 + i d)/2 if d ≡ 1(mod 4). Then ( √ e−2π d if d ≡ 2 or 3(mod 4) q = √ −e−π d if d ≡ 1(mod 4).

163

We have alreday seen that Let us take ( √ i d if d = 2 or 3 (mod 4) τ = √ (1 + i d)/2 if d ≡ 1(mod 4). Then ( √ e−2π d if d ≡ 2 or 3(mod 4) q = √ −e−π d if d ≡ 1(mod 4).

163

We have alreday seen that j(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREE EQUALS h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). If h = h(d√) = 1 then√ the transcendental number 1/q which is either e2π d or −eπ d is close to the rational integer J(q) − 744. The error is ∼ 196884q which is small if d is large. And this is what is happening when d = 163 since h(163) = 1. This good approximation can be used to give effective improvement over Liouville’s theorem for some numbers.

163

We have alreday seen that j(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREE EQUALS h− THE CLASS NUMBER OF THE QUADRATIC FIELD k = Q(τ). Let us take ( √ i d if d = 2 or 3 (mod 4) τ = √ (1 + i d)/2 if d ≡ 1(mod 4). Then ( √ e−2π d if d ≡ 2 or 3(mod 4) q = √ −e−π d if d ≡ 1(mod 4). And this is what is happening when d = 163 since h(163) = 1. This good approximation can be used to give effective improvement over Liouville’s theorem for some numbers.

163

If h = h(d√) = 1 then√ the transcendental number 1/q which is either e2π d or −eπ d is close to the rational integer J(q) − 744. The error is ∼ 196884q which is small if d is large. This good approximation can be used to give effective improvement over Liouville’s theorem for some numbers.

163

If h = h(d√) = 1 then√ the transcendental number 1/q which is either e2π d or −eπ d is close to the rational integer J(q) − 744. The error is ∼ 196884q which is small if d is large. And this is what is happening when d = 163 since h(163) = 1. This good approximation can be used to give effective improvement over Liouville’s theorem for some numbers. We get a large integral solution (x, y) to the equation 163y 2 = x3 + 1728 viz., x = −(j(τ))1/3 = 640320; y = 23337.11.19.127 = 40133016. Dividing by 26 we get a large solution to 163y 2 = x3 + 27. Dividing by 33, we get 3 · 163y 2 = x3 + 1. Dividing by 63, we get 6 · 163y 2 = x3 + 8.

Diophantine equations Dividing by 26 we get a large solution to 163y 2 = x3 + 27. Dividing by 33, we get 3 · 163y 2 = x3 + 1. Dividing by 63, we get 6 · 163y 2 = x3 + 8.

Diophantine equations

We get a large integral solution (x, y) to the equation 163y 2 = x3 + 1728 viz., x = −(j(τ))1/3 = 640320; y = 23337.11.19.127 = 40133016. Diophantine equations

We get a large integral solution (x, y) to the equation 163y 2 = x3 + 1728 viz., x = −(j(τ))1/3 = 640320; y = 23337.11.19.127 = 40133016. Dividing by 26 we get a large solution to 163y 2 = x3 + 27. Dividing by 33, we get 3 · 163y 2 = x3 + 1. Dividing by 63, we get 6 · 163y 2 = x3 + 8. There are 9 imaginary quadratic fields having class number 1.They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. The Schläfli modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ Then f ( d) √is an algebraic integer having degree 3h(d). It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisfies √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1 They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. The Schläﬂi modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ Then f ( d) √is an algebraic integer having degree 3h(d). It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisﬁes √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1

There are 9 imaginary quadratic fields having class number 1. The Schläfli modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ Then f ( d) √is an algebraic integer having degree 3h(d). It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisfies √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1

There are 9 imaginary quadratic ﬁelds having class number 1.They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. √ Then f ( d) √is an algebraic integer having degree 3h(d). It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisﬁes √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1

There are 9 imaginary quadratic fields having class number 1.They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. The Schläfli modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisfies √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1

There are 9 imaginary quadratic fields having class number 1.They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. The Schläfli modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ Then f ( d) is an algebraic integer having degree 3h(d). What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisfies √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2

Class number 1

There are 9 imaginary quadratic fields having class number 1.They are d = −3, −4, −7, −8, −11, −19, −43, −67, −163. The Schläfli modular function is given by ∞ Y f (z) = q−1/48 (1 + qn−1/2). n=1 √ Then f ( d) √is an algebraic integer having degree 3h(d). It is found that f ( −163) is the real root of the equation f 3 − 6f 2 + 4f − 2 = 0. What is the connection to the j function? √ Let F(z) = f 8(z). Then F = F( d) satisfies √ !1/3 −3 + d F 3 + j F 2 − 256 = 0. 2 Similarly it was found that √ f ( −3) satisfies x3 − 2 = 0 √ f ( −11) satisfies x3 − 2x2 + 2x − 2 = 0 √ f ( −19) satisfies x3 − 2x − 2 = 0 √ f ( −43) satisfies x3 − 2x2 − 2 = 0 √ f ( −67) satisfies x3 − 2x2 − 2x − 2 = 0 √ f ( −163) satisfies x3 − 6x2 + 4x − 2 = 0. And some of the partial quotients in their continued fraction expansions are very large giving rise to very good rational approximations to these numbers. This was first noticed by Brillhart (1964) and the connection to Class number 1 was analyzed by Stark (1970). Using the good approximation property Chudnovsky (1983) could give effective improvement over Liouville’s theorem for these numbers.

More equations And some of the partial quotients in their continued fraction expansions are very large giving rise to very good rational approximations to these numbers. This was ﬁrst noticed by Brillhart (1964) and the connection to Class number 1 was analyzed by Stark (1970). Using the good approximation property Chudnovsky (1983) could give effective improvement over Liouville’s theorem for these numbers.

More equations

Similarly it was found that √ f ( −3) satisfies x3 − 2 = 0 √ f ( −11) satisfies x3 − 2x2 + 2x − 2 = 0 √ f ( −19) satisfies x3 − 2x − 2 = 0 √ f ( −43) satisfies x3 − 2x2 − 2 = 0 √ f ( −67) satisfies x3 − 2x2 − 2x − 2 = 0 √ f ( −163) satisfies x3 − 6x2 + 4x − 2 = 0. This was first noticed by Brillhart (1964) and the connection to Class number 1 was analyzed by Stark (1970). Using the good approximation property Chudnovsky (1983) could give effective improvement over Liouville’s theorem for these numbers.

More equations

Similarly it was found that √ f ( −3) satisfies x3 − 2 = 0 √ f ( −11) satisfies x3 − 2x2 + 2x − 2 = 0 √ f ( −19) satisfies x3 − 2x − 2 = 0 √ f ( −43) satisfies x3 − 2x2 − 2 = 0 √ f ( −67) satisfies x3 − 2x2 − 2x − 2 = 0 √ f ( −163) satisfies x3 − 6x2 + 4x − 2 = 0. And some of the partial quotients in their continued fraction expansions are very large giving rise to very good rational approximations to these numbers. This was first noticed by Brillhart (1964) and the connection to Class number 1 was analyzed by Stark (1970). Using the good approximation property Chudnovsky (1983) could give effective improvement over Liouville’s theorem for these numbers. In his famous paper on " On certain Arithmetical Functions" Ramanujan (1916) introduced the following functions which were later recognized as special cases of normalized Eisenstein series. ∞ ∞ X nqn X n3qn P(q) = 1 − 24 , Q(q) = 1 + 240 , 1 − qn 1 − qn n=1 n=1

∞ X n5qn R(q) = 1 − 504 . 1 − qn n=1

Thus P(q) = E2(τ), Q(q) = E4(τ), R(q) = E6(τ) where 1 X 1 E (τ) = for k ≥ 2. 2k 2ζ(2k) (mτ + n)2k (m,n)6=(0,0)

Ramanujan Functions ∞ ∞ X nqn X n3qn P(q) = 1 − 24 , Q(q) = 1 + 240 , 1 − qn 1 − qn n=1 n=1

∞ X n5qn R(q) = 1 − 504 . 1 − qn n=1

Thus P(q) = E2(τ), Q(q) = E4(τ), R(q) = E6(τ) where 1 X 1 E (τ) = for k ≥ 2. 2k 2ζ(2k) (mτ + n)2k (m,n)6=(0,0)

Ramanujan Functions

In his famous paper on " On certain Arithmetical Functions" Ramanujan (1916) introduced the following functions which were later recognized as special cases of normalized Eisenstein series. Thus P(q) = E2(τ), Q(q) = E4(τ), R(q) = E6(τ) where 1 X 1 E (τ) = for k ≥ 2. 2k 2ζ(2k) (mτ + n)2k (m,n)6=(0,0)

Ramanujan Functions

In his famous paper on " On certain Arithmetical Functions" Ramanujan (1916) introduced the following functions which were later recognized as special cases of normalized Eisenstein series. ∞ ∞ X nqn X n3qn P(q) = 1 − 24 , Q(q) = 1 + 240 , 1 − qn 1 − qn n=1 n=1

∞ X n5qn R(q) = 1 − 504 . 1 − qn n=1 Ramanujan Functions

In his famous paper on " On certain Arithmetical Functions" Ramanujan (1916) introduced the following functions which were later recognized as special cases of normalized Eisenstein series. ∞ ∞ X nqn X n3qn P(q) = 1 − 24 , Q(q) = 1 + 240 , 1 − qn 1 − qn n=1 n=1

∞ X n5qn R(q) = 1 − 504 . 1 − qn n=1

Thus P(q) = E2(τ), Q(q) = E4(τ), R(q) = E6(τ) where 1 X 1 E (τ) = for k ≥ 2. 2k 2ζ(2k) (mτ + n)2k (m,n)6=(0,0) Taking τ = ω2/ω1, the invariants g2, g3 satisfy

4 π 4 8 π 6 g2 = E4(τ), g3 = E6(τ). 3 ω1 27 ω1 The discriminant ∆ and J invariant function are related to the Ramanujan functions as

Q3 − R2 Y ∆ = = q (1 − qn)24 and J = Q3/∆. 123

J in terms of Ramanujan Functions The discriminant ∆ and J invariant function are related to the Ramanujan functions as

Q3 − R2 Y ∆ = = q (1 − qn)24 and J = Q3/∆. 123

J in terms of Ramanujan Functions

Taking τ = ω2/ω1, the invariants g2, g3 satisfy

4 π 4 8 π 6 g2 = E4(τ), g3 = E6(τ). 3 ω1 27 ω1 J in terms of Ramanujan Functions

Taking τ = ω2/ω1, the invariants g2, g3 satisfy

4 π 4 8 π 6 g2 = E4(τ), g3 = E6(τ). 3 ω1 27 ω1 The discriminant ∆ and J invariant function are related to the Ramanujan functions as

Q3 − R2 Y ∆ = = q (1 − qn)24 and J = Q3/∆. 123 Manin’s Conjecture(1971): Let a be an algebraic number distinct from 0 and 1. Then for any ξ ∈ H at least one of the numbers aξ, j(ξ) is transcendental. This is still open. Mahler proposed this in 1969 when a = e2πi . Mahler’s conjecture was resolved by K. Barré- Sirieix, G. Diaz, F. Gramain and G. Philibert in 1996. Thus we have For any τ ∈ C with Im τ > 0, at least one of the numbers e2πiτ , j(τ) is transcendental. In other words, For any algebraic q with 0 < |q| < 1, the value J(q) is transcendental.

Conjecture of Mahler and Manin Mahler proposed this in 1969 when a = e2πi . Mahler’s conjecture was resolved by K. Barré- Sirieix, G. Diaz, F. Gramain and G. Philibert in 1996. Thus we have For any τ ∈ C with Im τ > 0, at least one of the numbers e2πiτ , j(τ) is transcendental. In other words, For any algebraic q with 0 < |q| < 1, the value J(q) is transcendental.

Conjecture of Mahler and Manin

Manin’s Conjecture(1971): Let a be an algebraic number distinct from 0 and 1. Then for any ξ ∈ H at least one of the numbers aξ, j(ξ) is transcendental. This is still open. Mahler’s conjecture was resolved by K. Barré- Sirieix, G. Diaz, F. Gramain and G. Philibert in 1996. Thus we have For any τ ∈ C with Im τ > 0, at least one of the numbers e2πiτ , j(τ) is transcendental. In other words, For any algebraic q with 0 < |q| < 1, the value J(q) is transcendental.

Conjecture of Mahler and Manin

Manin’s Conjecture(1971): Let a be an algebraic number distinct from 0 and 1. Then for any ξ ∈ H at least one of the numbers aξ, j(ξ) is transcendental. This is still open. Mahler proposed this in 1969 when a = e2πi . Mahler’s conjecture was resolved by K. Barré- Sirieix, G. Diaz, F. Gramain and G. Philibert in 1996. Thus we have For any τ ∈ C with Im τ > 0, at least one of the numbers e2πiτ , j(τ) is transcendental. In other words, For any algebraic q with 0 < |q| < 1, the value J(q) is transcendental. Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . √ for any natural number d, the numbers π and eπ d are algebraically independent.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that Results of Nesterenko

Their method was used by Nesterenko to prove algebraic independence results for the values of the Ramanujan functions. For instance, it can be shown that If q ∈ C with 0 < |q| < 1, then at least three of the numbers q, P(q), Q(q), R(q) are algebraically independent over Q. From this result one can derive that If τ ∈ H not equivalent to i or e2πi/3 then at least three numbers among q, J(q), δJ(q), δ2J(q) are algebraically independent over d Q. Here δ = z dz . Further he also showed that √ for any natural number d, the numbers π and eπ d are algebraically independent. Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Coefficients of the j function Surprisingly, little is known about the behaviour of the coefficients of J(q) modulo a prime. Lehner (1949): made an extensive study on the congruence properties of the coefficients. In particular he showed that

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1 Lehner (1949): made an extensive study on the congruence properties of the coefﬁcients. In particular he showed that

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

Surprisingly, little is known about the behaviour of the coefﬁcients of J(q) modulo a prime. In particular he showed that

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

Surprisingly, little is known about the behaviour of the coefﬁcients of J(q) modulo a prime. Lehner (1949): made an extensive study on the congruence properties of the coefﬁcients. Thus for all even integers n, we have c(n) even. In fact, it is well known (we will see in a while) that c(n) is even whenever n 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? There are some contributions by Ono, Ramsey and Alfes.

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p). In fact, it is well known (we will see in a while) that c(n) is even whenever n 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? There are some contributions by Ono, Ramsey and Alfes.

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Thus for all even integers n, we have c(n) even. What can we say when n ≡ 7(mod 8)? There are some contributions by Ono, Ramsey and Alfes.

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Thus for all even integers n, we have c(n) even. In fact, it is well known (we will see in a while) that c(n) is even whenever n 6≡ 7(mod 8). There are some contributions by Ono, Ramsey and Alfes.

Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Thus for all even integers n, we have c(n) even. In fact, it is well known (we will see in a while) that c(n) is even whenever n 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? Coefﬁcients of the j function

Let us write ∞ X J(q) = c(n)qn with c(−1) = 1, c(0) = 744 and so on. n=−1

c(n) ≡ 0(mod p) for p =, 2, 3, 5, 7, 11 and n ≡ 0(mod p).

Thus for all even integers n, we have c(n) even. In fact, it is well known (we will see in a while) that c(n) is even whenever n 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? There are some contributions by Ono, Ramsey and Alfes. The following result is due to Ram Murty and Thangadurai (2015): There are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is odd; there are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is even. Further, the number of such n ≤ x is log log x in each case. It is expected that for half of n ≡ 7(mod 8) we should have c(n) odd.

A recent result there are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is even. Further, the number of such n ≤ x is log log x in each case. It is expected that for half of n ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai (2015): There are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is odd; Further, the number of such n ≤ x is log log x in each case. It is expected that for half of n ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai (2015): There are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is odd; there are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is even. It is expected that for half of n ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai (2015): There are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is odd; there are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is even. Further, the number of such n ≤ x is log log x in each case. A recent result

The following result is due to Ram Murty and Thangadurai (2015): There are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is odd; there are inﬁnitely many integers n ≡ 7(mod 8) for which c(n) is even. Further, the number of such n ≤ x is log log x in each case. It is expected that for half of n ≡ 7(mod 8) we should have c(n) odd. From the expression for the j− function,

J(q)∆(q) = Q3(q) ≡ 1(mod 2).

Now ∞ ∞ Y Y ∆(q) = q (1 − qn)24 ≡ q (1 + q8n + q16n + q24n)(mod 2). n=1 n=1 Hence c(n) ≡ 0(mod 2) if n + 1 6≡ 0(mod 8) proving that c(n) is even whenever n 6≡ 7(mod 8).

Proof of the result of Ram and Thanga Now ∞ ∞ Y Y ∆(q) = q (1 − qn)24 ≡ q (1 + q8n + q16n + q24n)(mod 2). n=1 n=1 Hence c(n) ≡ 0(mod 2) if n + 1 6≡ 0(mod 8) proving that c(n) is even whenever n 6≡ 7(mod 8).

Proof of the result of Ram and Thanga

From the expression for the j− function,

J(q)∆(q) = Q3(q) ≡ 1(mod 2). Hence c(n) ≡ 0(mod 2) if n + 1 6≡ 0(mod 8) proving that c(n) is even whenever n 6≡ 7(mod 8).

Proof of the result of Ram and Thanga

From the expression for the j− function,

J(q)∆(q) = Q3(q) ≡ 1(mod 2).

Now ∞ ∞ Y Y ∆(q) = q (1 − qn)24 ≡ q (1 + q8n + q16n + q24n)(mod 2). n=1 n=1 Proof of the result of Ram and Thanga

From the expression for the j− function,

J(q)∆(q) = Q3(q) ≡ 1(mod 2).

Now ∞ ∞ Y Y ∆(q) = q (1 − qn)24 ≡ q (1 + q8n + q16n + q24n)(mod 2). n=1 n=1 Hence c(n) ≡ 0(mod 2) if n + 1 6≡ 0(mod 8) proving that c(n) is even whenever n 6≡ 7(mod 8). Q∞ 8n 3 We again consider ∆(q) ≡ q n=1(1 − q ) (mod 2).The well known Jacobi identity: ∞ ∞ Y X (1 − qn)3 = (−1)k (2k + 1)qk(k+1)/2. n=1 k=0 Hence ∞ ∞ X X 2 ∆(q) ≡ q q8n(n+1)/2 ≡ q(2n+1) (mod 2). n=0 n=0 Thus we get ∞ X X c(n − (2k + 1)2)qn ≡ 1(mod 2). n=0 k≥0 Comparing the coefﬁcients of qn we get X c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1. m≥0

Use of Jacobi Identity The well known Jacobi identity: ∞ ∞ Y X (1 − qn)3 = (−1)k (2k + 1)qk(k+1)/2. n=1 k=0 Hence ∞ ∞ X X 2 ∆(q) ≡ q q8n(n+1)/2 ≡ q(2n+1) (mod 2). n=0 n=0 Thus we get ∞ X X c(n − (2k + 1)2)qn ≡ 1(mod 2). n=0 k≥0 Comparing the coefﬁcients of qn we get X c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1. m≥0

Use of Jacobi Identity

Q∞ 8n 3 We again consider ∆(q) ≡ q n=1(1 − q ) (mod 2). Hence ∞ ∞ X X 2 ∆(q) ≡ q q8n(n+1)/2 ≡ q(2n+1) (mod 2). n=0 n=0 Thus we get ∞ X X c(n − (2k + 1)2)qn ≡ 1(mod 2). n=0 k≥0 Comparing the coefﬁcients of qn we get X c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1. m≥0

Use of Jacobi Identity

Q∞ 8n 3 We again consider ∆(q) ≡ q n=1(1 − q ) (mod 2).The well known Jacobi identity: ∞ ∞ Y X (1 − qn)3 = (−1)k (2k + 1)qk(k+1)/2. n=1 k=0 Thus we get ∞ X X c(n − (2k + 1)2)qn ≡ 1(mod 2). n=0 k≥0 Comparing the coefﬁcients of qn we get X c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1. m≥0

Use of Jacobi Identity

Q∞ 8n 3 We again consider ∆(q) ≡ q n=1(1 − q ) (mod 2).The well known Jacobi identity: ∞ ∞ Y X (1 − qn)3 = (−1)k (2k + 1)qk(k+1)/2. n=1 k=0 Hence ∞ ∞ X X 2 ∆(q) ≡ q q8n(n+1)/2 ≡ q(2n+1) (mod 2). n=0 n=0 Thus we get ∞ X X c(n − (2k + 1)2)qn ≡ 1(mod 2). n=0 k≥0 Comparing the coefﬁcients of qn we get X c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1. m≥0 Take n = 4a(a + 1) − 1 with a ≥ 1. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0

Here when m = a we get the term c(−1) = 1. When m < a, observe that 4a(a + 1) − 4m(m + 1) > a. Thus if for all the integers m in the interval (a, 4a(a + 1) − 1], c(m) is even, the above sum gives a contradiction. Hence for some m in this interval, c(m) is odd. As already seen, m ≡ 7(mod 8).

Proof contd.. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0

Proof contd..

Take n = 4a(a + 1) − 1 with a ≥ 1. Here when m = a we get the term c(−1) = 1. When m < a, observe that 4a(a + 1) − 4m(m + 1) > a. Thus if for all the integers m in the interval (a, 4a(a + 1) − 1], c(m) is even, the above sum gives a contradiction. Hence for some m in this interval, c(m) is odd. As already seen, m ≡ 7(mod 8).

Proof contd..

Take n = 4a(a + 1) − 1 with a ≥ 1. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0 When m < a, observe that 4a(a + 1) − 4m(m + 1) > a. Thus if for all the integers m in the interval (a, 4a(a + 1) − 1], c(m) is even, the above sum gives a contradiction. Hence for some m in this interval, c(m) is odd. As already seen, m ≡ 7(mod 8).

Proof contd..

Take n = 4a(a + 1) − 1 with a ≥ 1. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0

Here when m = a we get the term c(−1) = 1. Thus if for all the integers m in the interval (a, 4a(a + 1) − 1], c(m) is even, the above sum gives a contradiction. Hence for some m in this interval, c(m) is odd. As already seen, m ≡ 7(mod 8).

Proof contd..

Take n = 4a(a + 1) − 1 with a ≥ 1. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0

Here when m = a we get the term c(−1) = 1. When m < a, observe that 4a(a + 1) − 4m(m + 1) > a. Hence for some m in this interval, c(m) is odd. As already seen, m ≡ 7(mod 8).

Proof contd..

Take n = 4a(a + 1) − 1 with a ≥ 1. Then the previous congruence becomes X X c(n−(2m+1)2) ≡ c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2). m≥0 m≥0

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x. Then each one of the intervals [a0, a1], (a1, a2], ··· , (a`, x] has an integer m such that c(m) is odd. Thus there are at least ` integers ≤ x for which c(m) is odd. Thus we need to get a lower bound for `. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd Let ` be the largest integer such that a` ≤ x. Then each one of the intervals [a0, a1], (a1, a2], ··· , (a`, x] has an integer m such that c(m) is odd. Thus there are at least ` integers ≤ x for which c(m) is odd. Thus we need to get a lower bound for `. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2. Then each one of the intervals [a0, a1], (a1, a2], ··· , (a`, x] has an integer m such that c(m) is odd. Thus there are at least ` integers ≤ x for which c(m) is odd. Thus we need to get a lower bound for `. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x. Thus there are at least ` integers ≤ x for which c(m) is odd. Thus we need to get a lower bound for `. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x. Then each one of the intervals [a0, a1], (a1, a2], ··· , (a`, x] has an integer m such that c(m) is odd. Thus we need to get a lower bound for `. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x. Then each one of the intervals [a0, a1], (a1, a2], ··· , (a`, x] has an integer m such that c(m) is odd. Thus there are at least ` integers ≤ x for which c(m) is odd. Using the recurrence relation satisﬁed by ak ’s, we get 2 ak < 8ak−1 for k ≥ 0. By iterating, we get

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0. The proof that there are log log x number of n ≤ x with c(n) even is similar.

Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. Counting n ≤ x with c(n) odd

Put a0 = 1, a1 = 7 and

ak = 4ak−1(ak−1 + 1) − 1 for k ≥ 2.

k 2k−1 2k ak ≤ 8 a1 ≤ 8 for k ≥ 0.

Since a`+1 ≥ x we get ` log log x. The proof that there are log log x number of n ≤ x with c(n) even is similar.