Arithmetica of Diophantus

0.13 Diophantus

0.13.1 Arithmetica II.8 To divide a given square number into two squares. Solution. Given square number 16. x2 one of the required squares. Therefore 16 x2 must be equal to a square. Take a square of the form (mx 4)2, m being any integer− 1 and 4 the number which is the square root of 16, for example− , take (2x 4)2, and 2 2 2 2 − 16 equate it to 16 x . Therefore, 4x 16x +16= 16 x , or 5x = 16x, and x = 5 . − 256− 144 − The required squares are therefore 25 , 25 . This example is the historical origin of the famous Fermat Last Theorem.

0.13.2 Lemma to Arith. VI.12 Given two numbers the sum of which is a square, an infinite number of squares can be found such that, when the square is multiplied by one of the given numbers and the product is added to the other, the result is a square. (Exercise) Imitate the previous example to work out a Solution. for 3, 6.

Theorem 0.8. Let f(x, y) be an irreducible quadratic polynomial in x, y with integer (rational) coefficients. If f(x, y)=0 has a rational Solution. (x0,y0), then f(x, y)= 0 has infinitely many rational solutions.

Proof. For each rational number m, the line through (x0,y0) with slope m, namely, y − y0 = m(x x0), intersects the curve f(x, y)=0 in two points whose x-coordinates are − the roots of the quadratic equation f(x, y0 +m(x x0))=0 with rational coefficients. − One of the two roots is the rational number x0. The second root must be a rational number x1, since the sum of the two roots is rational. This yields a rational point (x1,y1) on the curve.

0.13.3 Arithmetica IV.24 To divided a given number into two parts such that the product is a cube minus its side.

1Since Diophantus accepts rational solutions, m can be an arbitrary rational number. 26 CONTENTS

Solution. Given number 6. First part x; therefore second = 6 x, and 6x x2 = a cube minus its side. Form a cube from a side of the form mx 1−, say, 2x −1, and equate 6x x2 to this cube minus its side. Therefore, 8x3 12x2−+4x =6x −x2. Now, if the coefficient− of x were the same on both sides, this− would reduce to a− simple equation, amd x would be rational. In order that this may be the case, we must put m for 2 in our assumption, where 3m m = 6 (the 6 being the given number in the original hypothesis). Thus, m =3. We− therefore assume (3x 1)3 (3x 1)=6x x3, or 27x3 27x2 +6x =6x x2, and x = 26 . The parts− are 26−and 136− . − − − 27 27 27 Remark. This problem seeks rational Solution. of the equation y3 y = 6x x2. Clearly, (0, 1) is a rational Solution.. The line through (0, 1) with slope− m (assumed− rational) has equation y = mx 1. This line cuts the curve E : y3 y = x2 6x at 3 points, one of which is (0,−1). In general, there is no guarantee− that any of− the remaining two points is rational. However, if this line is tangent to E at (0, 1, ), then (0, 1) being counted twice, it is clear that the remaining point is rational. The tangent 26 136 to E at (0, 1) turns out to be y = 3m 1 (exercise), and we obtain ( 27 , 27 ) for the third point. −

Exercise VI.1 Diophantus of Alexandria was believed to have lived in the 3rd century A.D. His main work Arithmetica consists of 7 books of worked examples on rational solutions of indeterminate equations.

God granted him to be a boy for the sixth part of his life, and adding a twelfth part to this, He clothed his cheeks with down. He lit him the light of wedlock after a seventh part, and five years after his marriage, He granted him a son. Alas ! late born wretched child, after attaining the measure of half of his father’s life, chill Fate took him. After consoling his grief by this science of numbers for four years he ended his life.

How long has Diophantus lived? 2

VI.2 Find all rational points on the ellipse 4x2 +9y2 = 36. VI.3 Find all rational Solution. of x2 5y2 = 1. − − VI.4 Use the method of Arithmetica IV.24 to find a rational point on y2 = x3 +2, starting with ( 1, 1). 3 −

2Answer: If this information is correct, then Diophantus married at 33, had a son who died at 42, four years before Diophantus himself died at 84. (Stillwell, Mathematics and Its History, p.36, Springer - Verlag, 1989). 3 17 71 Answer: ( 4 , 8 ).