International Journal of Algebra, Vol. 7, 2013, no. 16, 749 - 753 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2013.3984

A Note on Congruent Numbers

Umm¨¨ ug¨uls¨um O˘¨ g¨ut and Refik Keskin

Sakarya University, Mathematics Department, Sakarya, Turkey [email protected], [email protected]

Copyright c 2013 Umm¨¨ ug¨uls¨um O˘¨ g¨ut and Refik Keskin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this study, by showing that the systems of simultaneous equations

a2 − b2 = x2

and

a2 + b2 = y2

have no solutions in positive integers, we proved that a congruent num- ber can not be a perfect square. Moreover we proved Fermat’s last theorem for n =4.

Keywords: Congruent number, primitive Pythagorean triple, Fermat’s last theorem

1 Introduction

The problem of studying positive integers n which occur as areas of rational right triangle was of interest to the Greeks. The congruent number problem was first discussed systematically by Arab scholars of the tenth century. By the way recall that a positive integer n is a congruent number if it equals to the area of right triangle with rational sides. Since tenth century, some well-known mathematicians have devoted con- siderable energy of the congruent number problem. For example Euler showed n 24 35 337 that = 7 is a congruent number with sides of lenght 5 , 12 and 60 .Itis 750 Umm¨¨ ug¨uls¨um O˘¨g¨ut and Refik Keskin known that Leonardo Pisano (Fibonacci) was challenged around 1220 by Jo- hannes of Palermo to find a rational right triangle of area 5. He found the 3 20 41 right triangle with sides of lenght 2 , 3 and 6 . Notice that the definition of a congruent number does not require the sides of the triangle to be integer, only rational. While n = 6 is the smallest possible area of a right triangle with inte- ger sides of lenght 3,4,5 , n = 5 is the area of right triangle with rational sides 3 20 41 n of lenght 2 , 3 and 6 .So = 5 is the smallest congruent number. In 1225, Fibonacci wrote a general treatment about the congruent number problem, in which he stated out without proof that if n is a perfect square then n cannot be a congruent number. The proof of such a claim had to wait until Pierre de Fermat. He showed that n = 1 and so every square number is not a congruent number by using his method of infinite descent[6]. One can look at [4] and [7] for Fermat’s descent method. In the present study we will show that if n is a congruent number then n can not be a perfect square by using the same method. Moreover, we proved Fermat’s last theorem for n =4, which states that the equation x4 + y4 = z4 has no solutions in positive integers.

2 Main Theorems

Definition 1 Let x, y and z be positive integers. We say (x, y, z) is a Pythagorean triple if x2 + y2 = z2.If(x, y, z) is a Pythagorean triple and gcd(x, y, z)=1, then we say that (x, y, z) is a primitive Pythagorean triple.

The following theorem is well known and can be found many textbook on number theory(see [1]).

Lemma 2 Let (x, y, z) be a primitive Pytagorean triple. Then there exist op- posite parity natural numbers u and v such that x =2uv, y = u2 − v2 and z = u2 + v2 with (u, v)=1.

From the above lemma it follows that if (x, y, z) is a primitive Pytagorean triple, then z and one of x and y are odd. Now, we will give the following lemma which will be needed during the proof of the main theorem.

Lemma 3 There is no positive integers a, b, x, y such that a2 − b2 = x2 and a2 + b2 = y2.

Proof. Assume that b is the minimum integer satisfying the equations a2−b2 = x2 and a2 + b2 = y2. Then, it follows that (a, b, x)=(a, b, y) = 1. Since a2 − b2 = x2 and a2 + b2 = y2, it is clear that both (a, b, x) and (a, b, y) are A note on congruent numbers 751 primitive Pytagorean triples. Hence, a is odd and therefore b is even. Let z = xy. Then we may write    2 a4 − b4 = a2 − b2 a2 + b2 = x2y2 =(xy) = z2.

2 2 That is, (a2) = z2 +(b2) . Furthermore, it can be easily seen that (z,b2,a2)= 1. Thus, there exist opposite parity natural numbers u and v with (u, v)=1 such that b2 =2uv, a2 = u2 +v2,z = u2 −v2. Since a2 = u2 +v2 and (u, v)=1, it is seen that (u, v, a) is a primitive Pythagorean triple. Without loss of generality, we may suppose that u is odd and v is even. Then, there exist positive integers r and s with (r, s) = 1 such that a = r2 + s2, v =2rs, u = r2 − s2. Substituting these values of u and v into b2 =2uv gives

b2 =2uv =2(r2 − s2)2rs =4rs(r2 − s2), which implies that   b 2 = rs(r2 − s2). 2 Since (r, s)=1, it is easy to see that (rs, r2 − s2) = 1. Here both rs and r2 − s2 are perfect square. Since r and s are opposite parity and (r, s) = 1, it follows that (r − s, r + s) = 1. Using the fact that r2 − s2 =(r − s)(r + s), it is seen that r − s and r + s are perfect square. Then, there exist integers t, k, c, d such that r = t2,s= k2,r− s = c2 and r + s = d2. So, we immediately have t2 − k2 = c2 and t2 + k2 = d2. Besides, using v =2rs, b2 =2uv and s = k2, we see that b2 =4urk2. This implies that k

Theorem 4 Let n be a congruent number. Then n can not be a perfect square.

Proof. Assume that n is a congruent number and n = k2 for some k ∈ Z. Since n is a congruent number, there is a right triangle with rational sides x, y, z n xy . such that = 2 We may suppose that x = a/m, y = b/m, z = c/m for some positive integers a, b, c, m. By using x2 + y2 = z2, we get    a 2 b 2  c 2 , m + m = m which implies that a2 + b2 = c2. Since xy (a/m)(b/m) ab k2 = n = = = , 2 2 2m2 752 Umm¨¨ ug¨uls¨um O˘¨g¨ut and Refik Keskin we get

ab =2m2k2.

2 By using the equations a2 +b2 = c2 and ab =2(mk) , we can easily write that

2 2 (a + b) = a2 + b2 +2ab = c2 +(2mk) and

2 2 (a − b) = a2 + b2 − 2ab = c2 − (2km) .

But this is impossible by Lemma 3. This completes the proof. The proof of the following lemma is easy and will be omitted.

Lemma 5 Let n = s2m with m squarefree. Then n is a congruent number iff m is a congruent number.

By using Theorem 4 and Lemma 5 we can give the following.

Corollary 6 1 is not a congruent number.

Corollary 7 The equation x4 + y4 = z4 has no solutions in positive integers.

Proof. Assume that x4 + y4 = z4 for some positive integers x,y, z. Let d = gcd(x, y, z). Then x = da, y = db, and z = dc for some positive integers a, b, c with gcd(a, b, c)=1. Then it follows that a4 + b4 = c4. That is, (a2)2 +(b2)2 = (c2)2. This shows that (a2,b2,c2) is a a primitive Pythagorean triple. Therefore, b2 = u2 − v2 and c2 = u2 + v2 for some positive integers u and v by Lemma 2. But this is impossible by Lemma 3. This completes the proof.

References

[1] A. Adler and J. E. Cloury, The Theory of Numbers: A Text and Source Book of Problems, Jones and Bartlett Publishers, Boston, MA, 1995.

[2] J. S. Chahal, Congruent Numbers and Elliptic Curves, Amer. Math. Monthly, 113 (2006), 308-317.

[3] J. Coates, Congruent Number Problem, Pure and Appl. Math. Quaterly, 1 (2005), 14-27.

[4] K. Conrad, The Congruent Number Problem, http://www.thehcmr.org/issue2 2/congruent number.pdf A note on congruent numbers 753

[5] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Second Edition, Springer Verlag, 1993

[6] A. L. Robledo, Elliptic Curves, Modular Forms and their L-Functions, Student Mathematical Library, 2010

[7] S. W. Zhang, Congruent Numbers and Heegner Points, Asia Pacific Newsletter, 3 (2013), 12-15.

Received: October 24, 2013