8

Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y \in \mathbb{Z}^+$.

I think that $x^2 + 2xy + y^2$ and $x^2 + y^2$ are not consecutive squares since the difference is even. I think it has some relation with other squares like $(x+y)^2$ and $(x-y)^2$.

How should I proceed? I would love some hints.

TheRandomGuy
  • 4,014
  • 2
  • 20
  • 56

2 Answers2

12

Suppose $x^2+y^2=a^2$ and $x^2-y^2=b^2$, then by multiplying them you get $x^4-y^4=(ab)^2$. This last equation has no non-trivial solution; see e.g. Solving $x^4-y^4=z^2$.

Community
  • 1
user148212
  • 592
  • 2
    This proves a stronger result - that $x^2+y^2$ and $x^2-y^2$ can't have their product equal to a square (except for trivial cases). – user236182 Apr 22 '16 at 14:57
  • @user236182 Exactly. – user148212 Apr 22 '16 at 15:03
  • I think that this proof can be modified for Fermat's Last Theorem where $n=4$. This question itself can be a different form of asking for a proof of the FLT where $n=4$. No doubt my problem sheet says this question was asked by Fermat. – TheRandomGuy Apr 23 '16 at 13:50
0

Hint:

If $ x^2+y^2=c^2$ is a Pythagorean triple then there are two integers $m,n$ such that $$ x=m^2-n^2 \qquad y=2mn \qquad c=m^2+n^2 $$

so $$ x^2-y^2=(m^2-n^2)^2-4m^2n^2=m^4+n^4-6 m^2n^2 $$

J. W. Tanner
  • 60,406
Emilio Novati
  • 62,675
  • Provided $x$ and $y$ are coprime, but this is not an essential restriction. You should also justify why $x$ is odd and $y$ is even (after reduction to a primitive Pythagorean triple). – egreg Apr 22 '16 at 15:36
  • 4
    Why can't $m^4+n^4-6 m^2n^2$ be a perfect square? – TheRandomGuy Apr 22 '16 at 15:38
  • Let m^2=x & n^2=y Then the equation reduces to x^2-6xy+y^2 In which the discriminant is 0 if and only if y=8 or n=2\sqrt 2 So this would not have any integer solution – Atul Mishra Dec 08 '16 at 06:27