There don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares

Question

I need to prove that there don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares.

I suopose that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ with $c, d$ positive integers, so $$(a^2 + b^2)(a^2 - b^2) = c^2d^2$$ Therefore $$(a^2 + b^2)(a^2 - b^2) = (cd)^2$$ and $$a^4 - b^4 = (cd)^2$$ but this equation doesn't have solution in positive integers.

Is that right?

"And also, can you help with this?" That is a completely unrelated question to the first and deserves its own separate post. Do not edit it into this post. — JMoravitz, May 31 '17 at 18:30
The last equation doesn't have solutions in the positive integers about Fermat Theorem. — BpZ, May 31 '17 at 18:34
@Macavity The case when $a$ and $b$ are both odd is not so easily reduced. — Harald Hanche-Olsen, May 31 '17 at 18:40
@Macavity Ah, of course. Still, “not so easily” still stands, sort of. I was just thinking of splitting off common powers of $2$ as far as possible, not the next step. — Harald Hanche-Olsen, May 31 '17 at 18:49
No, the equation $a^4-b^4=x^2$ is not violating Fermat's theorem. — Harald Hanche-Olsen, May 31 '17 at 18:50
@HaraldHanche-Olsen I invite you to show the case n=4, https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#n.C2.A0.3D.C2.A04 — BpZ, May 31 '17 at 19:01
Oh, that Fermat theorem. Okay, then. I thought you meant Fermat's last theorem. — Harald Hanche-Olsen, May 31 '17 at 19:42
Hmmm. you have a point. This problem is equivalent to Fermats Right Triangle theorem, but I think referring to it and not proving FRRT is avoiding the point of the excercise. Your proof is legitimate (if you refer to FRRT) but I might call upon you to then prove FRRT. — fleablood, May 31 '17 at 21:49
Fermat's Right Triangle Theorem. That $x^4 - y^4 = z^4$ has no integer solutions. You were the one who quoted it. — fleablood, May 31 '17 at 22:58
Thank you. Well, in my class I did the proof of FRTT. But, I am really nervous because I don't know if my proof is ok. — BpZ, May 31 '17 at 23:23

score 3 · Accepted Answer · edited Jun 03 '17 at 10:21

3

If $a^4-b^4=(cd)^2$ then that yields $b^4+(cd)^2=a^4$ . @awllower proved here that there can never exist a right triangle with integer sides such that a leg and the hypotenuse can be perfect squares.

edited Jun 03 '17 at 10:21

Oscar Lanzi

39,403

answered May 31 '17 at 18:40

user451220

46

You mean $b^4+(cd)^2=a^4$, but it still works – John Doe May 31 '17 at 18:45

There don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares

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