# Do there exist any Pythagorean triples that share two out of three numbers? [duplicate]

The following question occurred to me while sketching out vector diagrams earlier today: Do there exist positive integers $a, b, c, d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$? In other words, do there exist any Pythagorean triples that share a "leg", and for which the "leg" of one is the "hypotenuse" of the other? If not, can it be proven that such triples do not exist?

What I've tried: I've played around a bit with Euclid's formula, and it's easy enough to show that if such a pair of triples exists, it must be the case that there are integers $m$ and $n$ such that $$d^2 = m^4 - 6 m^2 n^2 + n^4$$ is a perfect square. ($m$ and $n$ are the integers on which the triple $(a,b,c)$ is based.) Similarly, if the triple $(a,d,b)$ is based on the integers $m'$ and $n'$, we end up with $$c^2 = 2\left({m'}^4 + {n'}^4\right)$$ This latter equation looks a little more plausible as something that doesn't have integer solutions. But I don't really know where to go from here, or even if this is the best way to think about it.

## marked as duplicate by Michael Seifert, Community♦Sep 15 '15 at 16:54

This is Fermat's right triangle theorem, since $a^2-b^2,a^2,a^2+b^2$ would form an arithmetic progression with common difference a square.