Pythagorean triple with hypotenuse a power of $2$ Is there a Pythagorean triple whose largest element is a power of 2? That is: are there solutions to $a^2 + b^2 = 2^{2k}$ in the positive integers?
 A: A power of 2 does not have an odd factor. To explain WHY the hypotenuse cannot be a power of 2, Pythagorean triplets can typically be represented by the formulas $m^2-n^2$ and $2mn$ for the legs, and $m^2+n^2$ for the hypotenuse. But $2mn$ is always even when the triplet is primitive. Now you can conclude. 
A: An extended (but not much more complicated) answer, after Matthew Daly has pointed the way:
Suppose that $a^2 + b^2 = c^2$ with $a,b,c > 1$ having no factors common to all three. It follows that $a$, $b$, and $c$ are pairwise coprime. Let $p$ be a prime factor of $c$: we then have ${a^2 + b^2 \equiv 0 \pmod{p^2}}$. It also follows that $a$ and $b$ are coprime to $p$, from which we may infer that ${i^2 + 1 \equiv 0 \pmod{p^2}}$ has solutions, from which we may infer $p \ne 2$. As the same congruence $a^2 + b^2 \equiv 0$ holds modulo $p$ as well, we can in fact infer that $p = 4n+1$ for some integer $n$.
Thus the factors of $c$ are all odd integers congruent to $1$ modulo $4$. It follows that if $c = d^k$ for some positive integers $d,k > 1$, then  $d \ne 2$ (and also $d \notin \{3, 6, 7, 11, 12, 14, \ldots\}$).
A: No.  Even if it were not a primitive Pythagorean triple, you would need a reduced value to have a hypotenuse that is a power of 2.  And the hypotenuse of a primitive Pythagorean triple is an odd integer larger than  1, as its square is the sum of an odd and an even number.
