6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I assume you mean non-zero integers $\endgroup$ – J. W. Tanner Aug 23 at 12:37
  • 1
    $\begingroup$ @J.W.Tanner: yes, duly edited. $\endgroup$ – Niel de Beaudrap Aug 23 at 12:39
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A very simple answer. Thanks for the quick response! $\endgroup$ – Niel de Beaudrap Aug 23 at 12:41
  • 1
    $\begingroup$ Is that a very condensed and significantly more elegant version of what I arrived at? The even hypotenuse squared must be the sum of two even numbers or two odd numbers. If even, then all three sides are even, divide the base triangle by 2 and restart. If odd, they're both of the form (4n + 1), meaning the hypotenuse squared has the form (4x + 2) which is impossible, since all even squares are of the form 4x. Therefore we have proof by contradiction it's not possible. $\endgroup$ – dgnuff Aug 23 at 23:13
  • 1
    $\begingroup$ @dgnuff Yeah, that's the same logic of it. Generally speaking, "... and restart" is describing an algorithm and not a proof. You can see how I got around that by focusing on the primitive triple, which would be known to not be all even. $\endgroup$ – Matthew Daly Aug 24 at 7:51
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Fixed, thanks........ $\endgroup$ – imranfat Aug 23 at 12:55
3
$\begingroup$

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 $d \notin \{3, 6, 7, 11, 12, 14, \ldots\}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.