4
$\begingroup$

All primitive Pythagorean triples $(a, b, c) : \{ a^2 + b^2 = c^2 \} \wedge \{ a \equiv 0 \pmod{2} \}$ can be expressed in the form:$$\{ a = 2pq, b = p^2 - q^2, c = p^2 + q^2 \}$$ for positive integers $p, q : \{ \gcd(p,q) = 1 \} \wedge \{ p \not\equiv q \pmod{2} \}$.


I conjectured that this also holds for imprimitive Pythagorean triples (in this case $p,q$ are not necessarily relatively prime and of opposite parity).

However, I could not find any counterexamples and currently I am stuck in the developing of a proof.

That is why I am appealing to you. I would really appreciate any counterexamples, proofs, ideas, etc.

Thank you.

$\endgroup$
2
$\begingroup$

Sadly, it's not true for the general case. Easiest counterexample is to take the $3-4-5$ right triangle and multiply each side by $3$. $15$ cannot be written as the sum of $2$ squares. The sum of $2$ squares cannot be congruent to $3\pmod4$.

$\endgroup$
  • 1
    $\begingroup$ Just to provide a reference, the last statement by Mike is from Fermat's Two Square Theorem which might be of interest to the OP. $\endgroup$ – Sandeep Silwal Apr 13 '14 at 0:06
  • $\begingroup$ That certainly is sad. A mistake on my force-brute algorithm was bypassing all these triples. I did not know that theorem, as far as I have been reading, it is very interesting. Thank you for your answers. $\endgroup$ – Toto Apr 13 '14 at 13:41
  • $\begingroup$ Only primitive triples are guaranteed to have the oft-quoted form. $\endgroup$ – Oscar Lanzi Dec 29 '16 at 11:18

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.