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.


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 $3\pmod4$.

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    $\begingroup$ 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

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