2
$\begingroup$

Let $(a,b,c)$ be a Pythagorean triple, which means $c^2=a^2+b^2$.

If $c$ is odd and $a$ & $b$ are relatively prime, then there exist integers $m$ and $n$ such that $c=m^2+n^2, ~a=m^2-n^2, ~b=2mn$.

One can easily check the above by proving $\gcd(\frac{a+c}{2},\frac{a-c}{2})=1$.

My question is whether the converse of the above also holds or not. ;Whenever the odd number $c$ is a sum of 2 squares, then there exist integers $a$ and $b$ satisfying $c^2=a^2+b^2$ and $\gcd(a,b)=1$.

Any help will be appreciated.

$\endgroup$
2
$\begingroup$

The odd number $45$ is a sum of two squares, but not of two relatively prime squares.

More generally, let the odd number $n$ be of the form $p_1^{2a_1}\cdots p_k^{2a_k}m$, where the $p_i$ are primes congruent to $3$ modulo $4$, and $m$ is a product of primes congruent to $1$ modulo $4$. Then $n$ is a sum of two squares, but is not the sum of two relatively prime squares.

$\endgroup$

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.