Find two pairs of relatively prime positive integers $(a,c)$ so that $a^2 + 5929 = c^2$. Can you find additional pairs with $gcd(a,c) > 1$?
What I know:
$gcd(a,c) = 1$ implies that there are some $x$ and $y$ such that $ax + cy = 1$. Since $a d$oes not divide $c$, I'm guessing that $a^2$ does not divide $c^2$ as well (need confirmation). In that case we then have $gcd(a^2, c^2) = 1$ so there are some x and y such that $a^2 x + c^2 y = 1$. I'm not 100% sure if that leads us anywhere but it does give an equation that is somewhat matching the question.
Let $c^2 = d$ We know that a number $d$ can be written as a sum of two squares if all its prime factors are either 2 or congruent to $1 (mod 4)$. We have $\sqrt(5929) = 77$. So we have that if $d = a^2 + 5929$, $d$ must be a product of distinct primes that are congruent to $1 (mod 4)$ or 2.
What am I missing from here that is keeping me back from answering this? It doesn't seem like a very difficult question yet I'm having trouble with it. Maybe its midnight speaking :(.