Proving $\gcd(a, c) = \gcd(b, c)$ for $a + b = c^2$ I am trying to prove that, given positive integers $a, b, c$ such that $a + b = c^2$, $\gcd(a, c) = \gcd(b, c)$. I am getting a bit stuck.
I have written down that $(a, c) = ra + sc$ and $(b, c) = xb + yc$ for some integers $r, s, x, y$. I am now trying to see how I can manipulate these expressions considering that $a + b = c^2$ in order to work towards $ra + sc = xb + yc$ which means $(a, c) = (b, c)$. Am I starting off correctly, or am I missing something important? Any advice would help.

I don't see a way to proceed using the approach you suggest - this doesn't mean that there isn't one (it's often a good method to work through). But I don't see it yet. But you can directly show that the same primes to the same powers divide each:
Consider a prime $p$. Suppose that $p^\beta \mid \mid \gcd(a,c)$, so that in particular $p^\beta \mid c^2 - a = b$. And so $p^\beta | a$.
Suppose now that $p^\alpha \mid \mid a$. Then we know that $\alpha \geq \beta$ from this argument. Reversing the roles of $a$ and $b$ in the same argument shows that $\beta \geq \alpha$. Thus $\alpha = \beta$, and so the same primes to the same powers maximally divide $\gcd(a,c)$ and $\gcd(b,c)$.
Further, it seems this argument generalizes to cases that look like $a + b = c^n$ for any $n \geq 1$.
