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.

Hint $\ \ (b,c) = (c^2-a,c) = (-a,c),\:$ since $\:(n,c) = (n{\rm\: mod\:} c, c)\ $ by Euclid.
