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.

Here is essentially André Nicholas's answer in a different form.
Borrowing the proof style and structure and properties $(0)$ and $(1)$ from another answer of mine (https://math.stackexchange.com/a/451278/11994), the original statement translates to 
$$\langle \forall d :: d|a \land d|c \;\equiv\; d|b \land d|c \rangle$$
or equivalently (by extracting the common conjunct)
$$\langle \forall d : d|c : d|a \equiv d|b \rangle$$
which we must prove under the assumption $\;a+b=c^2\;$.
This is easily done, for any $\;d\;$, as follows:
\begin{align}
& d|a \;\equiv\; d|b \\
\Leftarrow & \;\;\;\;\;\text{"property of divisibility: numbers are equally divisible if their sum is} \\
& \;\;\;\;\;\phantom{"}\text{-- suggested by our assumption"} \\
& d|(a+b) \\
\equiv& \;\;\;\;\;\text{"our assumption"} \\
& d|c^2 \\
\Leftarrow& \;\;\;\;\;\text{"property of divisibility: a divisor of a factor also divides the product"} \\
& d|c \\
\end{align}
This completes the proof, and the last step shows that the assumption can be generalized to $\;a+b=c\cdot n\;$ for any integer $\;n\;$, or in other words, $\;c|(a+b)\;$.
