<s>[INST] $\frac{x^2+y^2}{x+y}$ is a divisor of $1978$ 
Two nonzero integers $x,y$ (not necessarily positive) are such that $x+y$ is a divisor of $x^2+y^2$, and the quotient $\dfrac{x^2+y^2}{x+y}$ is a divisor of $1978$. Prove that $x = y$.

Let $A = \dfrac{x^2+y^2}{x+y}$ where $A$ is an integer divisor of $1978$. Then $A \mid x^2+y^2$ and if $p$ is a prime divisor of $A$ we must have $x^2+y^2 \equiv 0 \pmod{p}$. 
We first deal with the case $x,y \equiv 0 \pmod{q}$. Otherwise $x,y \not \equiv 0 \pmod{q}$ where $q \equiv 3 \pmod{4}$ and is a prime divisor of $A$. 
I didn't find an easy way of dealing with the case that $x,y \equiv 0 \pmod{q}$. Is there an easier way of solving this?
 [/INST] $1978=2\cdot23\cdot43$.
As shown in this answer:

a prime $p\equiv3\pmod{4}$ divides $a^2+b^2$ ... implies that $p\,|\,a$ and $p\,|\,b$.

If $\left.23\,\middle|\,\frac{x^2+y^2}{x+y}\right.$, then $x=23u$ and $y=23v$ and $\frac{x^2+y^2}{x+y}=23\frac{u^2+v^2}{u+v}$. Then $\frac{u^2+v^2}{u+v}$ is a factor of $2\cdot43$.
If $\left.43\,\middle|\,\frac{x^2+y^2}{x+y}\right.$, then $x=43u$ and $y=43v$ and $\frac{x^2+y^2}{x+y}=43\frac{u^2+v^2}{u+v}$. Then $\frac{u^2+v^2}{u+v}$ is a factor of $2\cdot23$.
Via some combination of the preceding, we get an $s$ and $t$ so that
$$
\frac{s^2+t^2}{s+t}\in\{1,2\}
$$
the only solutions to which are $s=t\in\{1,2\}$. Multiplying by the appropriate combination of $23$ and $43$, we get that $x=y$.
</s>