$\frac{x^2+y^2}{x+y}$ 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?

For the equation.  
$$A=\frac{x^2+y^2}{x+y}$$
It may be better to write a decision?
$$x=(p-s)(k-s)$$
$$y=(p-s)(p-k)$$
$$A=p^2+s^2+2k(k-p-s)$$
