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

If a prime $p$ divides $x^2+y^2$ then $x^2\equiv -y^2$ modulo $p$ and thus, if $p$ does not divide both $x$ and $y$ we find that $-1$ is a quadratic residue modulo $p$. This cannot happen unless $p$ is of the form $4k+1$.
First let's suppose that $x,y$ are coprime. If they are not, then simply divide by their $gcd$ and find some $x',y'$ which are coprime and still satisfy the hypothesis. 
Now you know that $\frac{x^2+y^2}{x+y}$ divides $2\cdot 23\cdot 43$. If $23$ or $43$ divides $x^2+y^2$ then bot $x$ and $y$ have $23$ or $43$ as a factor, which contradicts the fact that they are coprime. This leaves us with
$x^2+y^2=2(x+y)$ which is equivalent to $(x-1)^2+(y-1)^2=2$. From here we see that $x=y=2$.
There is another case: $x^2+y^2 = x+y$ which implies $(2x-1)^2+(2y-1)^2 = 2$. This implies $x=y=1$.
