Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a multiple of 11?
Dont know where to progress after this. All help appreciated.

Since $y$ is an integer,$\frac{-11x}{x+11}$ will be an integer. When $\frac{x}{x+11}$ is an integer, it refers that $\frac{11}{x+11}$ is an interger, or $x=0$. Then you can find all such $x$ in this situation. Otherwise $\frac{x}{x+11}$ is not an integer but $\frac{-11x}{x+11}$ is. Thus $x$ is a multiple of $11$, write $x=11k$ where $k$ is an integer and solve $k$ in the same way.
