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.
Hint: First, dividing your second line by $2$ gives
$$xy + 11x + 11y = 0 \tag{1}\label{eq1A}$$
Next, using Simon's favorite factoring trick gives
$$(x + 11)(y + 11) = xy + 11x + 11y + 121 \tag{2}\label{eq2A}$$
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.
Let $x=a-11,y=b-11$ in $xy+11x+11y=0$ to get $$ab=121$$ Since the change of variables is completely reversible in the integers, every divisor of $121$ corresponds one-to-one to a solution: $a=\pm1,\pm11,\pm121$, correspinding to $$(x,y)=(0,0),(-22,-22),(-10,110),(110,-10),(-12,-132),(-132,-12)$$
Hint for another way:Very useful the "Simon's trick" cited above by @John Omielan to solve this kind of diophantine equations.We want to give another kind of solution.
From $11=\dfrac{-xy}{x+y}$ we have $-xy=11K$ and $x+y=K$ so we can put $y=11M$ from which $$-xM=K\text { and }x+11M=K\Rightarrow \begin{cases}x=\dfrac{-11M}{M+1}\\y=11M\end{cases}$$ a parametrization of the variables from which immediately the solutions corresponding to $M=0$ and $M=10$ the other being less immediate but not hard to find.
