<s>[INST] Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$ 
Find the set of integer solutions $(x,y)$ to
  $$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$

I can't seem to multiply both sides by the expression in the denominator. 
Nor can I simplify and cancel any terms. How should I it?
 [/INST] $$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y \iff \frac{x+y}{x-y}> x+y$$
Case 1: $x+y > 0$
In this case, we have $\dfrac1{x-y}> 1 \iff 0 < x-y< 1$ which is not possible.
Case 2: $x+y < 0$
In this case, we have $\dfrac1{x-y} < 1 \iff x-y < 0$ or $x-y > 1$.   
Now $x+y < 0, \; x-y < 0$ gives $x < 0, \;x < y < -x$
and $x+y < 0, \; x-y > 1$ gives $y < 0, \;1+y< x < -y$ 
Hence we can write the possible solutions also as integer pairs satisfying $\{|y| < -x\} \cup \{|x| < -y-\frac12\}$.
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