<s>[INST] $(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$ I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.

Find $x$, $y$ integers such that
  $$ (x+1)^2 + (y+1)^2 + xy(x+y+3)=2 $$

Some basic algebra will lead to:
$$
(x+y)^2 + x(xy+2) + y(xy+2) + xy = 0$$
 $$(x+y)(xy+x+y+2)=-xy$$
Now, some divisibility properties should end the problem.
Well, maybe my approach isn't the right one, but I'm just stuck at this point.
A piece of advice or a hint would be apreciated.
 [/INST] Replace $x+1$ by $a$ and $y+1$ by $b$ and you get
$$2=a^2+b^2 +(a-1)(b-1)(a+b+1)+1=a^2b+b^2a-ab =ab(a+b-1)+1.$$
Since $x$ and $y$ are integers, $a,b$ are integers too. Therefore $a,b,a+b-1 \in \{-1,1\}$ and so $a,b \in \{-1,1\}$ and $a+b \in \{0,2\}$. If $a=b=-1$ we get a contradiction. Since the starting equation was symmetric in $a$ and $b$, lets assume $a=1$. Then $b=1$ and $b=-1$ are still valid.
So the only solutions for $(x,y)$ are $(0,0), (2,0), (0,2)$.
</s>