TITLE: Solving $x^3-y^3=xy+61$ in integers
QUESTION [10 upvotes]: I'm trying to solve the following equation:
$$x^3-y^3=xy+61$$
I got:
$$(x-y)(x^2+xy+y^2)=xy+61$$
But I can't go any further. I am looking for a solution in integers.
I need some hints to proceed, please!

REPLY [21 votes]: Put $x+y=:u$, $x-y=:v$; then $u$ and $v$ are integers of the same parity. In the new variables the equation becomes
$${1\over8}(6u^2v+2v^3)={1\over4}(u^2-v^2)+61\ ,$$
which can be written as
$$27 u^2+(3v+2)^2={6584\over3 v-1}\ .$$
It follows that $v>1$ and that $3v-1$ must divide $6584=8\cdot 823$; whence  $v\in\{1,3,549,2195\}$. $549$ and $2195$ are obviously too large, and $v:=3$ leeds to $u^2=26$ which has no integer solutions. It remains $v=1$ which leads to $u^2=121$ or $u=\pm11$. Therefore we only have the two solutions $(x_1,y_1):=(6,5)$ and $(x_2,y_2):=(-5,-6)$.