TITLE: Find all $(x,y)$ of positive integers s.t. $x^3-y^3=xy+61$
QUESTION [5 upvotes]: Given 
$x^3-y^3=xy+61$.
Find all pairs (x,y) of positive integers which satisfies the given equation.
I tried solving this by following method-
$(x-y)(x^2+y^2+xy)-xy=61$
$(x-y)[(x-y)^2+3xy]-xy=61$
$(x-y)^3+3xy(x-y)-xy=61$
I don't know if doing this was in any means fruitful or not. 
Remember $x>y$, only then a solution is possible.

REPLY [5 votes]: $x\geq y+1$ so:
$61 = x^3-y^3-xy=x(x^2-y)-y^3\geq (y+1)(y^2+y+1)-y^3=2y^2+2y+1$
$y^2+y\leq 30$, so $1\leq y\leq 5$ - just check this numbers