TITLE: Solve this Diophantine Equation: $a^3-b^3 = ab+61$
QUESTION [9 upvotes]: "solve this diophantine equation: $a^3-b^3 = ab+61$"
For Diophantine Equations I have always had the question in the form $ax+by = c$ and solved it via Extended Euclidean Algorithm. However, this question is not formatted in such a way. 
What I am thinking is that we move $ab$ to the LHS so that we can have a single constant on the RHS. Maybe then we could expand? I'm not sure. Hints/Advice is what I am looking for. 
Maybe we can substitute $a$ and $b$ with something else of a form I am more comfortable with? I'm not really sure.

REPLY [11 votes]: $$(a-b)(a^2+ab+b^2)=ab+61$$
$$\implies a^2+ab+b^2 \le \mid ab+61\mid$$
The negative case inside the absolute value readily leads to contradiction, so now we have
$$a^2+b^2\le61$$
These are finitely many cases to check(having even less if we note that $a\ge b$ and that $(a,b)=1$ (why?)), so we end up with $a=6,b=5;a=-5,b=-6$.