<s>[INST] Solving $x^3-y^3=xy+61$ in integers 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!
 [/INST] If we restrict ourselves to positive solutions, we can do it by brute force. In that case clearly $x>y>0$.
If $x-y=1$, we have $x^2+y^2=61$, which has the unique solution $x=6,y=5$.
If $x-y=2$, we have $2(x^2+xy+y^2)=xy+61$, or $2(x^2+y^2)+xy=61$; clearly $x$ and $y$ must both be odd, with $x^2+y^2<30$. The only candidate is $x=3,y=1$, which clearly doesn't work.
If $x-y=3$, we have $3(x^2+y^2)+2xy=61$; here $x$ and $y$ must have opposite parity, with $x^2+y^2<20$, so $x=4,y=1$ is the only candidate, and it doesn't (quite) work.
If $x-y=4$, we have $4(x^2+y^2)+3xy=61$, so $x$ and $y$ must be odd, and $x^2+y^2<15$; there are no solutions in this case.
Suppose that $x-y\ge 4$. Then $(x-y)(x^2+xy+y^2)\ge 4(5^2+5\cdot1+1^2)>61$, so there are no solutions.
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