<s>[INST] Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :- 
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is clearly rejected as $x,n$ are both positive.
CASE $2$ : $mn<0$ and $m^3+y^3=8+2my$
For this we apply AM - GM to get,
$$9 + 2my = m^3 + y^3 + 1 \geq 3my \implies my \le 9 $$
Case bash (sadly) gives that the solutions to this case are
$$(m,y)=(0,2), \enspace (2,0), \enspace (2,2)$$

Now we are left with two cases
CASE $3$ : $mn>0$ and $m^3-n^3=8-2mn$
CASE $4$ : $mn>0$ and $x^3-y^3=2xy-8$
Now how to finish off these last $2$ cases?
Thanks in advance!
 [/INST] Observe that $x^3 - (x-1)^3 > 2x(x-1) + 8$ for all $x$ except $-2 ≤ x ≤ 3$ and that $x^3 - x^3 < 2x(x)+ 8$ for all $x$. From this, the equation lies between $n = m-1$ and $n = m$, hence the only integer solutions are when $-2 ≤ m ≤ 3$.
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