A Random Diophantine

The 4 solutions to this question are: $(3,-4), (4,-3), (-5,-6), (6,5)$

First, the factor of ${x^3-y^3}$ is ${(x-y)(x^2+xy+y^2)}$

Next, the factors of $91$ are $\small\begin{aligned} & 1*91 \\ & 91*1 \\ & 13*7 \\ & 7*13 \end{aligned}$

Now, assume $\begin{cases} x-y=7 \\ {x^2+xy+y^2}=13 \end{cases}$ and we will solve it as a system of equations.

$\begin{aligned} x-y & =7 \\ -y & =7-x \\ y & =x-7 \end{aligned}$

Then

$13={x^2+xy+y^2} \\ 13={x^2+x(x-7)+(x-7)^2} \\ 13={x^2+x^2-7x+x^2-14x+49} \\ 13={3x^2-21x+49} \\ 0={3x^2-21x+36} \\ 0={x^2-7x+12} \\ 0=(x-3)(x-4) \\ x=3 \cup\, 4$

If we plug $y$ in, we should get $\color{#302B94}{(3,-4), (4,-3)}$

$\text{WLOG,}$ with $\begin{cases} x-y=1 \\ {x^2+xy+y^2}= 91 \end{cases},$ we get $\color{#302B94}{(6,5), (-5,-6)}$

For the other 2 equations, we will only get (ir)rational solutions for $x$ and $y$ .

Therefore, there are $\color{#EC7300}{\boxed{4}}$ solutions to this question. $\blacksquare$