<s>[INST] All integer solutions of $x^3-y^3=2020$. 
Find all integer pairs  $(x,y)$ satisfying $$x^3-y^3=2020\,.$$

First,
$x^3-y^3=(x-y)(x^2+xy+y^2)=2020$
and $2020=2^2\cdot 5 \cdot 101$.
But what next? Can it be worked out by using modulo? Or how?
Any idea? Thanks in advance.
 [/INST] Let $d = x - y$. Then we want $x^3 - (x-d)^3 = 2020$, which is a quadratic equation in $x$. The discriminant is $24240d - 3d^4 = 3d (8080 - d^3)$, which is nonnegative only for $d \in [0, 8080^{1/3}]$, and since $d$ is an integer it must be between $0$ and $20$.
Moreover, since $(x-y)^3 = d(x^2 + xy + y^2) = 2020$, $d$ must be a divisor of $2020$. This leaves only $6$ possibilities: $d = 1, 2, 4, 5, 10, 20$. We already had a finite problem to solve before, but this observation reduces the number of cases.
For each value of $d$, the solutions to the quadratic equation have irrational solutions, so there are no integer solutions $(x,y)$.
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