<s>[INST] Can we find $x,y \in \mathbb{N} $ such that $x^y -y^x=2020$? Happy new year. 

Can we find $x,y \in \mathbb{N} $ such that : $$x^y -y^x=2020$$ 

One solution is $(x,y) = (2021,1)$.
It is clear that both of $x,y$ are odd or even . 
Note that $2020=2^2 \times 5 \times 101 $
 [/INST] If $x$ or $y$ is $1$, then $y=1$ and $x=2021$.
If $x$ or $y$ is even then both are even. Since $2020$ is not divisible by $8$, either $x$ or $y$ is $2$. For $ t\in \mathbb{N}$,$2^t-t^2\ge 0 $ if $t\ne 3$ and is an increasing function of $t$ for $t\ge 4$. Therefore $x=2$ and $y\approx 11$ but $2^{11}-11^2=1927\ne 2020. $
We now have $x,y\ge 3$ and therefore $x^y>y^x$  only for $y > x$.Then
$$2020=x^x\left(x^{y-x}-\left(1+\frac{y-x}{x}\right)^x\right)\ge x^x\left(x^{y-x}-e^{y-x}\right) $$ and so $x<5$.
Therefore  $x=3$ and $$2020\ge 27\left(3^{y-3}-e^{y-3}\right) $$
then $y\le 7$ and the values $y=5$ and $7$ are easily checked.
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