Find the ( $\mathrm{x}, \mathrm{y}$ ) natural integers such that $x^{5}=y^{5}+10 y^{2}+20 y+1$
$$
y^{5}y^{5}+10 y^{2}+20 y+1=x^{5}
$$

So $x=y+1$

It remains to solve:

$$
y^{5}+5 y^{4}+10 y^{3}+10 y^{2}+5 y+1=y^{5}+10 y^{2}+20 y+1
$$

Which is:

$$
y\left(y^{3}+2 y^{2}-3\right)=0
$$

The solutions are therefore the pairs $(y=0, x=1)$ and $(y=1, x=2)$