902*. Solve the equation in natural numbers

$$
y^{2}+y=x^{4}+x^{3}+x^{2}+x
$$
$\triangle$ Multiply the equation by 4 and add 1 to both sides of the resulting equation so that its left side becomes a perfect square.

$$
(2 y+1)^{2}=4 x^{4}+4 x^{3}+4 x^{2}+4 x+1
$$

The right side of the last equation is, for almost all natural $x$, between $\left(2 x^{2}+x\right)^{2}$ and $\left(2 x^{2}+x+1\right)^{2}$.

$$
\left(2 x^{2}+x\right)^{2}2$.

The remaining cases are: $x=1$ and $x=2$.

For $x=1$, the original equation becomes $y^{2}+y=4$, and this equation has no solutions in natural numbers.

If $x=2$, then $y^{2}+y=30$, from which $y=5$.

Answer: $(2 ; 5)$.