Solve the equation $x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}$ in natural numbers.
When $y=1$, the equation becomes quadratic, having only one solution $x=3$ (the second root $x=-1$ is negative).

Let $y>1$. Set $t=x+1$. Note that the numbers $x^{2 y}$ and $(x+2)^{2 y}$ have the same parity, so the number $x+1=t$ is even. Substitute into the equation and expand the brackets: $(t-1)^{2 y}+t^{2 y}=(t+1)^{2 y}, t^{2 y}-2 y t^{2 y-1}+\ldots-2 y t+1+t^{2 y}=t^{2 y}+$ $2 y t^{2 y-1}+\ldots+2 y t+1$.

Move $2 y t$ to the right side, and all other terms to the left. Since $2 y>3$, all the non-cancelled terms on the left side are divisible by $2 t^{2}$. Therefore, the right side, equal to $4 y t$, is also divisible by $2 t^{2}$. Thus, $2 y \geq t$.

Therefore, $(t+1)^{2 y}=t^{2 y}+2 y t^{2 y-1}+\ldots>2 t^{2 y}>t^{2 y}+(t-1)^{2 y}$. Contradiction.

## Answer

$x=3, y=1$.