Solve the equation $x^{2 y-1}+(x+1)^{2 y-1}=(x+2)^{2 y-1}$ in natural numbers.

#
$x^{2 y-1} \equiv 1(\bmod x+1)$. Since the number $2 y-1$ is odd, then $x^{2 y-1}=-1(\bmod x+1)$. Therefore, $0=(x+2)^{2 y-1}-x^{2 y-1}$ $-(x+1)^{2 y-1}=1+1=2(\bmod x+1)$, that is, $x+1=2$. Hence, $1+2^{2 y-1}=3^{2 y-1} \Leftrightarrow 2 y-1=1 \Leftrightarrow y=1$.

## Answer

$x=y=1$.