18.4.15 ** Find the positive integer solutions to the equation $x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}$.
When $y=1$, we get $x=3$.
When $y>1$, since $x^{2 y}$ and $(x+2)^{2 y}$ have the same parity, $(x+1)^{2 y}=(x+2)^{2 y}-x^{2 y}$ must be even, so $x+1$ is even. Let $x+1=2 x_{1}$, then
$$
\begin{aligned}
\left(2 x_{1}\right)^{2 y} & =(x+1)^{2 y}=(x+2)^{2 y}-x^{2 y} \\
& =\left(2 x_{1}+1\right)^{2 y}-\left(2 x_{1}-1\right)^{2 y} .
\end{aligned}
$$

Expanding the right side of (1), we get
$$
\begin{aligned}
\left(2 x_{1}\right)^{2 y} & -2\left[\mathrm{C}_{2 y}^{1}\left(2 x_{1}\right)^{2 y-1}+\mathrm{C}_{2 y}^{3}\left(2 x_{1}\right)^{2 y-3}+\cdots+\mathrm{C}_{2 y}^{1}\left(2 x_{1}\right)\right] \\
& =2 \cdot 2 y \cdot\left(2 x_{1}\right)^{2 y-1}+2 \cdot \mathrm{C}_{2 y}^{3}\left(2 x_{1}\right)^{2 y-3}+\cdots+2 \cdot 2 y \cdot 2 x_{1},
\end{aligned}
$$

which simplifies to $\left(2 x_{1}\right)^{2 y}-2 \cdot 2 y \cdot\left(2 x_{1}\right)^{2 y-1}-2 \mathrm{C}_{2 y}^{3}\left(2 x_{1}\right)^{2 y-3}-\cdots-2 \cdot \mathrm{C}_{2 y}^{3}\left(2 x_{1}\right)^{3}=2 \cdot 2 y \cdot 2 x_{1}$. Dividing both sides by $\left(2 x_{1}\right)^{3}$, we get $x_{1}^{2} \mid y$.
Therefore, $y \geqslant x_{1}^{2}$. Thus, the right side of (2) $>2 \cdot 2 y \cdot\left(2 x_{1}\right)^{2 y-1} \geqslant\left(2 x_{1}\right)^{2 y+1}>$ the left side, a contradiction! Therefore, the original equation has only the solution $x=3, y=1$.