$12 \cdot 93$ Find the positive integer solutions of the equation $x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}$.
(21st Moscow Mathematical Olympiad, 1958)
【Sol】When $y=1$, the given equation becomes

that is,
$$
\begin{array}{c}
x^{2}+(x+1)^{2}=(x+2)^{2}, \\
x^{2}-2 x-3=0 .
\end{array}
$$

Solving this, we get $x=3, x=-1$ (discard $x=-1$).
Thus, $x=3, y=1$ is a positive integer solution to the given equation.
When $y>1$, since $x$ and $x+2$ have the same parity, then $x+1$ is even. Let $x+1=2k$, where $k$ is a positive integer.
Thus, we have
$$
(2 k-1)^{2 y}+(2 k)^{2 y}=(2 k+1)^{2 y} .
$$

Expanding by the binomial theorem, we get
$$
(2 k)^{2 y}=2\left[C_{2, y}^{1}(2 k)^{2 y-1}+C_{2 y}^{3}(2 k)^{2 y-3}+\cdots+C_{2 y-1}^{2 y-1}(2 k)\right] \text {. }
$$

From this, it is not hard to see that $y$ is a multiple of $k$.
On the other hand, dividing both sides of (1) by $(2 k)^{2 y}$, we get
$$
2>\left(1-\frac{1}{2 k}\right)^{2 y}+1=\left(1+\frac{1}{2 k}\right)^{2 y}>1+\frac{2 y}{2 k} \text {. }
$$

Thus,
$$
\frac{2 y}{2 k}<1.
$$

When $y>1$, there are no integer solutions.
Therefore, the given equation has a unique positive integer solution: $x=3, y=1$.