, 2002

Solution:

One easily sees that neither $x$ nor $y$ can be zero.
For $y=1$ one obtains from $x^{2} + (x+1)^{2} = (x+2)^{2}$ the equation $x^{2} - 2x - 3 = 0$, of which only the solution $x=3$ comes into question.

Now let $y > 1$.
Since $x$ and $x+2$ have the same parity, $x+1$ is even and hence $x$ is odd.
With $x = 2k-1$ ($k \in \mathbb{N}$) the equation becomes

$\quad (2k-1)^{2y} + (2k)^{2y} = (2k+1)^{2y}$

from which, by expanding, one obtains the following:

$$(2k)^{2y} - 2y (2k)^{2y-1} + \ldots - 2y 2k + 1 + (2k)^{2y} = (2k)^{2y} + 2y (2k)^{2y-1} + \ldots + 2y 2k + 1$$

Since $y > 1$, we also have $2y \geq 3$. If one now collects all terms in $yk$ on one side and factors out $(2k)^{3}$ on the other side, one obtains:

$$8yk = (2k)^{3} \left[2 \binom{2y}{3} + 2 \binom{2y}{5} (2k)^{2} + \ldots - (2k)^{2y-3}\right]$$

from which it follows that $y$ is a multiple of $k$.

Dividing the equation by $(2k)^{2y}$ gives:

$$\left(1 - \frac{1}{2k}\right)^{2y} + 1 = \left(1 + \frac{1}{2k}\right)^{2y}$$

where the left side is less than $2$. The right side, however, is greater than $1 + \frac{2y}{2k} \geq 2$, which cannot be, since $y$ is a multiple of $k$.

The given equation therefore has only the solution $x=3$ and $y=1$.