Find all solutions of the equation $\quad x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}$ with $x, y \in N$.
One can easily see that neither $x$ nor $y$ can be zero.
For $y=1$, from $x^{2}+(x+1)^{2}=(x+2)^{2}$, we obtain the equation $x^{2}-2 x-3=0$, of which only the solution $x=3$ is valid.
Now let $y>1$.
Since $x$ and $x+2$ have the same parity, $x+1$ is an even number and thus $x$ is an odd number.
With $x=2 k-1(k \in \mathbb{N})$, we get the equation
(\#) $\quad(2 k-1)^{2 y}+(2 k)^{2 y}=(2 k+1)^{2 y}$
from which, by expanding, we obtain:

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

Since $y>1$, it follows that $2 y \geq 3$. If we now move all terms in $yk$ to one side and factor out $(2k)^{3}$ on the other side, we get:

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

from which it follows that $y$ is a multiple of $k$.
By dividing equation (\#) by $(2k)^{2 y}$, we get:

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

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

Therefore, the given equation has only the solution $x=3$ and $y=1$.
That there are no solutions for $y > 1$ was clear to most participants, as it represents a special case of Fermat's Last Theorem, which has since been proven. Unfortunately, only a few managed to provide a complete proof of this special case.