For which natural numbers $x, y$ does the following equality hold:

$$
x^{2}\left(x^{2}+y^{2}\right)=y^{100}
$$
I. solution. The equation in the problem is quadratic in $x^{2}$, solving this (with $y$ as a parameter):

$$
x^{2}=\frac{-y^{2} \pm \sqrt{y^{4}+4 y^{100}}}{2}
$$

For $x^{2}$ to be an integer, it is necessary that $y^{4}+4 y^{100}$ is a perfect square. Since $\left(2 y^{50}\right)^{2} \leq y^{4}+4 y^{100}y^{2}$ and $y^{50}+x^{2}>x^{2}$, therefore

$$
x^{2} y^{2}=\left(y^{25}-x\right)\left(y^{25}+x\right)\left(y^{50}+x^{2}\right)>1 \cdot y^{2} \cdot x^{2}
$$

which is a contradiction. Therefore, $x$ or $y$ must be zero, and thus the only solution is $x=y=0$.

Gergely Naszódi (Fazekas M. Fóv. Gyak. Gimn., 12th grade)

III. solution. Similarly to the previous solution, we will also deal with the case $x y \neq 0$. Let the greatest common divisor of $x$ and $y$ be $d$; then $x=d a, y=d b$, where $a$ and $b$ are relatively prime positive integers. Dividing both sides of the equation by $d^{4}$:

$$
a^{2}\left(a^{2}+b^{2}\right)=d^{96} b^{100}
$$

Since $a$ and $b$ are relatively prime, $b^{100}$ is also relatively prime to both $a^{2}$ and $\left(a^{2}+b^{2}\right)$. However, according to the equation, $b^{100}$ must divide $a^{2}\left(a^{2}+b^{2}\right)$; this is only possible if $b^{100}=1$, i.e., $b=1$. In this case, $a^{2}+1=\frac{d^{96}}{a^{2}}=\left(\frac{d^{48}}{a}\right)^{2}$, which is impossible, since the successor of a positive perfect square cannot be a perfect square.