1. Determine all ordered pairs of natural numbers $(x, y)$ such that

$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2}
$$

(Moldova)
1. Let $x=d x_{1}$ and $y=d y_{1}$, where $d=(x, y)$. The given equation can be written as

$$
d\left(x_{1}+y_{1}\right)\left(x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}\right)=x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}+43 x_{1} y_{1}
$$

from which it follows that $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2} \mid 43 x_{1} y_{1}$. Since $\left(x_{1}, x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}\right)=\left(y_{1}, x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}\right)=$ $\left(x_{1}, y_{1}\right)=1$, it follows that $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2} \mid 43$. Note that $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}>0$.

If $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}=1$, then $x_{1}=y_{1}=1$, and (*) gives $x=y=d=22$.

Let $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}=43$, i.e., $\left(2 x_{1}-y_{1}\right)^{2}+3 y_{1}^{2}=172$, and let without loss of generality $x_{1} \leqslant y_{1}$. Then $3 y_{1}^{2} \leqslant 172 \leqslant 4 y_{1}^{2}$, which holds only for $y_{1}=7$. We get $x_{1}=1$ and $d=1$, or $x_{1}=6$ and $d=\frac{43}{13}$. Only the first possibility makes sense and gives the solution $(x, y)=(1,7)$. All solutions $(x, y)$ are pairs $(1,7),(7,1)$ and $(22,22)$.