1. Find all ordered pairs $(x, y)$ of positive integers 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 rewrites as

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

which implies that $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2} \mid 43 x_{1} y_{1}$. Since $\left(x_{1}, x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}\right)=\left(y_{1}, x_{1}^{2}-\right.$ $\left.x_{1} y_{1}+y_{1}^{2}\right)=\left(x_{1}, y_{1}\right)=1$, 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 $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}=1$, we must have $x_{1}=y_{1}=1$, so $(*)$ yields $x=y=d=22$.

The other possibility is $x_{1}^{2}-x_{1} y_{1}+y_{1}^{2}=43$, i.e. $\left(2 x_{1}-y_{1}\right)^{2}+3 y_{1}^{2}=172$. Assuming w.l.o.g. that $x_{1} \leqslant y_{1}$, we get $3 y_{1}^{2} \leqslant 172 \leqslant 4 y_{1}^{2}$, which holds only for $y_{1}=7$. Then $x_{1}=1$ and $d=1$, or $x_{1}=6$ and $d=\frac{43}{13}$. Only the first possibility gives a solution: $(x, y)=(1,7)$.

Thus, the solutions $(x, y)$ are pairs $(1,7),(7,1)$ and $(22,22)$.