Senderov B.A.

Find all pairs $(a, b)$ of natural numbers such that for any natural $n$, the number $a^{n}+b^{n}$ is a perfect $(n+1)$-th power.
Let there be a sequence of natural numbers $c_{n}$ such that $a^{n}+b^{n}=c_{n}^{n+1}$. Clearly, $c_{n} \leq a+b$, because otherwise $c_{n}^{n+1}>(a+b)^{n+1}>a^{n}+b^{n}$. Therefore, in the sequence $c_{n}$, at least one number $c \leq a+b$ repeats infinitely many times, meaning the equation

$a^{n}+b^{n}=c^{n+1}$ (with fixed $a, b, c$) has infinitely many solutions in natural $n$.

Let $a \geq b$ for definiteness. Rewrite the equation $a^{n}+b^{n}=c^{n+1}$ as $\left(\frac{a}{c}\right)^{n}+(b / c)^{n}=c$.

If $a>c$, then ${ }^{a / c}=1+d$, where $d>0$. But then $\left({ }^{a} / c\right)^{n}+(b / c)^{n}>\left({ }^{a} / c\right)^{n}=(1+d)^{n}>1+n d$, which is greater than $c$ for $n>$ $c / d$, meaning the number of solutions to the equation is finite.

If $a \leq c$, then $\left({ }^{a} /\right)^{n} \leq 1$ and $\left({ }^{b} / c\right)^{n} \leq 1$, from which $c=2$ or $c=1$. The first case is possible only if $a / c=b / c=1$, that is, if $a=b=2$. This case satisfies the condition of the problem. The second case is clearly

impossible.

## Answer

$(2,2)$.