Senderov V.A.

Find all pairs $(a, b)$ of natural numbers such that for any natural $n$, the number $a^{n}+b^{n}$ is an exact $(n+1)$-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 $\left(\frac{a}{c}\right)^{n}+\left(\frac{b}{c}\right)^{n}=c$.

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

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

## Answer

$(2,2)$.