
Problem 3. Find all prime numbers $p$, for which the number $p\left(2^{p-1}-1\right)$ is a $k$-th power $(k>1)$ of a positive integer.


Solution. Let $p\left(2^{p-1}-1\right)=x^{k}(x>0$ is an integer $)$. It is clear that $p \neq 2$, i.e. $p=2 q+1$ is an odd number. Since $p / x$, then $x=p \cdot y\left(y\right.$ is a positive integer) and $\left(2^{q}-1\right)\left(2^{q}+1\right)=p^{k-1} y^{k}$. At least one of the numbers $2^{q}-1$ and $2^{q}+1$ is a $k$-th power of an integer, because they are relatively prime numbers.

1. Let $2^{q}-1=z^{k}$, i.e. $2^{q}=z^{k}+1$. If $k$ is even, then $z^{k}+1$ is not divisible by 4 . Hence $q=1, p=3$ and $p\left(2^{p-1}-1\right)=3^{2}$.

If $k=2 l+1$ then $2^{q}=(z+1)\left(z^{2 l}-z^{2 l-1}+\cdots-z+1\right)$, i.e. $z+1=2^{\alpha}$, where $0 \leq \alpha<q$. On the other hand

$$
2^{q}=\left(2^{\alpha}-1\right)^{2 l+1}+1=2^{2 \alpha} \cdot A+2^{\alpha}(2 l+1)
$$

( $A$ is an integer). The last equality contradicts with $\alpha<q$.

2. Let $2^{q}+1=z^{k}$, i.e. $2^{q}=z^{k}-1$. If $k$ is odd, we obtain a contradiction as in the previous case.

If $k=2 l$, then $\left(z^{l}-1\right)\left(z^{l}+1\right)=2^{q}$ and since $\operatorname{GCD}\left(z^{l}-1, z^{l}+1\right)=2$, we have $z^{l}-1=2$, i.e. $q=3, p=7, p\left(2^{p-1}-1\right)=7.63=21^{2}$.

The sought numbers are $p=3$ and $p=7$.

## Grade 11 .
