The Problems of Ukrainian Authors

Consider number $n=2^{p-1}$, where $p$ is a prime number. Then $\sigma (n)=2^p-1$. Let's prove that $\varphi (2^p-1)>2^{p-1}$ for infinitely many primes $p$.

Let $m=2^p-1=p_1^{{\alpha }_1}\dots p_k^{{\alpha }_k}$. Then $$\varphi (m)=(2^p-1)\frac{p_1-1}{p_1}\dots \frac{p_k-1}{p_k}.$$ Notice that every $p_i>p$, $i=1,\dots ,k$. Really, as it follows from the little Fermat theorem, $2^{p_i-1}\equiv 1(\mathrm{mod}{p}_i)$, and $2^p-1\equiv 0(\mathrm{mod}{p}_i)$ as well, therefore $(p_i-1)\nmid p$, because $p$ is a prime number.

Hence $m=2^{p-1}=p_1^{{\alpha }_1}\dots p_k^{{\alpha }_k}>p^k$ and so $k\le p{\log }_{p}2$. I.e. $k\le \frac{p}{2}$ for sufficiently large $p$.

Then $$\frac{\varphi (2^p-1)}{2^{p-1}}\ge \frac{2^{p-1}}{2^p}{\left(1-\frac{1}{p}\right)}^p.$$ But ${\left(1-\frac{1}{p}\right)}^p\to \frac{1}{e}$ as $p\to \infty$, therefore the number $n=2^{p-1}$ satisfies the condition for sufficiently large $p$.