Mongolian National Mathematical Olympiad

Let $k$ be a positive integer. We choose a prime divisor $p_j$ of $2^{2^j}+1$ for each $0\le j<k$. Then the product $n_k=p_0{p}_1\dots p_{k-1}$ is a divisor of $2^{2^k}-1={\prod }_{j=0}^{k-1}(2^{2^j}+1)$.

Since $n_k$ is odd, $\sigma (n_k)=(p_0+1)\dots (p_{k-1}+1)$ is divisible by $2^k$. Hence $2^{2^k}-1$ divides $2^{\sigma (n_k)}-1$. It follows that $2^{\sigma (n_k)}\equiv 1(\mathrm{mod}{n}_k)$.