Show that there exist a prime divisor of $\sigma{((2^k)!)}$ which is greater than $2^k$

Question

Let $k$ be a positive integer. Show that there exists a prime divisor of $\sigma{((2^k)!)}$ which is greater than $2^k$, where $\sigma{(n)}$ is the sum-of-divisors function.

Answer 1

Put together the following pieces:

• $\sigma$ is multiplicative in the sense that if $\gcd(m,n)=1$ the $\sigma(mn)=\sigma(m)\sigma(n)$. This is just a consequence of unique factorization of integers.
• The highest power of two that is a factor of $(2^k)!$ is $2^{2^k-1}$. You have probably seen the formula for the $p$-adic value of a factorial. Can be found on our site as well.
• Thus $\sigma(2^{2^k-1})=2^{2^k}-1=(2^{2^{k-1}}-1)(2^{2^{k-1}}+1)$ is a factor of $\sigma((2^k)!).$
• From the context of prime factors of Fermat numbers we know that all the prime factors $p$ of $2^{2^{k-1}}+1$ satisfy the congruence $p\equiv1\pmod{2^{k+1}}$ if $k\ge3$, and $p\equiv1\pmod{2^k}$ for all $k$. See this question for a local explanation. Actually this question suffices here.