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

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.

• What if $k=1$ ? Aug 17, 2015 at 14:53
• $3\mid \sigma((2^1)!)$ and $3>2^1$ Aug 17, 2015 at 14:56

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.
• How to get here? Don't you think that because we were asked to look at $(2^k)!$ the prime factor $2$ may become important? Let's see, where that takes us! Aug 17, 2015 at 15:32
• Nice answer!+1 Thank you
– user246688
Aug 17, 2015 at 15:45
• if we find this $\sigma{((3^k)!)}$ prime divisor is greater than $3^{k}?$
– user246688
Aug 17, 2015 at 15:53