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.
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What if $k=1$ ? – Martín Forsberg Conde Aug 17 '15 at 14:53
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1$3\mid \sigma((2^1)!)$ and $3>2^1$ – Paolo Leonetti Aug 17 '15 at 14:56
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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.
Jyrki Lahtonen
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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! – Jyrki Lahtonen Aug 17 '15 at 15:32
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