# Abstract Nonsense

## Number Theory Problem

Problem: Define $\sigma$ by

$\displaystyle \sigma:\mathbb{N}\to\mathbb{N}:n\mapsto\sum_{d\mid n}d$

Prove that if $2^{m}+1$ is prime then $\sigma(n)=2^{m}+1$ has no solution for $n\in\mathbb{N}$ except for $\sigma(2)=3$

Proof: We need a quick lemma.

Lemma: Suppose that $n=ab$ where $(a,b)=1$, then $\sigma(ab)=\sigma(a)\sigma(b)$

Proof: Let $\text{div}(a),\text{div}(b)$ denote the set of divisors of $a$ and $b$ respectively. For notational convenience let

$\text{div}(a)=\left\{1,a_1,\cdots,a_j,a_{j+1}\right\}$

and

$\text{div}(b)=\left\{1,b_1,\cdots,b_k,b_{k+1}\right\}$

Note then that if $d\mid n$ then by the assumption that $(a,b)=1$ we have that $d=a_pb_q$ for some $a_q\in\text{div}(a)$ and some $b_q\in\text{div}$. But, not only is it true that if $d\mid n$ that $d=a_pb_q$ but in fact every number of that form also divides $n$. It follows that

$\displaystyle \text{div}(n)=\bigcup_{p=1}^{j+1}\left\{a_p\cdot 1,\cdots,a_p\cdot b_k,a_p\cdot b_{k+1}\right\}$

So that

$\displaystyle \sigma(n)=\sum_{p=1}^{j+1}\left(a_p\cdot 1+\cdots+a_p\cdot b_{k+1}\right)=\left(1+\cdots+b_{k+1}\right)\sum_{p=1}^{j+1}a_p=\sigma(b)\sigma(a)$

From where the conclusion follows. $\blacksquare$

Noting then that if $n=p^{m}$ for some prime $p$ we have that

$\text{div}(n)=\left\{1,\cdots,p^{m-1},p^m\right\}$

So that

$\displaystyle \sigma(n)=\sum_{j=0}^{m}p^j$

We can combine this with the above lemma to conclude that if

$n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$

that

$\displaystyle \sigma(n)=\prod_{k=1}^{m}\sum_{j=0}^{\alpha_k}p_k^j$

So, getting to the actual problem. If $m=1$ then $2^m+1=3$ and $\sigma(2)=1+2=3$. Thus, we may assume WLOG that $m>1$.

So, suppose that

$\displaystyle 2^{m}+1=\sigma(n)$

then by previous comment

$\displaystyle 2^{m}+1=\prod_{k=1}^{m}\sum_{j=0}^{\alpha_k}p_k^j$

But, since $2^{m}+1$ is prime and the RHS of the above equation is a factorization of it it follows that

$\displaystyle 2^{m}+1=\sum_{j=0}^{\alpha_{k_0}}p_k^j$

for some $k_0$. But, expanding the RHS and subtracting one gives

$2^{m}=p_{k_0}+\cdots+p_{k_0}^{\alpha_{k_0}}=p_{k_0}\left(1+\cdots+p_{k_0}^{\alpha_{k_0}-1}\right)\quad (1)$

and thus $p_{k_0}\mid 2^m$ and thus $p_{k_0}=2^\ell$ for some $\ell\in\mathbb{N}$. But, remembering that $p_{k_0}$ is prime it follows that $p_{k_0}=2$. Thus, dividing both sides of $(1)$ by $2$ gives

$2^{m-1}=1+\cdots+2^{\alpha_{k_0}-1}\quad (2)$

Now, since $m>1$ we see that $2^{m-1}>1$ from where it follows that $\alpha_{k_0}-1>0$ and thus we see that the LHS of $(2)$ is even whereas the RHS is odd, contradiction. The conclusion follows. $\blacksquare$