The way to prove that $σ(a) = 3^k$ has no solution?

Question

$\sigma(n)$ = sum of divisors of n is a divisors function.

How to prove there are no such $a$ and $k \ge 2$ satisfy $\sigma(a) = 3^k$.

This proplem can be simplify to the case when $a$ is a power of prime ($a=p^\alpha$) because

if $a = p_0^{\alpha_{0}}p_1^{\alpha_{1}}...p_n^{\alpha_{n}}$, then $$\sigma(a) = \sigma(p_0^{\alpha_{0}})\sigma(p_1^{\alpha_{1}})...\sigma(p_n^{\alpha_{n}})=\prod_{i=0}^n \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}$$

Answer 1

As you stated, since $\sigma()$ is a multiplicative function, we only need to check the prime powers, i.e., for prime $p$, $e \ge 1$ and $j \ge 1$, that

$$\sigma(p^e) = \sum_{i=0}^{e}p^{i} = \frac{p^{e + 1} - 1}{p - 1} = 3^{j} \tag{1}\label{eq1A}$$

First, $p = 2$ and $e = 1$ gives $\sigma(2) = 2 + 1 = 3$, so $j = 1$ works. For $e \gt 1$, then

$$\sigma(2^e) = 2^{e + 1} - 1 = 3^{j} \; \; \to \; \; 2^{e + 1} - 3^{j} = 1 \tag{2}\label{eq2A}$$

Here, $j \gt 1$, but Mihăilescu's theorem shows there's no such solution (also, more simply, with $2^{e+1} = 3^j + 1$, then $e + 1 \ge 3$ means $3^j \equiv 7 \pmod{8}$, but $3^j \equiv 1 \pmod{8}$ or $3^j \equiv 3 \pmod{8}$ only). This means $a$ has at most $1$ factor of $2$ and there must be odd prime factors. Since $p = 3$ gives $\sigma(3^e) \equiv 1 \pmod{3}$, this means $p \ge 5$.

With the summation in \eqref{eq1A}, as there are $e + 1$ terms, getting an odd sum requires $e + 1$ to be odd, i.e., $e + 1 = 2r + 1$ for some integer $r \ge 1$. Using the fraction part in \eqref{eq1A}, if $p \equiv 2 \pmod{3}$, then $p^{e + 1} - 1 \equiv (2^2)^r(2) - 1 \equiv (1)^r(2) - 1 \equiv 1 \pmod{3}$ and $p - 1 \equiv 1 \pmod{3}$, so the result would be $\equiv 1 \pmod{3}$. Thus, this means $p \equiv 1 \pmod{3}$.

Once again from the summation in \eqref{eq1A}, as each term is $\equiv 1 \pmod{3}$, there must be a multiple of $3$ terms, i.e., $e + 1 = 3m$ for some integer $m \ge 1$. The fractional part of \eqref{eq1A} becomes

$$\frac{p^{e + 1} - 1}{p - 1} = \frac{p^{3m} - 1}{p - 1} = \left(\frac{p^{m} - 1}{p - 1}\right)(p^{2m} + p^{m} + 1) \tag{3}\label{eq3A}$$

Note that $\frac{p^{m} - 1}{p - 1} = \sum_{i=0}^{m-1}p$ is an integer. Also, since $p \equiv 1 \pmod{3}$, then $p^m = 3q + 1$ for some integer $q \ge 1$. Thus,

$$\begin{equation}\begin{aligned} p^{2m} + p^{m} + 1 & = (3q + 1)^2 + (3q + 1) + 1 \\ & = (9q^2 + 6q + 1) + (3q + 1) + 1 \\ & = 9q^2 + 9q + 3 \\ & = 3(3q^3 + 3q + 1) \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

However, $3q^3 + 3q + 1 \gt 1$ and is not a power of $3$, so \eqref{eq1A} cannot hold. This therefore proves what is requested, i.e., there's no integer $a$ such that $\sigma(a) = 3^k$ for $k \ge 2$.

Answer 2

Let $(p,q)$ be primes. Suppose we want to prove whether or not there exists at least one pair $(r,k)$ of positive integers with $\sigma(p^r)=q^k$.

Assume that at least one solution exists, and let $(r,k)$ be a solution with $r$ minimal. Let $d>1$ divide $r+1$, then $$q^k=\sigma(p^r)=\frac{p^{r+1}-1}{p-1}=\frac{p^{r+1}-1}{p^d-1}\cdot \frac{p^d-1}{p-1}.$$ Hence, there exists a positive integer $\ell$ such that $(d,\ell)$ is also a solution. By minimality, $d=r+1$ and $r+1$ is prime.

Next, let $j:=\operatorname{ord}_q(p-1)$. Note that $$\operatorname{ord}_q(p^{r+1}-1)=\operatorname{ord}_q((p-1)q^k)=j+k\ge j+1.$$ Hence, $p\not\equiv 1\pmod {q^{j+1}}$ while $p^{r+1}\equiv 1\pmod{q^{j+1}}$. This means that $r+1$ is not coprime to the order of the multiplicative group $\left(\mathbb{Z}/q^{j+1}\mathbb{Z}\right)^\times$, which is $q^j(q-1)$. Therefore, $r+1$ is not coprime to $q(q-1)$. Because $r+1$ is prime, we find that $$r+1\mid q(q-1).$$

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Now let's apply this knowledge to the case $q=3$. We find that $r\in \{1,2\}$. If $r=1$, then $p+1=3^k$, whence $p$ is even, whence $p=2$ and $k=1$.

If $r=2$, then $p^2+p+1=3^k$, so $p^2+p+(1-3^k)=0$. As a quadratic polynomial in $p$, this has discriminant $$1-4(1-3^k)=4\cdot 3^{k}-3=3(4\cdot 3^{k-1}-1).$$ In order for this to be a perfect square, we need $3\mid 4\cdot 3^{k-1}-1$, which is only the case for $k=1$. However, this gives $p=1$, which is a contradiction.

Therefore, all solutions have $p=2$. In this case, we may rewrite $\sigma(2^r)=3^k$ as $2^{r+1}-1=3^r$, which has only the solution $(r,k)=(1,1)$, by Mihăilescu's theorem. We are done.