Prove that if $\sigma_x(n) = \sigma_x(m)$ and $\sigma_{-x}(n) = \sigma_{-x}(m)$, then $n=m$.

Question

Only thing I've gotten so far is writing out each individual expression as a product of primes. What do I do next? Also, the function I mean is this one

https://en.wikipedia.org/wiki/Divisor_function

Also, $n,m\in \mathbb{Z}^+$ and $x\in \mathbb{R}$

Here is the expression I have :

\begin{gather} n = \prod_{i = 1}^r p_i^{a_i} \\ \sigma_{x}(n) = \prod_{i = 1}^r \bigl(1 + p_i^{x} + p_i^{2x} + \dotsc + p_i^{a_ix}\bigr). \end{gather}

Answer 1

We need $x \neq 0$. For $x = 0$ we have $\sigma_x(n) = \sigma_{-x}(n) = \tau(n)$ and $\tau(m) = \tau(n)$ does not imply $m = n$.

For $x \neq 0$ the involution $d \mapsto \frac{n}{d}$ of the set of divisors of $n$ yields

$$\sigma_{-x}(n) = \sum_{d\mid n} \biggl(\frac{n}{d}\biggr)^{-x} = n^{-x}\sum_{d\mid n} d^x = n^{-x}\sigma_x(n),$$

so $\sigma_x(n) = \sigma_x(m)$ and $\sigma_{-x}(n) = \sigma_{-x}(m)$ yields

$$n^x = \frac{\sigma_x(n)}{\sigma_{-x}(n)} = \frac{\sigma_x(m)}{\sigma_{-x}(m)} = m^x.$$

Since the power function $t \mapsto t^x$ is injective on $(0,+\infty)$ for $x\neq 0$, it follows that $n = m$.