prove the formula where $n=\prod_{i=1}^k p_{i}^{\alpha_i}$

Question

I have problems with fourth part of the exercise. Here we must prove the formula where $n=\prod_{i=1}^k p_{i}^{\alpha_i}$. How to prove, that $$\sigma_{-4}=\prod_{i=1}^k \frac{p_{i}^{4\alpha_i+4}-1}{p_{i}^{4\alpha_i+4}(p_i^4-1)}$$ I know that function $\sigma$ multiplicative, so $\sigma_{-4}$ also will be multiplicative. $\sigma(n)$ means sum of positive divisors of n and here $\sigma_{-t}=\sum_{d|n, n>0}d^t$

Answer 1

It seems your formula is a bit off; with your definition of $\sigma_{-t}$ it should in fact be $$\sigma_{-4}=\prod_{i=1}^k \frac{p_{i}^{4\alpha_i+4}-1}{p_i^4-1}.$$ Though I should add, the usual definition differs from your by a sign; it is normally $$\sigma_t:=\sum_{d\mid n,\ d>0}d^t.$$ With this definition you get the similar (but slightly different!) formula $$\sigma_{-4}=\prod_{i=1}^k\frac{p_{i}^{-4\alpha_i-4}-1}{p_i^{-4}-1}=\prod_{i=1}^k \frac{p_{i}^{4\alpha_i+4}-1}{p_i^{4\alpha_i}(p^4-1)}.$$

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Hint: Prove that for any integer $t$ you have $$\sigma_{-t}=\prod_{i=1}^k\sum_{j=0}^{\alpha_i}(p_i^t)^j$$

Alternative hint: If you know that $\sigma_{-4}$ is multiplicative, then it suffices to show that $$\sigma_{-4}(p^{\alpha})=\frac{p^{4\alpha+4}-1}{p^{4\alpha}(p^4-1)},$$ for any prime $p$ and natural number $\alpha$. Of course by definition $$\sigma_{-4}(p^{\alpha})=\sum_{j=0}^{\alpha}p^{4j}.$$