Prove that exist finitely many positive integers $n$ satisfying $\tau (n)=a$ and $n|\phi (n)+\sigma (n)$

Question

Given a fixed positive integer $a\geq 9$. Prove there exist finitely many positive integers $n$, satisfying:

1. $\tau (n)=a$
2. $n|\varphi (n)+\sigma (n)$

My ideal is if i write the factorization $ n = \prod_{i = 1}^{m}p_i^{a_i}, $ then since $ \prod_{i = 1}^{m}(a_i + 1) = a, $ there are only a finite number of suitable $ (m, (a_1, a_2, \dots, a_m)) \in \mathbb{Z}^{+} \times \mathbb{N}^m. $ Therefore, we can fix a $ (m, (a_1, a_2, \dots, a_m)) \in \mathbb{Z}^{+} \times \mathbb{N}^m $. For the remainder of the proof, it suffices to show there are a finite number of sets of distinct primes $ (p_1, p_2, \dots, p_m) $ such that $ n = \prod_{i = 1}^{m}p_i^{a_i} $ and $ n \vert \phi(n) + \sigma(n)$. But how can I show there are finite $ (p_1, p_2, \dots, p_m) $?

Answer 1

Continue it like the following: Assume for the sake of contradiction there are infinitely many such sets and let $ T $ be the set of these sets. Pick an arbitrary set $S\in T$. It's clear that $ \phi(n) = \prod_{i = 1}^{m}(p_i^{a_i - 1}(p_i - 1)) $ and $ \sigma(n) = \prod_{i = 1}^{m}\frac{p_i^{a_i + 1} - 1}{p_i - 1} $ therefore $ \frac{\phi(n) + \sigma(n)}{n} = \prod_{i = 1}^{m}\frac{p_i - 1}{p_i} + \prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right). $ Clearly these pimes have an upper bound (we'll call it $L$) because otherwise we would have $ 1 < \frac{\phi(n) + \sigma(n)}{n} < 2. $ Therefore it's clear that there are an infinite number of sets in $ T $ that all have the same prime $ p<L. $ Now going through the same procedure over and over for all sets, we have that every element of a set $ S\in T $ is bounded by some number so $ T $ is finite and our proof is complete.

P.S. If you're wondering about the part I said that if the prime factors don't have an upper bound, $\frac{\phi(n) + \sigma(n)}{n} < 2$ , here's the proof:
As we said, $ \frac{\phi(n) + \sigma(n)}{n} = \prod_{i = 1}^{m}\frac{p_i - 1}{p_i} + \prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right). $ Now clearly $\prod_{i = 1}^{m}\frac{p_i - 1}{p_i}$ is always less than one so assume that $\prod_{i = 1}^{m}\frac{p_i - 1}{p_i}=1-s$. On ther other hand $\prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right)$ can be arbitrarily close to one if $p$ doesn't have an upper bound so we'll choose one of the prime factors such that $\prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right)<1+s$. Then we'll have $\frac{\phi(n) + \sigma(n)}{n} < 2$ and the proof is complete.