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

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 $$(p_1, p_2, \dots, p_m)$$?

## 1 Answer

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 $$p Now going through the same procedure over and over for all sets, we have that every element of a set $$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.

• How can i prove that $\dfrac{\varphi(n)+\sigma(n)}{n}<2$ if $p_i$ goes to infinitely, i am stuck in here. – Zootopia Feb 3 at 10:35
• I added details in the solution – Aryaaaaan Feb 3 at 13:48