Given a positive integer $a$, prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$.
($\sigma(n)$ is the sum of all positive divisors of the positive integer number $n$.)
I have two problem when trying to use the hint.
- Let $p_1,p_2,...,$ be the primes which do not divide $a$. Let $k$ be the minimum integer such that $\dfrac{2\sigma(a)}{a} < \prod_{i=1}^k \left (1 + \dfrac{1}{p_i} \right )$. How can I prove $\prod_{i=1}^\infty \left (1 + \dfrac{1}{p_i}\right )$ diverges to show $k$ exist.
- By Dirichlet's Theorem there exists infinitely many primes $q$ such that $q \equiv -a^{-1} \pmod{p_i}$ for $1 \le i \le k$. Choose $m=q$, then $\sigma(aq+1) \ge (aq+1) \cdot \prod_{i=1}^k \left (1 + \dfrac{1}{p_i}\right ) > \dfrac{2(aq+1)\sigma(a)}{a} > 2q\sigma(a) > \sigma(aq)$.
The problem is done after that, but how can I prove $q$ exists without using Dirichlet's Theorem (which are too much for a High School Olympiad problem).