Find infinitely many n such that $\sigma (n) \le \sigma (n-1)$

Question

This number theory problem has me confused. It is finding infinitely many n such that $\sigma (n) \le \sigma (n-1)$. I know that n should be prime by the examples I have tried, but I'm not sure where to go after I assume that n has to be prime.

Answer 1

Hint: If $n$ is an odd prime, then $\sigma(n) = 1 + n$. What divisors of $n-1$ do you know?

Answer 2

$\sigma(p-1) \geq (p-1)+1+2>p+1=\sigma(p)$ for all prime $p \geq 5$.