Q:

Prove that $n \equiv \pm 1 \bmod 6 \implies \sigma(n)<2n$ i.e. $n$ is deficient.

Prove that $n \equiv \pm  1 \bmod 6 \implies \sigma(n)<2n$. Is there an easy proof of this or of the same inequality but including the perfect numbers (so that $\sigma(n)\leq 2n$)? 

A:

To see why it's not true without considering a specific example, suppose $n=p_1p_2...p_k$ with all $p_i$ distinct primes and greater then $5$. Then $n$ is of the desired form and $\sigma(n)=n(1+1/p_1)(1+1/p_2)...(1+1/p_k)$. Clearly, the product $(1+1/p_1)...(1+1/p_k)$ grows without bound, and in particular, it can be made greater than $2$.

