Exercise 4.2.18 (If $\sigma(q) = q+k$ where $k \mid q$ and $k<q$, then $k=1$)

Proof. Here $k>0$. Since $\sigma (q)>q$, $q\ne 1$.

Suppose, for the sake of contradiction, that $k\ne 1$. Then $1<k<q$, so $q$ has at least three distinct divisors, i.e. $1,q,k$. Therefore

thus $q+1>q+1+k$: this is a contradiction. This proves $k=1$.

If $\sigma (q)=q+k$ where $k\mid q$ and $k<q$, then $k=1$. □