If $ n=\prod_{i=1}^k p_i^{x_i}$ is the prime factorization of $ n$, then $ \sigma(n)=\prod_{i=1}^k \frac{p_i^{x_i+1}-1}{p_i-1}$ We know that $ \frac{t^{u+1}-1}{t-1}=t^u+t^{u-1}+\ldots+1$ So, in our problem, when we apply the above, we get that $ p$ must divide at least one of the factors. So there are $ a,b$ such that $ a^b\mid p-1,a\ge 2,b\ge 1,a$ is prime. Also, $ A=a^b+a^{b-1}+\ldots+1\le a^b+\frac{a^b}{2}+\ldots+\frac{a^b}{2^b}=a^b\frac{2^{b+1}-1}{2-1}<2a^b$ Now, $ A=mp,m\in\mathbb{N}$ If $ m\ge 2$, then $ A\ge 2p>2p-2\ge 2a^b$, impossible, so $ A=p$ $ A\equiv 1\pmod a^b$, because $ a^b\mid p-1$ $ a(a^{b-2}+\ldots+1)\equiv 0\pmod a^b$, if $ b\ge 2$. Left side isn't divisible by $ a^2$ so $ b=1$ $ p-1=a$, $ p$ and $ a$ are primes, so $ p=3$, and $ a=2$