Problem
NT-B2-M10-P010 Prime Sum of Divisors
#10
★★★☆☆ Level 3 of 5
Prove that if \(\sigma(n)\) is prime, then \(\tau(n)\) is also prime.
1001 Problems in Classical Number Theory (method inspiration) · Problem 486
First show that \(n\) must be a prime power.
If \(n\) has at least two coprime prime-power factors \(p^a\) and \(q^b\), then \(\sigma(n)\) factors as a product of two integers greater than \(1\), so it is not prime. Hence \(n=p^a\). Then \(\sigma(n)=1+p+\cdots+p^a\). If \(a+1\) is composite, say \(a+1=uv\) with \(u,v>1\), then \(1+p+\cdots+p^a=\frac{p^{uv}-1}{p-1}\) factors as \(\frac{p^u-1}{p-1}(1+p^u+\cdots+p^{u(v-1)})\), composite. Thus \(a+1\) is prime, and \(\tau(n)=a+1\).
A strong problem using factorisation of a geometric sum.