Problems & Puzzles: Puzzles

Puzzle 455. 1+σ0(n).σ(n) = n*rad(n) Juan López proposes the following puzzle: Prove or refute that if n>3 and satisfies: 1+σ0(n).σ(n) = n*rad(n) then n=q^p where p and q=p+2 are twin primes. References: Divisor function & Rad(n) Note: JL has verified his statement from 4 to 105.

---

Contributions came from Antoine Verroken and Luke Pebody.

***

Antoine wrote:

Suppose n = q ^ p q is prime p = q – 2 and is prime or not, then:

Sigma0(n) = p + 1
Sigma1(n) = (( q ^ (p+1)) – 1 ) / ( q – 1 )
Rad(n) = q

If we put these figures with (p + 1) = ( q – 1 ) in the formula we become an identity.

***

Luke wrote:

There is no need for q to be prime. For p a prime >=3, p^(p-2) is a
solution. In particular, 11^9 is a solution. It is easy to see that
these are the only prime power solutions...The only other numbers n below 630,000,000 for which n is a factor of 1+sigma_0(n)*sigma(n) are 77 and 128807153

***