Problems & Puzzles:
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.
Note: JL has verified his statement from 4 to 105.
Contributions came from Antoine Verroken and Luke Pebody.
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.
There is no need for q to be prime. For p a prime >=3, p^(p-2) is
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