Prove if sigman 3n and n is odd then n must be a squareSolu

Prove: if sigma(n) = 3n and n is odd then n must be a square

Solution

If that\'s the divisor sum, I\'ll use s(n). There are two properties that help:

1. If gcd(a,b)=1 then s(ab) = s(a)s(b).

2. If p is prime then s(p^m) = 1 + p + p^2 + ... + p^m

Those imply that if n=p1^m1 * p2^m2 * ... is the prime factorization of n, then:

s(n) = (1 + p1 + p1^2 + ... + p1^m1)(1 + p2 + p2^2 + ...+ p2^m2)....

If n is odd, then each prime pk is odd.

Each factor has (mk + 1) terms, all odd when pk is odd.

So the factor for pk is odd iff mk is even. The whole product is then odd only if all exponents mk are even; and that means n is a square.

SInce 3n is odd when n is odd, then s(n)=3n means s(n) is odd, and that means that n must be a square.