NT-B2-M10-P009 Odd \(\sigma(n)\)

Look at \(\sigma(p^a)\) for odd \(p\) and for \(p=2\).

Let \(n=2^a\prod p_i^{\alpha_i}\), where \(p_i\) are odd. For odd \(p\), the sum \(1+p+\cdots+p^\alpha\) consists of \(\alpha+1\) odd terms, so it is odd if and only if \(\alpha\) is even. For \(2^a\), the sum \(1+2+\cdots+2^a\) is always odd. Therefore \(\sigma(n)\) is odd exactly when all exponents of odd primes are even. This means the odd part of \(n\) is a square; the exponent of \(2\) is arbitrary. If \(a\) is even, \(n\) is a square; if \(a\) is odd, \(n\) is twice a square.

A neat problem on parity of the divisor sum.