Don't know how to prove that sum of all divisors of a square number is always odd.
ex: $27 \vdots 1,3,9,27$; $27^2 = 729 \vdots 1,3,9,27,81,243,729$; $\sigma_1 \text{(divisor
function)} = 1 + 3 + 9 + 27 + 81 + 243 + 729 = 1093$ is odd;
I think it somehow connected to a thing that every odd divisor gets some kind of a pair when a number is squared and 1 doesn't get it, but i can't formalize it. Need help.
Best Answer
Yes, you can assign a pair to every odd divisor, except one of them.
So, we have a square number $n^2$. Extract the largest power of $2$ from $n$ to get $n = 2^k m$ where $m$ is odd. Odd divisors of $n^2$ are the same as all divisors of $m^2$. If $d$ is a divisor of $m^2$, then $m^2/d$ is also its divisor. So they all come in pairs, except for $m$ which is paired with itself.