prove sum of divisors of a square number is odd 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.

The formula for $\sigma_1(n)$ for $\displaystyle{n=\prod_{i=1}^kp_i^{\alpha_i}} $ is $\displaystyle{\sigma_1(n)=\prod_{i=1}^k\frac{p_i^{\alpha_i+1}-1}{p_i-1}}$ (see this).
Therefore $\displaystyle{\sigma_1(n^2)=\prod_{i=1}^k\frac{p_i^{2\alpha_i+1}-1}{p_i-1}}$. But $\frac{p_i^{2\alpha_i+1}-1}{p_i-1}=1+p_i+p_i^2+\ldots+p_i^{2\alpha_i}$ is always odd. Thus $\sigma_1(n^2)$ is odd as product of odd numbers.
