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 divisors $1=d_1<d_2<\cdots <d_k=n^2$ can be partitioned into pairs $(d, \frac {n^2}d)$, except that there is no partner for $n$ itself.
Therefore, the number of divisors is odd.
Thus if $n$ itself is odd (and  so are all its divisors), we have the sum of an odd number of odd numebrs, hence the result is odd.
But if $n$ is even, we can write it as $n=m2^k$ wit $m$ odd. The odd divisors of $n^2$ are precisely the divisors of $m^2$. Their sum is odd and all other (even) divisors do not affect the parity of the sum.
