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.
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.
$144 = 12^2 = 12 \times 12$ = $2^4 \times 3^2$

so every square number is of the form $a^{2p} \times a^{2q}$

$\therefore$ total number of divisors = $(2p + 1)(2q + 1)$

$\because (2p + 1) \wedge (2q + 1)$ are both odd numbers $\forall(p \wedge q) \in \mathbb{Z}$, then every squared number $n^2$ will have an odd number of divisors.