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.
 A: 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.
A: Sum of divisors =$\prod_{p_i\mid n}(1+p_i+p_i^2+\cdots+p_i^{2P_i})$ where $p_i$s are distinct primes and $p_i^{P_i}\mid\mid n$
Now, $p_i^r+p_i^{2P_i+1-r}$ is even for all the prime divisor $p_i$ of $n$ for $0< r\le 2P_i+1-r\implies 0<r\le P_i$ leaving $p_i^0=1$ unpaired.

Alternatively, $p_i^k\equiv p_i\pmod 2$ for $k\ge 1$
So, $1+p_i+p_i^2+\cdots+p_i^{2P_i}\equiv 1+\sum_{1\le s\le 2P_i}p_i\pmod 2\equiv 1+2P_is\pmod 2\equiv 1$
A: 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.
A: 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.
A: If $n=2^{k_{1}}...p_{m}^{k_{m}}$, then the odd divisors of $n^{2}$ are exactly the divisors of 
$f=\frac{n^{2}}{2^{2k_{1}}}$ which, because $f$ is a square, has an odd number of divisors. Since the sum of an odd number odd's is odd, it follows that the sum of the divisors of $n$ is odd.
A: $144 = 12^2 = 12 \times 12$ = $2^4 \times 3^2$
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.
