Sum of divisors is odd(own)

[Masum]

Prove that the sum of divisors of a perfect square is odd.
This problem only requires the idea: the sum of divisors of a positive integer $n$ is the sum of all positive integers dividing $n$ including $1$ and $n$ itself. The hint sounds great, huh??

[Corei13]

Let, $n^2 = \prod_i { p_i ^{2e_i} }$ and note that, $a^r \equiv a$ (mod $2$ ) for $r \in \mathbb{N}$.
So, $\sigma ( n^2 ) = \sigma \left ( \prod_i { p_i ^{2e_i} }\right ) = \prod_i { \sigma \left ( p_i ^{2e_i} \right ) }$
$= \prod_i { \sum_{j=0}^{2e_i} { p_i^j} } \equiv \prod_i { \left ( 1+ \sum_{j=1}^{2e_i} { p_i} \right ) }$
$\equiv \prod_i { \left ( 1+ 2e_i p_i \right ) } \equiv 1$ (mod $2$ )

[Nadim Ul Abrar]

If
n=p^a x p1^b... [p is prime]
Then the number of divisors of n^2 be (2a+1)(2b+1)......

Its an odd number .

And n^2=p^2a x p1^2b.....

So number of even divisors of n^2

is 2a x 2b x ..... If 2|n .

Its even .
So sum of divisor's of n^2 will be odd for this case .

Again . If 2 don't divide n then
every divisor of n^2 be odd.

So for this case the sum of divisor's of n^2 is odd tooooo

[Masum]

See this topic as well.

[nayel]

If the perfect square is odd, then the sum of its divisors is a sum of an odd number of odd numbers, and is therefore odd. If the perfect square is even, let it be $4^tq^2$, where $q$ is odd. The sum all even factors is even. The remaining odd factors are factors of $q^2$ whose sum is odd by the previous reasoning. even+odd=odd, hence the conclusion.

[Masum]

That was my solution too.

[*Mahi*]

And my solution was the Corei13 one ... direct use of the $\sigma (n)$ formula...

[Phlembac Adib Hasan]

And my solution was the Corei13 one ... direct use of the $\sigma (n)$ formula...

The same here.