Number theory

Let S be a perfect square.

We can express S as

$S={p_1}^{x_1}{p_2}^{x_2}{p_3}^{x_3}...$

$\text{Where }p_1, p_2, \text{etc are prime numbers}$ .

$\forall{}x\in{}\{x_1, x_2, x_3,...\}, 2|{x}$

$\text{The number of divisors is given by}$ $\prod_{i=1}^n{(x_i+1)}$

As all factors are odd numbers, the product will be odd.

$\text{As the product is the number of divisors, the number of divisors is odd. }\blacksquare{}$

"My first attempt at a mildly decent proof."

The number is always odd as one of the pair is n × n so you have to count one time n only.

e.g. 16 divisors are 1, 2, 4, 4, 8,16 count 4 for 1 time.