Total number of factors of any perfect square number is always
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Let S be a perfect square.
We can express S as
S = p 1 x 1 p 2 x 2 p 3 x 3 . . .
Where p 1 , p 2 , etc are prime numbers .
∀ x ∈ { x 1 , x 2 , x 3 , . . . } , 2 ∣ x
The number of divisors is given by i = 1 ∏ n ( x i + 1 )
As all factors are odd numbers, the product will be odd.
As the product is the number of divisors, the number of divisors is odd. ■
"My first attempt at a mildly decent proof."