Number theory

Total number of factors of any perfect square number is always

Odd or Even depending on the number None of these Cannot be determined Odd

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2 solutions

Bernardo Sulzbach
Jun 29, 2014

Let S be a perfect square.

We can express S as

S = p 1 x 1 p 2 x 2 p 3 x 3 . . . S={p_1}^{x_1}{p_2}^{x_2}{p_3}^{x_3}...

Where p 1 , p 2 , etc are prime numbers \text{Where }p_1, p_2, \text{etc are prime numbers} .

x { x 1 , x 2 , x 3 , . . . } , 2 x \forall{}x\in{}\{x_1, x_2, x_3,...\}, 2|{x}

The number of divisors is given by \text{The number of divisors is given by} i = 1 n ( x i + 1 ) \prod_{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. \text{As the product is the number of divisors, the number of divisors is odd. }\blacksquare{}

"My first attempt at a mildly decent proof."

Sunil Pradhan
Jun 3, 2014

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.

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