Factors of squares of primes

Postulate: "The square of any prime number has 3 factors"

Proof: If we express the square of a prime as $p^2$, we can factor the square of a prime as follows:

$p^2 = p*p = (p*1)*(p*1)$

We can do this because any prime $p$ has only 2 factors - itself and 1.

So, all the factors of $p^2$ will be all the distinct products which can be formed from the above expression:

$1$ , $p$ , and $p^2$

These are the only products which can be formed. Thus, there are a total of $\boxed{3}$ factors for the square of any prime number.

Challenge: Generalize the relationship between the power a prime is raised to and the number of factors the entire expression has.

#NumberTheory

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

If p is a prime, then the number p raised to the power n has (n+1) factors........

Aaghaz Mahajan - 2 years, 7 months ago

Can you give a proof of this? Not much use just saying stuff if you aren't substantiating your claims.

P.S. It's quite obvious, but still...

A Former Brilliant Member - 2 years, 7 months ago

If I have a number 'n' such that $\boxed{ n = (p_{1})^{q_{1}} (p_{2})^{q_{2}} (p_{3})^{q_{3}} ....... } where p_{i}$ are all unique primes. Then, The number of factors of 'n' is $\boxed{(1+q_{1})(1+q_{2})(1+q_{3})...... }$

Avyukta Manjunatha Vummintala - 8 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$