The question is as follows:

"Show every positive integer is the product of a square (possibly 1) and a square free integer"

We begin by writing a positive integer n in its refined prime power factorization form,

$n=(p_{1}^{2\alpha _{1}+1}p_{2}^{2\alpha _{2}+1}...p_{k}^{2\alpha _{k}+1})\left ( q_{1}^{2\beta _{1}} q_{2}^{2\beta _{2}}...q_{j}^{2\beta _{j}}\right )$

$n=(p_{1}^{2\alpha _{1}+1}p_{2}^{2\alpha _{2}+1}...p_{k}^{2\alpha _{k}+1})\left ( q_{1}^{\beta _{1}} q_{2}^{\beta _{2}}...q_{j}^{\beta _{j}}\right )^{2}$

My question is this: How do I show that, $(p_{1}^{2\alpha _{1}+1}p_{2}^{2\alpha _{2}+1}...p_{k}^{2\alpha _{k}+1})$ is square free?

The definition of square free is that it is not divisible by a perfect square other than 1, so I proposed to show it in two cases with an integer $m^{2}$:

case 1: $m^{2}$ = 1

Trivial case as clearly 1 | $(p_{1}^{2\alpha _{1}+1}p_{2}^{2\alpha _{2}+1}...p_{k}^{2\alpha _{k}+1})$

case 2: $m^{2} \neq $ 1

Prove by contradiction by supposing that $m^{2}$ | $(p_{1}^{2\alpha _{1}+1}p_{2}^{2\alpha _{2}+1}...p_{k}^{2\alpha _{k}+1})$

Then $m^{2}$ | $p_{1}^{2\alpha _{1}+1}$ and each other odd prime.. and here is where I am uncertain.

Your help is super appreciated!

  • 1
    $\begingroup$ Possible duplicate of math.stackexchange.com/questions/21282/…. $\endgroup$ – lhf Oct 6 '15 at 16:14
  • $\begingroup$ I'd prove it by strong induction. It's essentially the same as prime factorization, but cleaner without all the "$\dots$"s. $\endgroup$ – Thomas Andrews Oct 6 '15 at 16:19
  • $\begingroup$ I couldn't find a similar question when I searched! I apologize if it is a similar question, although I was stuck at a particular point in the proof. $\endgroup$ – Danika Oct 7 '15 at 5:10

Your product of prime powers with odd exponents is not square-free unless the $\alpha_i$ are all $0$. Use $p_i^{2\alpha_i+1} = p_i\cdot p_i^{2\alpha_i}$ to reduce it to the product of a square-free integer and a square.

  • $\begingroup$ Ah I see. So then I would have a product of primes and could prove by contradiction that my m squared term must divide one of the primes. Thank you! $\endgroup$ – Danika Oct 7 '15 at 5:19

The statement is obviously true if n = 1 or if n is a prime. Assume n is composite with a prime factor p, so n = x * p. By induction we can assume that x = a^2 * b where b is square free.

If p divides b, that is b = c * p, then n = a^2 * b * p = a^2 * c * p^2 = (ap)^2 * c; c is square free because it is the divisor of a square free number.

If p doesn't divide b, then bp is also square free, so n = a^2 * bp where bp is square free.


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