5
$\begingroup$

I'm trying to learn number theory on my own, and here's a proof I'm not quite sure I got right. It feels too simple(?), I'm thinking maybe I'm missing something. So the question is:

Prove that if $n$ is a square, then each exponent in its prime-power decomposition is even.

My proof:

Let $n=m^2$, with $m$ having prime factors $p_i$ with exponents $e_i$ so that

$m= p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n.$

When squared, this gives

$m^2 = (p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n)^2 = p^{2e_1}_1 p^{2e_2}_2 \ldots p^{2e_n}_n,$

where all the exponents are even.

  • 0
    Duplicate of http://math.stackexchange.com/questions/23360/prime-factorization-of-square-numbers2011-10-04

1 Answers 1

4

That is fine.

You can even use it in the other direction to prove that if each exponent in the prime-power decomposition of $n$ is even, then $n$ is a square by saying
$n = p^{2e_1}_1 p^{2e_2}_2 \ldots p^{2e_n}_n = (p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n)^2$ so $n = m^2$ where $m= p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n.$