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