If $n^2$ is a multiple of an even-exponent prime factorization, then is $n$ a multiple of the square root of the given prime factorization

Question

Inspired by an excerpt from pp. 209 of ETS's Official GRE Prep Guide (2017):

Since every value of $n^2$ is a multiple of $(2^4)(3^4)$, the values of $n$ are the positive multiples of $36$.

Generally, is the following true:

Let $p_1, p_2$ be prime numbers, and $n_1, n_2$ be integers divisible by 2. Then

$(\forall n^2, n \in \mathbb Z)(p_1^{n_1} p_2^{n_2} \mid n^2 \Rightarrow p_1^{n_1/2} p_2^{n_2/2} \mid n)$

How to prove (refute) this? Seems I essentially need to show that the square root of the scalar multiple assumed in the if statement is an integer? Seems I can use the fact that the prime factorization of a square must have even exponents to prove that it is indeed an integer?

Answer 1

More generally, if $p^{2e}$ divides $n^2$, then $p^{e}$ divides $n$.

Indeed, let $p^b$ be the largest power of $p$ that divides $n$. Then $2e \le 2b$ and so $e\le b$.