If $p^4 | z^2$, then $p^2 | z$ for $p$ a prime and $z$ some positive integer.

Question

Problem: Suppose that $p$ is a prime, and $z$ is some positive integer. If $p^4 | z^2$, then $p^2 | z$.

Thoughts:

If for some positive integer $a$, that $p^4 a = z^2$, then necessarily $p^2 \sqrt{a} = z$, so that if my desired conclusion is correct, then $a$ would have to be a perfect square. How can I show this is true?

Edit: I know that if $gcd(p^4, a) = 1$, then necessarily $a$ is a square.

Answer 1

I've tried to construct a proof based "prime"-arily on the defining property of primes:

$p \mid ab \Longrightarrow [p \mid a] \vee [p \mid b] \tag 0$

and the fact that cancellation applies in the integral domain $\Bbb Z$; here's what I got:

$p^4 \mid z^2 \Longrightarrow p \mid z^2; \tag 1$

$p \mid z^2 \Longrightarrow p \mid z \Longrightarrow z = py; \tag 2$

$z = py \Longrightarrow z^2 = p^2y^2 \tag 3$

$p^4 \mid p^2y^2 \Longrightarrow p^2y^2 = p^4x \Longrightarrow y^2 = p^2x; \tag 4$

$y^2 = p^2x \Longrightarrow p \mid y^2 \Longrightarrow p \mid y \Longrightarrow y = wp; \tag 5$

$y = wp \Longrightarrow z = p^2w \Longrightarrow p^2 \mid z. \tag 6$

$OE\Delta$.

Answer 2

Since $p^4 \mid z^2$, then $p \mid z^2$, so $p \mid z$. Let $z = p^a b$ where $a \ge 1$ and $p \not{\mid} \; b$. Note that $a = 1$ doesn't work, as $z^2$ would only have $2$ factors of $p$ meaning that $p^4 \not{\mid} \; z^2$, so $a \ge 2$. Thus, $p^2 \mid z$.

Answer 3

If you're allowed to use the fundamental theorem of arithmetic then $z = p_1^{e_1} \cdots p_k^{e_k}$ so $z^2 = p_1^{2e_1} \cdots p_k^{2e_k}$. That $p^4 \mid z^2$ implies $p = p_i$ for some $1 \leq i \leq k$ and $2e_i \geq 4$. (If a prime power $p^k$ divides a natural number $n$ then the exponent of that prime in the prime factorsiation of $n$ is at least $k$.) Then $e_i \geq 2$ so $p^2 \mid z$.

Answer 4

$\dfrac{z^2}{p^4} = n\in\Bbb Z\ $ so $\ x = \dfrac{z}{p^2}\Rightarrow\,x^2= n\,$ $\Rightarrow\,x\in\Bbb Z\,$ by the Rational Root Test.