Prove that if $p$ is a prime, then $p^3\mid y^2\implies p\mid y^2\implies p\mid y$.

Question

Please note that I am requesting a formal proof, using any technique.

There are two parts in the implication:
(i) if $p$ is a prime, then $p^3\mid y^2\implies p\mid y^2$,
(ii) if $p$ is a prime, then $p\mid y^2 \implies p\mid y$.

I know that, as by the post asked by me : if $p$ is a prime, and $p\mid a^k$ for any natural number $a,k$, then $p\mid a$, but the first part is not obvious from this.

Answer 1

Partial Hint

It can be shown that if $p^m | a^2$, then $p^{\lceil m/2 \rceil} | a$. Your consideration here, is the case where $m=1$.

Edit (Sketch proof)

Let \begin{align} n &= \prod p_i^{n_i} \\ m &= \prod p_i^{m_i}\\ a &=\prod p_i^{a_i} \end{align} Then $n | a^2$ means that $n_i \le 2a_i$ or $a_i\ge \lceil \frac{n_i}{2} \rceil$. On the other hand $m | a$ means that $m_i \le a_i$.

If $m$ is as large as possible, $m_i = a_i$ so $m_i = \lceil \frac{n_i}{2} \rceil $.

Answer 2

It is obvious from your first post, because $p\mid p^3\mid y^2$, so that $p\mid y\mid y^2$ is true. Indeed, $p\mid y^k$ implies $p\mid y$.

Answer 3

(i) Since $p\mid p^3$, $p\mid y^2$

(ii) Suppose by contradiction that $ p \nmid y$, but $p \mid y^2$.This implies that $yk=p$ for some integer $k$. But this is not possible since

• We know that $y \neq 1$, it is obvious.
• $k$ is not $1$, since if $k=1$ then $p\mid y$

then $y$ and $k$ are both integers greater than 1, which is not possible since $p$ is a prime. Contradiction! Hence $p$ is a prime. $\ _\square$