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

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.

## closed as off-topic by TheGeekGreek, JonMark Perry, Jimmy R., Brian Borchers, DanMar 22 '18 at 15:23

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## 3 Answers

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 $m_i = \lceil \frac{n_i}{2} \rceil$.

• Please give link in MSE, or elsewhere, for its proof. – jiten Mar 22 '18 at 11:17
• @jiten Could not find a link as such, but I have given a little sketch proof of the statement. – complexmanifold Mar 22 '18 at 11:29

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 $p\mid y$.

• Thanks a lot for that. So, it can be generalized to : for any prime $p, \exists k,n,i,j \in \mathbb{Z+},\,\, k\gt i, n\gt j,\,\, p^k\mid y^n\implies p^{k-i}\mid y^{n-j}\implies p\mid y.$ – jiten Mar 22 '18 at 10:57

(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. $\ _\square$

• Thanks, but you have proved the failure of converse indirectly (by contradiction), but have not stated this at all. You have actually stated the proof by contradiction of failure of : If $p\mid y^2$, then $p\nmid y$; which is converse of failure of : if $p \nmid y$, then $p\mid y^2$.The reason I am stating this is that you have taken the implication $yk=p$ which can 'directly' arise out of $p\mid y$ only. – jiten Mar 22 '18 at 11:41
• @jiten Thanks for the comment. I have stated that I proved by contradiction now. You said that I 'have taken the implication $yk=p$ which can 'directly' arise out of $p\mid y$ only.' Well, I am not sure what you mean but if you say that $p\mid y \implies yk=p$, I don't think that is true. $p\mid y$ means '$p$ is a factor of $y$' and means $y=pn$ for some integer $n$. What I am trying to say is that since $y$ is a factor of $p$ (since $y\mid p^2$ but $y\mid p$) hence $yk=p$ – Vee Hua Zhi Mar 22 '18 at 11:50
• Agreed, I was wrong. But, in the last line of your comment you have stated that : "(since $y\mid p^2$ but $y\mid p$) hence $yk=p$." This line is causing confusion, slrly. your answer for the same cause; as in the answer you have stated that : $p\nmid y$, but $p\mid y^2.$ I request elaborating the same. – jiten Mar 22 '18 at 12:09