Prove or disprove : if $x|y^2 $then $x|y$

Question

How can I prove such statement?

I ask this question as such template because I think that it doesn't matter which $x$ or $y$ represents.

However, the original question is: if $6|n^2$ then $6|n$ and if $12|n^2$ then $12|n$.

If $x = 4$ and $y = 2$, then $x|y^2$ but $x$ does not divide $y$ — Simon S, Feb 04 '15 at 17:43
@Stacknewbie. Just perhaps aren't you confused with the converse of the statement? If x|y then x|y² Because that is true... — imranfat, Feb 04 '15 at 17:57

score 2 · Accepted Answer · answered Feb 04 '15 at 17:43

2

False. $4|36$ but $4\!\! \not|\, 6$. You should require $x$ to be a prime number (or a product of distinct primes). $x=6$ will work, $x = 12$ won't.

answered Feb 04 '15 at 17:43

Umberto P.

52,165

Thanks for the explanation! – StackNewbie Feb 04 '15 at 18:10

Ivo Terek · Answer 2 · 2015-02-04T21:13:43.610

1

The claim $x \mid y^2 \implies x \mid y$ is clearly false. Take $x = y^2 > 1$ a perfect square and see for yourself. What is true is: $$\begin{cases} x \mid yz \\ \gcd(x,y)=1\end{cases} \implies x \mid z.$$ Having this in mind, if $\gcd(x,y)=1$, and taking $y = z$...

edited Feb 04 '15 at 21:13

answered Feb 04 '15 at 17:44

Ivo Terek

77,665

$x=y$ does not falsify the claim. – MJD Feb 04 '15 at 21:12
Had mistyped. Fixed. – Ivo Terek Feb 04 '15 at 21:14

score 0 · Answer 3 · edited Apr 13 '17 at 12:19

0

Hint $\ $ It fails for $\,x = y^2 > 1.\ $ It is true iff $x$ is squarefree.

edited Apr 13 '17 at 12:19

Community

1

answered Feb 04 '15 at 19:13

Bill Dubuque

272,048

Prove or disprove : if $x|y^2 $then $x|y$

3 Answers3