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Let $x, ~y, ~ z$ be three integers such that $$x|y^2 ~~\text{ and } ~~ y|z^2.$$ Then prove or disprove: $x$ divides $z^2$.

My attempt: There exist integers $k$ and $l$ such that $$y^2=kx~~\text{ and } ~~ z^2= ly.$$ Then we have $~ x| z^4$ but not $z^2$ also I am unable to find the counterexample. Please help me to solve this.

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3 Answers 3

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A simple example to disprove $x$ divides $z^2$ is with $x = 16$, $y = 4$ and $z = 2$. Then $x \mid y^2$ becomes $16 \mid 4^2 = 16$, and $y \mid z^2$ becomes $4 \mid 2^2 = 4$. However, $x \mid z^2$ becomes $16 \mid 2^2 = 4$, which is not true.

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    $\begingroup$ Wait. Isn't $|$ means divisibility? $4$ certainly divides $16$. $\endgroup$
    – Yuki.F
    Jan 29, 2021 at 8:14
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    $\begingroup$ @Yuki.F You have it the wrong way around. $a \mid b$ means $a$ divides into $b$. In other words, $b$ is a multiple of $a$, e.g., there's an integer $k$ such that $b = ka$. Thus, $16 \mid 4$ would mean $4$ is an integral multiple of $16$, which is not true. $\endgroup$ Jan 29, 2021 at 8:16
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Here is a generalized version of the previous counter example..

Let $x=t^4,y=t^2,z=t$, $\forall x \in Z^+$ then $x∣y^2$ becomes $t^4|(t^2)^2$ which is true, $y∣z^2$ becomes $t^2|(t)^2$ again true, and $x∣z^2$ becomes $t^4|(t)^2$ which is obviously false....

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  • $\begingroup$ sorry edited that. I had meant Here is the generalized version of the previous answer.. $\endgroup$
    – Aatmaj
    Jan 29, 2021 at 8:24
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John Omielan answered the question by giving a counterexample. To complement his answer, note that the statement is true if $x$ is a prime number. Indeed, a prime number $x$ has the property that if $x|ab$, then $x|a$ or $x|b$. Applied to the situation in the question, $x|y^2$ implies that $x|y$ if $x$ is prime. But $y|z^2$, so $x|z^2$.

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