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I am stuck in the follow problem:

State whether the following statement is true or false:

If $a$ and $b$ are positive integers, and $p$ is prime then $p$ divides $\gcd(a^2,b)$ if and only if p divides $\gcd(a,b^2)$

My work:

When $a=2$, and $b=5$, The $\gcd(4,5) = 1$ and $\gcd(2,25) = 1$, the only "number" that divides $1$, is $1$, but that is not a prime by convention.

But my feeling says: this is not the right answer.

J. W. Tanner
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Joe Goldiamond
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  • In your example, is it true that, if $p$ is prime, then:

    $p$ divides $gcd(a^2,b) \iff $ $p$ divides $gcd(a^2,b)$ ?

     – Adam Rubinson Nov 07 '22 at 20:52
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    Indeed, that doesn't show this general statement to be true or false, as it is an example of the general statement holding true (instead of a counterexample showing it to be false). In this case, the equivalence $p \mid \operatorname{gcd}(a^2, b) \iff p \mid \operatorname{gcd}(a, b^2)$ is vacuously true, as no prime $p$ divides $1$. – Theo Bendit Nov 07 '22 at 20:52
  • Note by Euclid's lemma that if $p$ is prime and $p \mid a^2$, then $p \mid a$. – John Omielan Nov 07 '22 at 20:53
  • As a hint, consider $\gcd(a,b)$ – jlammy Nov 07 '22 at 20:54
  • $a=6$ and $b=15$ would be a better example for you to try the statement – Tony Mathew Nov 07 '22 at 20:57
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    $p|(a^2,b)\Rightarrow p|a^2\text { and } p|b\Rightarrow p|a\text { and } p|b\Rightarrow p|a\text { and } p|b^2\Rightarrow p|(a,b^2)$. It is the same going from right to left. – Piquito Nov 07 '22 at 21:07
  • By the gcd universal property in the first dupe it is equivalent to $,p\mid a^2,b\iff p\mid a,b^2,,$ which is true by $,p\mid n^2\iff p\mid n,,$ by Euclid's Lemma (2nd dupe). – Bill Dubuque Nov 08 '22 at 03:08

1 Answers1

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Let $p$ be a prime.

$p \mid n \iff p\mid n^2 \tag*{$\dagger$}\label{eq1}$

Proof:
$\implies$

$p \mid n \implies p \mid n^2$ is trivially true.

$\impliedby$

Consider the contrapositive i.e $p\nmid n \implies p\nmid n^2$ which is again trivially true $\blacksquare$

Using this result, we have:

$$p \mid \gcd(a^2,b) \overset{\text{def}}{\Longleftrightarrow}(p \mid a^2) \land (p \mid b) \overset{\ref{eq1}}{\Longleftrightarrow} (p\mid a) \land (p\mid b^2) \overset{\text{def}}{\Longleftrightarrow} p\mid \gcd(a,b^2)$$

Sam
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