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Suppose that $gcd(ab,p^4) = p^3$ then $p^3 |ab$. $p$ is prime.

$p^3 |ab \implies p^2|a$ and $p|b$

Is this last statement true?

The converse is true i believe.

EDIT.

$gcd(a,p^2) = p, gcd(b,p^2) = p. $

(This is a part of a larger proof, i am trying to disprove the first statement.)

3 Answers3

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$a=8,b=3,p=2$ is a counter example

Extremal
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It's false: let $p,q$ be two distinct prime numbers. Let $a=p^3$; then $\gcd(p^3q,p^4)=p^3$, but $p\not\mid q$.

Bernard
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No, it's a long way from being true that $p^3 | ab \Rightarrow p^2 | a $ and $p | b$ (I think that's your question). For example, take $a = p$ and $b = p^2$.

Hew Wolff
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  • Thats great thanks. Could you suggest another method to disprove the supposition. Given the conditions in the edit? –  Mar 06 '15 at 17:29
  • I'm still not sure what you're trying to prove, but if $p^3$ is the greatest common divisor of $ab$ and $p^4$, then in particular $p^3$ must be a divisor of $ab$. So it seems clear that the first statement is true. – Hew Wolff Mar 06 '15 at 17:41
  • I'm trying to prove that if $gcd(a,p^2) = p, gcd(b,p^2) = p. $ then $gcd(ab,p^4) = p^2$. –  Mar 06 '15 at 17:50
  • I see. Well, if you write $a = px$ (where $p \nmid x$) and $b = py$ (where $p \nmid y$), then it looks pretty clear that $p^2 \mid ab$ but $p^3 \nmid ab$. See if that helps. – Hew Wolff Mar 06 '15 at 19:17