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In class today we were talking about proving the definition of 'divides' and the teacher never got to finish this proof.

if a divides b^2 , then a divides b.

First line was: Let a and b be integers. a > 0 since you cant divide by 0. If a divides b^2, then there exists an integer q such that b^2 = aq. I would assume the next lines may be trying to show that b^2 is a product of some integer and itself and then that integer can always be divisible by the same q? Not sure here..

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This is not true. Note that $4$ divides $2\times 2$, but $4$ does not divide $2$. Or if that is too simple, $4$ divides $60$, but it divides neither $6$ nor $10$. –  André Nicolas Nov 5 '13 at 5:44
    
$a > 0$ since you can't divide by $0$? What about $a<0$? –  Newb Nov 5 '13 at 5:44
    
$a$ and $b$ are integers? Are you supposing $|a|<b^2$? –  GIANCANE Nov 5 '13 at 8:45

1 Answer 1

The given statement is false. Indeed, if $a=4$ and $b=6$, then $a$ divides $b^2$ (since $4$ divides $36$) yet $a=4$ does not divide $b=6$.

One way to salvage the statement is to also assume that $a$ is prime.

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Square free would suffice I suppose though the prime case is simplest. –  Macavity Nov 5 '13 at 6:17

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