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I think this is false, a counter example could be:
$c = 100,$
$b = 10,$
$a = 5$

But the book answer is true :( ! Did I misunderstand the problem or the book's answer was wrong?

Thanks,
Chan

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    @yunone: Exactly.$c$needs some more information. If it is merely "there exists a c" then that hypothesis is meaningless since you can always pick $c=a^2$.2011-02-02

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With or without your edit, b does not divide a. I suspect the question you want is

If b is the largest square divisor of c (not a) and a^2|c then a|b?

Then the answer would be true.

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That is not a counterexample: $10$ is not a divisor of $4$ and $4^2$ does not divide $100$. You must have at least one typo.

Perhaps you are trying to ask this: if $b^2 | c$ and $b$ is the largest number with this property (NOTE NOT $b^2 | a$), and also $a^2|c$ then must $a|b$?

The answer to that would be yes, which you can prove by thinking about each prime that divides $a$ and showing it divides $b$ at least as many times.

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    @arron: You're right. I thought the book's answer was correct now. I was thinking 2^2 but then I took 4|10.2011-02-02