If $b$ is the largest square divisor of $a$ and $a^2|c$ then $a|b$? 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

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.
