Trying to think of different ways to prove this, besides method of sqrt. Help will be greatly appreciated.
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1This might be useful for you: If $a^2$ divides $b^2$, then $a$ divides $b$. – Martin Sleziak Sep 29 '12 at 14:55
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From the assumptions you have $b|m^2$ and $b|n^2$ then $b^2|\ldots$ – hmmmm Sep 29 '12 at 14:59
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$b \mid m^2$ and $b \mid n^2$, so $b^2 \mid m^2 n^2$, so $b \mid mn$.
To see why $a^2 \mid b^2$ implies $a \mid b$, see the link already provided by Martin in his comment to the question.
GeoffDS
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3You need to justify why for integers $\rm:b^2:|:c^2:\Rightarrow:b:|:c,:$ else the proof may be circular. $\ $ – Bill Dubuque Sep 29 '12 at 15:27
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Hint $\rm\ mn/b\ $ is a root of $\rm\: x^2 - ac\:$ so is integral by the Rational Root Test.
Bill Dubuque
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You can use prime factorizations of integers... You know that $ab^2c = m^2n^2 = (mn)^2$. If a prime $p$ appears to some power $e$ in the prime factorization of $(mn)^2$, then it appears to the power $e$ in $ab^2c$, so it appears to a power $d \leq e$ in the prime factorization of $b^2$. This is equivalent to the statement that if $p$ appears to the power $e' = e/2$ in the prime factorization of $mn$, then it appears to the power $d' = d/2\leq e'$ in the prime factorization of $b$.
Thus every prime appears to at least a large a power in the prime factorization of $mn$ as it does in the prime factorization of $b$. So $b$ divides $mn$.
Zarrax
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$ab=m^2$, $bc=n^2$ so $acb^2=(mn)^2$ and $b^2$ divides $(mn)^2$.
Thus $mn/b$ is a rational number whose square is an integer. By unique factorization into primes it now follows that $mn/b$ is an integer; any prime factor in the denominator of the lowest form of a rational number will also occur in the factorization of the denominator of its square.
i. m. soloveichik
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i.e. if $\rm:r = mn/b\not\in \Bbb Z:$ and $\rm:r = c/d:$ in lowest terms, then there is a prime $\rm:p:|:d,,\ p\nmid c,:$ therefore $\rm:r^2= c^2/d^2 \not\in \Bbb Z:$ since $\rm:p:|:d^2,, \ p\nmid c^2:$ (else $\rm:p:|:c^2\Rightarrow p:|:c:$ by $\rm,p,$ prime). Alternatively it is a special case of Euclid's Lemma $\rm: (d,c) = 1,,\ d:|:c,e:\Rightarrow:d:|:e,:$ viz. the case $\rm:e=c.:$ – Bill Dubuque Sep 29 '12 at 21:40