Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Trying to think of different ways to prove this, besides method of sqrt. Help will be greatly appreciated.

share|improve this question
1  
This might be useful for you: If $a^2$ divides $b^2$, then $a$ divides $b$. –  Martin Sleziak Sep 29 '12 at 14:55
    
From the assumptions you have $b|m^2$ and $b|n^2$ then $b^2|\ldots$ –  hmmmm Sep 29 '12 at 14:59

4 Answers 4

$b \mid m^2$ 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.

share|improve this answer
3  
You 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

Hint $\rm\ mn/b\ $ is a root of $\rm\: x^2 - ac\:$ so is integral by the Rational Root Test.

share|improve this answer

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 $mn$.

share|improve this answer

$ab=m^2$, $bc=n^2$ 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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.