Prove $m=[a,b]\Longleftrightarrow \left(\frac{m}{a},\frac{m}{b}\right)=1$ Was solving some exercise of Number theory, and used this theorem
$$m=[a,b]\Longleftrightarrow \left(\frac{m}{a},\frac{m}{b}\right)=1$$Remembered that the teacher showed it in class, but I do not remember how, and I think I may be in the evaluation, and I am also curious how to proof? $\;\;\;$ Please do not use modular arithmetic
 A: 
CLAIM1 $\hspace{5.5 cm }(a,b)[a,b]=ab$

P. Let $d=(a,b), e=\dfrac{ab}{[a,b]}$. We will prove that $d=e$. Recall we define $d$ as the (unique) positive number such that $d$ divides both $a$ and $b$, and if $d'$ is any other common divisor, $d'\mid d$. We wish to show $e$ has these two properties. First, note that $$\frac{a}{e} = \frac{{\left[ {a,b} \right]}}{b} \in {\Bbb Z}$$  $$\frac{b}{e} = \frac{{\left[ {a,b} \right]}}{a} \in {\Bbb Z}$$ since both $a,b$ divide $[a,b]$, so $e$ is a common divisor. Now suppose $d'$ is another common divisor. Recall that $[a,b]$ is the unique positive number such that $a,b$ both divide $[a,b]$ and whenever $a,b$ both divide some $f$, $[a,b]$ divides this $f$. We will use this to finish off the proof. Since $d'$ is a common divisor, $\dfrac{ab}{d'}$ is an integer. Moreover, both $a,b$ divide it, so it is a common multiple. It follows that $$\frac{f}{{\left[ {a,b} \right]}} = \frac{{ab}}{{\left[ {a,b} \right]d'}} = \frac{e}{{d'}}$$ is an integer, do $d'\mid e$, whence $d=e$. $\blacktriangle$

CLAIM2 $\hspace{4 cm }d=(a,b)\iff \left(\dfrac ad,\dfrac bd\right)=1$

P Suppose that $d=(a,b)$.  Suppose that $d'\mid \dfrac a{(a,b)},\dfrac b{(a,b)}$. Then $d'(a,b)\mid a$ and $b$, so that  $d'(a,b)\mid (a,b)\implies d'=1$. 
Now suppose $d\mid a,b$ and $\left(\dfrac ad,\dfrac bd\right)=1$. Let $e=(a,b)$. Then $d\mid e$, so $\dfrac e d $ is an integer. But $$\frac{d}{e}\frac{a}{d}=\frac{a}e\in\Bbb Z\\\frac{d}{e}\frac{b}{d}=\frac{b}e\in\Bbb Z$$
it follows that $$\frac{e}{d}\mid \frac ad\\\frac ed\mid \frac bd$$ and since $\left(\dfrac ad,\dfrac bd\right)=1$ we must have $\dfrac e d =1$ that is $e=d$.$\blacktriangle$
Now, we want to show

CLAIM3 $\hspace{ 4 cm}m=[a,b]\iff  \left(\dfrac ma,\dfrac mb\right)=1$

P Can you use the above now?
A: Let $A,B$ be the highest powers of prime $p$ in $a,b$ respectively.
WLOG we can set $A\ge B\ge0$
So, the highest powers of prime $p$ in  $m$ will be $A$
So, the highest powers of prime $p$ in  $\frac ma$ will be $A-B$
and the highest powers of prime $p$ in  $\frac ma$ will be $B-B=0$
$\implies $ the highest powers of prime $p$ that will divide $\left(\frac ma,\frac mb\right)$ will be $0$
This is true for any prime $p$ that divides $a$ or $b$ or both
A: Let
$a = \prod_p p^{a_i}$
and
$b = \prod_p p^{b_i}$
where the product is over the primes,
all the $a_i$
and $b_i$ are non-negative integers,
and
only a finite number of them
are positive.
Then
$m = \prod_p p^{\max(a_i, b_i)}
$,
so
$m/a = \prod_p p^{\max(a_i, b_i)-a_i}
$
and
$m/b = \prod_p p^{\max(a_i, b_i)-b_i}
$.
If
$\gcd(m/a, m/b) > 1$,
there is a $p$ such that
$\max(a_i, b_i)-a_i > 0$
and
$\max(a_i, b_i)-b_i > 0$,
or
$\max(a_i, b_i)>a_i$
and
$\max(a_i, b_i)>b_i$.
But $\max(a_i, b_i)$
must be equal to the
larger of $a_i$ and $b_i$,
so it
can not be greater than
both of them.
This contradiction shows that
$\gcd(m/a, m/b) = 1$.
A: $$m=[a,b]\Longleftrightarrow(\frac{m}{a},\frac{m}{b})=1$$
$\Longrightarrow$$$[a,b]=\frac{ab}{(a,b)}\Rightarrow m=\frac{ab}{(a,b)}\Rightarrow m(a,b)=ab\Rightarrow(ma,mb)=ab\Rightarrow$$$$\frac{(ma,mb)}{ab}=1\Rightarrow(\frac{m}{b},\frac{m}{a})=1$$$\Longleftarrow$$$(\frac{m}{a},\frac{m}{b})=1\Rightarrow(mb,ma)=ab\Rightarrow m(b,a)=ab\Rightarrow m=\frac{ab}{(a,b)}=[a,b]$$
