# For integers $a$ and $b$, $ab=\text{lcm}(a,b)\cdot\text{hcf}(a,b)$

I was reading a text book and came across the following:

Important Results
(This comes immediately after LCM:)

If 2 [integers] $a$ and $b$ are given, and their $LCM$ and $HCF$ are $L$ and $H$ respectively,
then $L \times H = a \times b$

#### Solutions Collecting From Web of "For integers $a$ and $b$, $ab=\text{lcm}(a,b)\cdot\text{hcf}(a,b)$"

Let $p$ be a prime. If $p$ occurs in $a$ with multiplicity $m$ and in $b$ with multiplicity $n$, then it will occur in the LCM of $a$ and $b$ with multiplicity $\mathrm{max(m,n)}$ and in their HCF with multiplicity $\mathrm{min(m,n)}$.

Hence, in the product of LCM and HCF the multiplicity of $p$ is $$\mathrm{max}(m,n)+\mathrm{min}(m,n)=m+n,$$
which is also the multiplicity of $p$ in $a\cdot b$. Since this holds for every $p$, the two products must be equal.

Below is a proof that works in any domain, using the universal definitions of GCD, LCM.

THEOREM $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\qquad d\mid (a,b)\iff d\mid a,b \iff a,b\mid ab/d \iff [a,b]\mid ab/d \iff d\mid ab/[a,b]$

You can also figure this out without using any use of unique factorization. Since you say this comes immediately after LCM, I assume you know that for $m\gt 0$, $[ma,mb]=m[a,b]$, where $[a,b]=\mathrm{lcm}(a,b)$ and $(ma,mb)=m(a,b)$ where $\gcd(a,b)=(a,b)$.

Now suppose $(a,b)=1$, and also assume they are positive, since if they are negative $[a,-b]=[a,b]$ anyway. Since $[a,b]$ is some multiple of $a$, let $[a,b]=ma$. Then $b\mid ma$, but $(a,b)=1$, so $b\mid m$. If this hasn’t be addressed yet, notice $(ma,mb)=m(a,b)=m$, so $b\mid ma$, and $b\mid mb$, so $b\mid m$ since any common divisor of $ma$ and $mb$ divides the greatest common divisor $m$ in this case.

So $b\leq m$ as they are both positive, which implies $ba\leq ma$. But $ba$ is a common multiple of $a$ and $b$, so cannot be strictly less than $ma$, so $ba=ma=[a,b]$.

More generally, if $(a,b)=g\gt 1$, then you have $(\frac{a}{g},\frac{b}{g})=1$. Then by the special case above,
$$\left[\frac{a}{g},\frac{b}{g}\right]\left(\frac{a}{g},\frac{b}{g}\right)=\frac{a}{g}\frac{b}{g}.$$
Multiply through by $g^2$, you have
\begin{align*} g^2\left[\frac{a}{g},\frac{b}{g}\right]\left(\frac{a}{g},\frac{b}{g}\right) &= g\left[\frac{a}{g},\frac{b}{g}\right]g\left(\frac{a}{g},\frac{b}{g}\right)\\ &= [a,b](a,b)\\ &= g^2\frac{a}{g}\frac{b}{g}=ab \end{align*}
so $[a,b](a,b)=|ab|$.