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prove that $l.c.m.(ab,ad)=a[l.c.m.(b,d)]$.

my work so far:

I know $l.c.m.(ab,ad)=a^2bd/g.c.d.(ab,ad)$ and ∃ $x,y\in \mathbb{Z}$ | $g.c.d.(ab,ad)=abx+ady$

∴ we now have $l.c.m.(ab,ad)=abd/(bx+dy)$

So we know that ∃ $x,y\in \mathbb{Z}$ | $bx+dy=c$ for some $c\in \mathbb{Z}$. If $c=g.c.d.(b,d)$ we are done. But, how do we know that the $x$ and $y$'s are | $c=g.c.d.(b,d)$?

here are the relevant corollaries:

1) if $d=g.c.d.(a,b)$, then ∃ $x,y\in \mathbb{Z}$ | $ax+by=d$

2)In order that ∃ $x,y\in \mathbb{Z}$ | $ax+by=c$ it is necessary and sufficient that d|c, where $d=g.c.d.(a,b)$.

Thank you in advance.

Vladimir Louis
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2 Answers2

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We write a proof along the lines in the OP. We have $$\text{lcm}(ab,ad)=\frac{a^2bd}{\gcd(ab,ad)}.$$ It is enough now to prove that $\gcd(ab,ad)=a\cdot\gcd(b,d)$. For if we have done that, we can conclude that $$\frac{a^2bd}{\gcd(ab,ad)}=\frac{a^2bd}{a\cdot\gcd(b,d)}=a\cdot\frac{bd}{\gcd(b,d)}=a\cdot\,\text{lcm}(b,d).$$

To show that $\gcd(ab,ad)=a\cdot\gcd(b,d)$ we can argue as follows. Certainly $ab$ and $ad$ divide $a\cdot\gcd(b,d)$. Nothing larger does. For there exist integers $x$ and $y$ such that $bx+dy=\gcd(b,d)$. Thus $abx+ady=a\cdot\gcd(b,d)$, and therefore any common divisor of $ab$ and $ad$ divides $a\cdot\gcd(b,d)$.

Remark: It is perhaps more natural to use the Fundamental Theorem of Arithmetic. The lcm of two numbers has a nice representation in terms of the prime power factorization of these numbers. If we use that, the proof will almost write itself.

André Nicolas
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$ {\rm lcm}(ab,ad)\mid n\color{#c00}\iff ab,ad\mid n\iff\begin{align} a &\mid n\\ b,d &\mid n/a\end{align}$ $\color{#c00}\iff \begin{align} a &\mid n\\ {\rm lcm}(b,d) &\mid n/a\end{align} \iff a\,{\rm lcm}(b,d) \mid n $

Remark $\ $ Above we used $\ x,y\mid z\color{#c00}\iff {\rm lcm}(x,y)\mid z,\,$ the universal property of $\,\rm lcm$

See here for a few proofs of the gcd distributive law.

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Bill Dubuque
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