If $l=lcm(n,m)$ and $x$ is a common multiple of $n$ and $m$, prove $l|x$ using quotient remainder thm

Question

lcm is the least common multiple and n,m are positive integers.
Do I somehow incorporate the fact that $n|x$ and $m|x$?

Answer 1

By remainder theorem.

For any $n,q$ there are integers $a, r$ with $0 \le r < q$ so that $n = aq + r$.

So if $k$ is a common multiple of $n,m$ then $k \ge lcm(n,m)$ because $lcm(n,m)$ is the least (non-zero) common multiple.

So there is an $a$ and $r; 0 \le r < lcm(m,n)$ so that $k = a*lcm(m,n) + r$. As $n|k$ and $n|lcm(m,n)$ then $n|r$. And as $m|k$ and $m|lcm(m,n)$ then $m|r$. So $r$ is a common multiple of $n$ and $m$.

But $r < lcm(m,n)$ which is the least common multiple of $m,n$. The only common multiples that can be less then than the least are the "trick" ones $r=0$ (or $r < 0$). So $r = 0$ and $k = a*lcm(m,n)$ and $lcm(m,n)$ divides all other common multiples.

=== old answer below ====

Let $\gcd(n,m) = d$. Let $n' = \frac nd; m'=\frac md$. Note: $\gcd(n',m') = 1$.

Proof? That's standard exercise. If $\gcd(n',m')= g\ge 1$ then $dg\ge d$ is a common divisor. But $d$ is greatest common divisor so $g = 1$.

Now if $k$ is a common multiple of $n$ and $m$ there are integers $a,b$ so that $k = an = bm$ so $k = an'd=bm'd$ so $\frac kd = an' = bm'$ is an integer. And $\frac k{dm'} = \frac {an'}{m'} = b$ is an inteer. So $m'|an'$ but $n'$ and $m'$ are relatively prime so $m'|a$ and if we let $a' = \frac a{m'}$ we have $\frac {k}{dm'} = \frac {a'm'n'}{m'}= a'n' = b$. So $n'| b$ and if $b' = \frac b{n'}$ we have $\frac {k}{dm'n'} = a' = b'$.

So $m'n'd$ divides all common multiples of $m,n$.

And $m'n'd = nm' = m'n$ is also a common multiple of $m,n$. And there can't be any smaller multiple as that would have to be divisible by $m'n'd$ and we can't have a smaller positive number divisible by a larger one.

So $m'n'd$ is the least common multiple of $m$ and $n$ and it divides all other multiples.