# More on coprime integers

• Feb 1st 2009, 07:06 AM
Louise
More on coprime integers
Let n and m be coprime positive integers. Show that if n divides a and m divides a then nm divides a. Deduce that nm is the least common multiple of n and m.

I have shown the first part by expressing a as a product of distinct prime numbers but not the deduction.

I'm also having trouble with the following;

Suppose that n and m are arbitrary positive integers. Show that if n divides a and m divides a then nm/(n,m) divides a. Deduce that nm/(n,m) is the least common multiple of n and m.

• Feb 1st 2009, 04:47 PM
star_tenshi
For the first part, I got this from Niven 5th edition:

There exist integers $\displaystyle x_{0}, y_{0}, x_{1}, y_{1}$ such that $\displaystyle 1 = mx_{0} + ay_{0} = nx_{1} + ay_{1}$

Thus,
$\displaystyle mx_{0} = 1 - ay_{0}$
$\displaystyle nx_{1} = 1 - ay_{1}$ and gives....
$\displaystyle (mx_{0})(nx_{1}) = (1 - ay_{0})(1 - ay_{1}) = 1 - a(y_{0} + y_{1} - ay_{0}y_{1})$

Rearranging, we get:
$\displaystyle 1 = mnx_{0}x_{1} + a(y_{0} + y_{1} - ay_{0}y_{1})$
which is in the form $\displaystyle 1 = mn (some integer) + a (some integer)$
and in the form of the gcd.

Thus $\displaystyle (nm,a) = 1$
• Feb 1st 2009, 05:30 PM
star_tenshi
For the second part, this may help:

Theorem: $\displaystyle gcd(m,n) \cdot lcm(m,n) = |mn|$

Proof:
$\displaystyle gcd(m,n) = gcd(|m|,|n|)$
and $\displaystyle lcm(m,n) = lcm(|m|,|n|)$ thus we can assume $\displaystyle m,n > 0$

Let $\displaystyle gcd(m,n) = d$
Thus $\displaystyle d|a$ and $\displaystyle d|b$...which gives... $\displaystyle m = dr$ and $\displaystyle n=ds$ for some integers $\displaystyle r,s$

Let $\displaystyle l=\frac{mn}{d}$
then claim $\displaystyle l = lcm(m,n)$
$\displaystyle m = dr \Rightarrow l = rn \Rightarrow n|l$
$\displaystyle n = ds \Rightarrow l = sm \Rightarrow m|l$
here we have shown that $\displaystyle l$ is a common multiple of $\displaystyle m$ and $\displaystyle n$.

Let $\displaystyle c$ be some common multiple of $\displaystyle m$ and $\displaystyle n$.
$\displaystyle \Rightarrow \exists u,v$ integers such that $\displaystyle c = mu$ and $\displaystyle c = nv$

Since $\displaystyle d = gcd(m,n)$ then $\displaystyle d = mx + ny$
$\displaystyle \frac{c}{l} = \frac{cd}{mn} = \frac{c(mx+ny)}{mn} = \frac{cmx}{mn} + \frac{cny}{mn} = \frac{cx}{n} + \frac{cy}{m}$ but $\displaystyle c = mu = nv$

Thus $\displaystyle \frac{c}{l} = vx + uy$ which is an integer.

This proves that any generic common multiple $\displaystyle c$ is divisible by $\displaystyle l$ and we can conclude that $\displaystyle l$ is the $\displaystyle lcm(m,n)$

Thus $\displaystyle l = \frac{mn}{d} \Rightarrow lcm(m,n) = \frac{mn}{gcd(m,n)}$