# a question about relatively prime numbers

Is it true that if $m, n$ are relatively prime integers, then $mn$, $m-n$ are also relatively prime? It seems intuitively true but I can't prove it...

Could anyone help me how to prove it?

If $d$ divides $mn,m-n;d$ must divide $mn+m(m-n)=m^2$

$d$ must divide $mn-n(m-n)=n^2$

$d$ must divide $(m^2,n^2)=(m,n)^2$

$$\gcd(mn,m-n)=d \implies \exists p:p\mid d\land (p\mid m \lor p\mid n)$$

$p\mid m \implies p\nmid n$

1. $p\mid m \implies p\mid mn$

2. $p\mid m \land p\nmid n \implies p\nmid m-n$

Here $p$ is a prime.

$(m,n)=1$ means there exist $a,b$ such that $$am+bn=1\tag{1}$$ Add and subtract $bm$ from $(1)$ to get $$(a+b)m+b(n-m)=1\tag{2}$$ Add and subtract $an$ from $(1)$ to get $$a(m-n)+(a+b)n=1\tag{3}$$ Multiply $(a+b)m=1+b(m-n)$ by $(a+b)n=1-a(m-n)$, then collect the multiples of $(m-n)$ from the right side and move them to the left $$(a+b)^2\color{#C00000}{mn}+(a-b+ab(m-n))\color{#C00000}{(m-n)}=1\tag{4}$$ $(4)$ says that $(mn,m-n)=1$.