2
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $(m^2,n^2)=(m,n)^2$

$\endgroup$
2
$\begingroup$

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

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

  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.

$\endgroup$
1
$\begingroup$

$(m,n)=1$ 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 $(mn,m-n)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.