1

Assume $\gcd(m,n)=1$. Then $d\mid mn$ if and only if $d=ab$, where $a\mid m$, $b\mid n$ and $\gcd(a,b)=1$.

So I know that $mx+ny=1$ for some integers $x,y$. Also that $m=aE$ and $n=bF$ for some integers $E,F$.

If $d\mid mn$ then $mn=dQ$.

So then you have that $mn=abEF$ so $dQ=abEF$. I'm not sure how to conclude that $Q=EF$ so you get $d=ab$. Now the reverse direction seems simpler if $d=ab$ then you have right away that $d\mid mn$.

Xam
  • 6,119
HighSchool15
  • 2,061
  • 14
  • 35

2 Answers2

1

Hint $\ \underbrace{(d,m)}_{\large a}\underbrace{(d,n)}_{\large b} = (dd,dm,dn,mn) = d(d,\color{#c00}{m,n},mn/d) = d\ $ by $\ (\color{#c00}{m,n}) = 1$

Bill Dubuque
  • 272,048
1

Solve $(d,m)=dx+my,$ and $(d,n)=du+nv$.

Use these to show that if $d\mid mn$ then $d\mid (d,m)(d,n)$.

But since $(d,m)\mid d$ and $(d,n)\mid d$ and $((d,m),(d,n))=1$, you have that $(d,m)(d,n)\mid d$.

(This last uses that if $a\mid d$ and $b\mid d$ and $(a,b)=1$ then $ab\mid d$.)

So $d=\pm(d,m)(d,n)$.

Thomas Andrews
  • 177,126