1

It's easy to use prime factorization to show:

If $m\mid n^2$ then $\gcd(m,n^2/m)\mid n$.

Can anybody find some other proof - perhaps a simple reduction of some sort? Maybe solving $m^2x + n^2y=mn$, for example?

I suppose the result would follow if you could prove that $\gcd(m^2,n^2)=\gcd(m,n)^2$, since $\gcd(m,n)^2\mid mn$ since $\gcd(m,n)\mid m$ and $\gcd(m,n)\mid n$. That result doesn't even need $m\mid n^2$.

Thomas Andrews
  • 177,126

3 Answers3

2

Since $m \mid n^2$, we have $n^2 = dm$ for some integer $d$. Inserting into the equation $m^2 x + n^2 y = mn$ gives $$ m^2x + dmy = mn\\ mx + dy = n\\ \gcd(m, d) \mid n $$ But $\gcd(m, d)^2 \mid md$, and we had $md = n^2$, so we're done.

Arthur
  • 199,419
  • Very nice. Much briefer than my answer. I had a feeling there was a "slick" answer :) – Thomas Andrews Jan 12 '16 at 09:02
  • 1
    (I suppose it uses $a^2\mid b^2$ implies $a\mid b$, which is often proved using prime factorization, too.) – Thomas Andrews Jan 12 '16 at 09:06
  • But of course, proofs that if $\sqrt{c}$ is rational, then it is an integer, also tend to use prime factorization. :) – Thomas Andrews Jan 12 '16 at 09:10
  • @ThomasAndrews So in the end it's difficult to avoid prime factorisation all together this way, but at least I've hidden it in a more general lemma :P – Arthur Jan 12 '16 at 09:15
  • 1
    Interestingly, the theorem that $\gcd(m^2,n^2)=\gcd(m,n)^2$ is equivalent to the statement that a square root that is rational is an integer. Because we can pick $r=m/n$ with $\gcd(m.n)=1$, then we get that $n^2$ must divide $m^2$, so $n^2=1$. – Thomas Andrews Jan 12 '16 at 09:15
  • @ThomasAndrews That's neat. Just to nitpick: any two proven theorems are equivalent. But yes, I know what you really meant. – Arthur Jan 12 '16 at 09:19
1

By here $\,\color{#0a0}{(m^2,n^2)} = (m,n)^2 = (m^2,\color{#c00}{mn},n^2)\Rightarrow \color{#0a0}{(m^2,n^2)}\mid \color{#c00}{mn} \overset{\!\div\,m}\Rightarrow (m,n^2/m)\mid n$

Bill Dubuque
  • 272,048
0

Oh, this all follows from the result that if $\gcd(a,b)=1$ then $\gcd(a^2,b^2)=1$. This is true since if $ax+by=1$ then $$1=(ax+by)^3 = a^2X+b^2Y.$$

So we deduce that $\gcd\left(\frac{n^2}{\gcd(n,m)^2}\,,\frac{m^2}{\gcd(n,m)^2}\right)=1$, and hence that $\gcd(m^2,n^2)=\gcd(m,n)^2$, and hence that $m^2x+n^2y=mn$ has a solution.

Thomas Andrews
  • 177,126