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I want to prove this theorem:

Let $n = n_1 n_2$, $(n_1, n_2) = 1$. Then for all $d^2 | n$, $d^2$ can be uniquely written as $d^2 = (d_1 d_2)^2$ where $d_1^2 | n_1$, $d_2^2 | n_2$.

I've already proved the following two results:

(1) If $d^2 | n$, then $(d^2, n_1) = d_1^2$, $(d^2, n_2) = d_2^2$ where $d = d_1 d_2$.

(2) If $d_1^2 | n_1$, $d_2^2 | n_2$ where $d = d_1 d_2$, then $d^2 | n$.

Do these results prove that theorem? Why? And are both the two results required to prove the theorem?

And similarly, can we prove the following theorem?

Let $n = n_1 n_2$, $(n_1, n_2) = 1$. Then for all $d | n$, $d$ can be uniquely written as $d = d_1 d_2$ where $d_1 | n_1$ and $d_2 | n_2$.

helloworld142857
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1 Answers1

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(2) is obvious and useless.

(1) proves the existence of $(d_1,d_2)$ in your theorem because $(d^2,n_i)\mid n_i$.

The uniqueness stems from the fact that if $b\mid n_2$ then $(b,n_1)=1$ hence $(ab,n_1)=(a,n_1)$. This proves that $d_1^2$ has to be equal to $(d^2,n_1)$. The uniqueness of $d_2^2$ can then be derived from that of $d_1^2$, or proved the same way.

An alternative method to prove your theorem is to show independently (in $\Bbb N$) that:

  • if $(n_1,n_2)=1$ then any divisor of $n_1n_2$ can be uniquely written $m_1m_2$ s.t. $m_i\mid n_i$;
  • if $(m_1,m_2)=1$ and $m_1m_2$ is a square, so are $m_1$ and $m_2$.

Edit. Details upon request about the uniqueness of $d_1^2,d_2^2$ in your theorem, or more generally of $m_1,m_2$ in the first point above.

If $(n_1,n_2)=1$ and $m=m_1m_2$ s.t. $m_i\mid n_i$ then $(n_1,m_2)=1$ hence $(n_1,m)=(n_1,m_1)=m_1$, and similarly $m_2=(n_2,m)$.

Anne Bauval
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  • Thank you for your response! Could you explain more about how the uniqueness is proved? How did you prove that $d_1^2$ has to be equal to $(d^2, n_1)$? I'm new to number theory, so I might need more steps. Thank you! – helloworld142857 Jan 04 '25 at 02:38
  • I'm still confused. Are you proving an easier version of the theorem: Let $n = n_1 n_2$, $(n_1, n_2) = 1$. Then for all $d | n$, $d$ can be uniquely written as $d = d_1 d_2$ where $d_1 | n_1$ and $d_2 | n_2$. ? – helloworld142857 Jan 04 '25 at 21:10
  • Yes, this (up to the notation) is exactly what I wrote in the first point ("if $(n_1,n_2)=1$ then any divisor of $n_1n_2$ can be uniquely written $m_1m_2$ s.t. $m_i\mid n_i$"). Though more general, the uniqueness in this statement is easier than in your theorem because the notations are simpler. – Anne Bauval Jan 04 '25 at 21:15
  • Ok, I see. But I still don't understand how the uniqueness is proved. – helloworld142857 Jan 04 '25 at 21:17
  • I cannot see what you are missing. I proved $m_i=(n_i,m)$, so $m_i$ is unique. – Anne Bauval Jan 04 '25 at 21:18
  • Do you mean that you let $m_i := (n_i, m)$, which proves that it's unique? – helloworld142857 Jan 04 '25 at 21:23
  • I do not let $m_i:=(n_i,m)$. I prove $m_i=(n_i,m)$ (if $m_i\mid n_i$ and $m=m_1m_2$ and $(n_1,n_2)=1$). – Anne Bauval Jan 04 '25 at 21:31
  • I think I'm confused about the logic. I want to prove that for any $m | n_1 n_2$, we have some unique $m_1, m_2$ such that $m = m_1 m_2$ and $m_i | n_i$. I think that only $m$ is given at the beginning and we do not have $m = m_1 m_2$, and you need to prove that if $m_i$ satisfy the conditions, then $m_i$ is unique. – helloworld142857 Jan 04 '25 at 21:44
  • Given $m\mid n_1n_2$, we prove the uniqueness of $(m_1,m_2)$ such that $m_i\mid n_i$ and $m=m_1m_2$ by showing that such $m_i$s cannot be anything else than $(m,n_i)$. – Anne Bauval Jan 04 '25 at 22:03
  • So do you mean this: For each $m | n_1 n_2$, any $m_1, m_2$ such that $m_1 m_2 = m$ and $m_i | n_i$ must be $m_i = (m, n_i)$. And this proves both the existence and uniqueness of the theorem, is that correct? – helloworld142857 Jan 04 '25 at 22:08
  • Yes this is proved in the edit of my answer. 2) No, only uniqueness.
    • – Anne Bauval Jan 04 '25 at 22:13
  • So basically, for uniqueness, you proved that $m_i$ must be equal to $(m, n_i)$, and for the existence, we just need to show that the $m_i = (m, n_i)$ actually satisfy $m_1 m_2 = m$. Is this correct? – helloworld142857 Jan 05 '25 at 02:37
  • Yes, and for the existence you must have done it, when you "already proved" your "result 1)". – Anne Bauval Jan 05 '25 at 07:14
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    Ok, now I understand it. Thank you very much! – helloworld142857 Jan 05 '25 at 19:39