Let $d=(m,n)$, then for any $q,r \in \mathbb{Z}$ there is a $k \in \mathbb{Z}$ such that $kd=mq+nr$

Question

This is a small conjecture I believe to be true. On Group Theory class we were asked to calculate the subgroups $HK$ of $\mathbb{Z}$, were $H=mZ$ and $K=nZ$. I decided to solve the problem by considering 3 cases:

($m=nq$ for some $q \in \mathbb{Z}$) or ($n=mt$ for some $t \in \mathbb{Z}$)
$(m,n)$=1
None of the conditions on 1) is satisfied and $(m,n)>1$.

The case I'm stuck at is 3). I found it easy to prove that $d\mathbb{Z} \subset (n\mathbb{Z})(m\mathbb{Z})$. I struggle to prove that $(n\mathbb{Z})(m\mathbb{Z}) \subset d\mathbb{Z}$. In order to prove this last part I think my conjecture is key.

Is there any elementary result in Number Theory that I'm missing in order to prove this?

This is just a thought, but can you use Bézout's identity for this? The conjecture looks a lot like it. — Accelerator, Sep 20 '22 at 03:18
@Accelerator: looks like we had the same thought at the same time : ) — user1090793, Sep 20 '22 at 03:19
@Accelerator you are absolutely right, it does help! I also used that $d$ divides both $m$ and $n$. So for any $nq$ and $mp$, $d$ divides $nq$ and $d$ divides $mp$. — Blue, Sep 20 '22 at 03:28
By dupe: for any common divisor $,\color{#c00}{d\mid m,n}\Rightarrow d\mid qm+rn,,$ with quotient $,k = q(\color{#c00}{m/d}) + r(\color{#c00}{n/d})$ — Bill Dubuque, Sep 20 '22 at 07:52
@user1090793 Bezout is a red herring - it is true for any common divisor - see the prior comment. — Bill Dubuque, Sep 20 '22 at 18:22
@BillDubuque: duly noted. OP, note that Bill’s comment provides a more general and simpler answer — user1090793, Sep 20 '22 at 22:49

score 3 · Accepted Answer · answered Sep 20 '22 at 03:25

3

Since $d = (m, n)$, we know $m = dx$ and $n = dy $, for some $x, y \in \mathbb{Z}$.

Then $mq + nr = (dx)q + (dy)r = d(xq + yr)$. Setting $xq + yr = k$, the result is proven.

answered Sep 20 '22 at 03:25

ani

54

1

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 20 '22 at 07:53

Let $d=(m,n)$, then for any $q,r \in \mathbb{Z}$ there is a $k \in \mathbb{Z}$ such that $kd=mq+nr$

1 Answers1