Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know, If $(m_1,m_2)=(a,m_1)=(b,m_2)=1, \iff (am_2+bm_1, m_1m_2)=1$

I tried to generalize now.

Let $(a,m_1)=d_1, (b,m_2)=d_2$ where $d_1,d_2$ need not to be 1.

$(am_2+bm_1, m_1m_2)$

$=(am_2+bm_1, m_1)(am_2+bm_1, m_2)\ as\ (m_1,m_2)=1$

$=(am_2, m_1)(bm_1, m_2)$

$=(a, m_1)(b, m_2)\ as\ (m_1,m_2)=1$

So, $(am_2+bm_1, m_1m_2)=(a, m_1)(b, m_2)$

Now, I want to make $(m_1,m_2)=D$ where D is not necessarily 1.

Let $\frac{a}{A}=\frac{m_1}{M_1}=d_1$ and $\frac{b}{B}=\frac{m_2}{M_2}=d_2$,

so $(A,M_1)=1$ and $(B,M_2)=1$

$(am_2+bm_1, m_1m_2)=d_1.d_2(AM_2+BM_1, M_1M_2)=(a,m_1)(b,m_2)(AM_2+BM_1, M_1M_2)$

Now let $\frac{M_1}{M_{11}}=\frac{M_2}{M_{22}}=D_{12}$ i.e., $(M_1,M_2)=D_{12}$

so, $(M_{11},M_{22})=1$

then $(am_2+bm_1, m_1m_2)$

$=(a,m_1)(b,m_2)D_{12}(AM_{22}+BM_{11}, M_{11}M_{22}D_{12})$

$=(a,m_1)(b,m_2)(M_1,M_2)(AM_{22}+BM_{11}, M_{11}M_{22}D_{12})$

But, this $D_{12}$ is not necessarily co-prime with $M_{11}$ or $M_{22}$.

So, I could not proceed any further.

share|cite|improve this question
@RossMillikan, thanks for your observation, now it's rectified. – lab bhattacharjee Jul 29 '12 at 17:02
up vote 1 down vote accepted

Part 1:

We have both $$ (am_2+bm_1,m_1m_2)=d_1d_2\,\left(\frac{a}{d_1}\!\!\frac{m_2}{d_2}+\frac{b}{d_2}\!\!\frac{m_1}{d_1} ,\frac{m_1}{d_1}\!\!\frac{m_2}{d_2}\right)\tag{1} $$ and $$ (am_2+bm_1,m_1m_2)=D\,\left(a\frac{m_2}{D}+b\frac{m_1}{D} ,m_1\frac{m_2}{D}\right)\tag{2} $$ Equations $(1)$ and $(2)$ show that $$ \mathrm{lcm}(d_1d_2,D)\,\vert\,(am_2+bm_1,m_1m_2)\tag{3} $$

Part 2:

Since $(m_1,m_2)=D$, let $$ m_1x+m_2y=D\tag{4} $$ Since $(a,m_1)=d_1$, let $$ au_1+m_1v_1=d_1\tag{5} $$ Since $(b,m_2)=d_2$, let $$ bu_2+m_2v_2=d_2\tag{6} $$ Now, $(4)$ and $(5)$ yield $$ \begin{align} (am_2+bm_1)y+m_1(ax-by)&=aD\\ (am_2+bm_1)u_1+m_1(m_2v_1-bu_1)&=m_2d_1 \end{align}\tag{7} $$ and $(4)$ and $(6)$ yield $$ \begin{align} (am_2+bm_1)x+m_2(by-ax)&=bD\\ (am_2+bm_1)u_2+m_2(m_1v_2-au_2)&=m_1d_2 \end{align}\tag{8} $$ Using $(7)$ and Bezout, we can write $$ (am_2+bm_1)w_1+m_1z_1=(aD,m_2d_1)\tag{9} $$ Using $(8)$ and Bezout, we can write $$ (am_2+bm_1)w_2+m_2z_2=(bD,m_1d_2)\tag{10} $$ Thus, taking the product of $(9)$ and $(10)$ yields $$ (am_2+bm_1)w+m_1m_2z_1z_2=(aD,m_2d_1)(bD,m_1d_2)\tag{11} $$ and therefore, $(11)$ and Bezout yield $$ (am_2+bm_1,m_1m_2)\,\vert\,(aD,m_2d_1)\,(bD,m_1d_2)=d_1d_2D^2\,\left(\frac{a}{d_1},\frac{m_2}{D}\right)\,\left(\frac{b}{d_2},\frac{m_1}{D}\right)\tag{12} $$


The best I have come up with so far is $$ \mathrm{lcm}(d_1d_2,D)\,\vert\,(am_2+bm_1,m_1m_2)\,\vert\,d_1d_2D^2\,\left(\frac{a}{d_1},\frac{m_2}{D}\right)\,\left(\frac{b}{d_2},\frac{m_1}{D}\right)\tag{13} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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