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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.

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@RossMillikan, thanks for your observation, now it's rectified. –  lab bhattacharjee Jul 29 '12 at 17:02

1 Answer 1

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 $$ (am_2+bm_1,m_1m_2)\,\vert\,(aD,m_2d_1)\,(bD,m_1d_2)=d_1d_2D^2\,\left(\frac{a}{d_1},\frac{m_2}{D}\right)\,\left(\frac{b}{d_2},\frac{m_1}{D}\right)\tag{12} $$


Conclusion:

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

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