Showing $(m,n)=(u,v)$ given initial conditions

Question

I am working on my proofs, and I think this is valid. Can anyone confirm?

For the initial conditions, we have that there are integers $a,b,c,d,m,n,u,v$ such that $$ad-bc=\pm1, u=am+bn, v=cm+dn$$ Now, I can state that suppose $ad>bc$. Then $ad-bc=1$ Thus we can say that both $a,b$ and $c,d$ are integers such that $(a,b)=(c,d)=1$ $$adu-bcu=u, adv-bcv=v $$ Letting $m=du=-bv, n=-cu=av$ Then $$(m,n)=(m,m,n,n)=(du,-bv,-cu,av)=(u(d,-c),v(a,-b))=(u,v)$$ We can argue that $bc>ad$ yields the same result.

Is this okay?

EDIT: I am not sure how to move forward...

$(m,n)=(m,m,n,n)$ is meaningless, and the next equality isn't true... — Thomas Andrews, Jul 02 '14 at 15:24
In particular, it is not true that $m=du=-bv$ and and it is not true that $n=-cu=av$. — Thomas Andrews, Jul 02 '14 at 15:25
Why would $(m,n)=(m,m,n,n)$ be meaningless? I was just shown that if I have $(a,b,c)$ then I can argue that $(a,b,c)=(a,b,a,c)=((a,b),(a,c))...$ — Lalaloopsy, Jul 02 '14 at 15:30
@ThomasAndrews, how should I procede? I'm not sure how to go from here then if these arguments fail... — Lalaloopsy, Jul 02 '14 at 15:41
@Thomas Why do you think that $\ (m,n) = (m,m,n,n)\ $ is "meaningless"? — Bill Dubuque, Jul 02 '14 at 16:30
Hint: If $q$ divides $m$ and $n$, then $q$ divides $u$ and $v$. For the converse, consider for example $d(am+bn)-b(cm+dn)=\pm m$. (You had this.) Forget about $\gt$, it is simpler than that. — André Nicolas, Jul 02 '14 at 17:51
@BillDubuque I was taking it to mean a $2$-tuple is a $4$-tuple, but I forgot about the GCD notation... — Thomas Andrews, Jul 02 '14 at 18:58

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

$\begin{eqnarray}{\bf Hint}\ \ \ \text{determinant of}\,\ \left[ \begin{array}{cr} a&\!\! b\\ c & d\end{array}\right] \left[ \begin{array}{cc} m & x\\ n & y\end{array}\right] &=& \left[ \begin{array}{cc} u & \bar u\\ v & \bar v\end{array}\right] \\ \Rightarrow\ \ \color{#0a0} D\!\!\!\!\!\!\!\!\!\! \smash[b]{\underbrace{\color{#0a0}{(my\! -\! nx)}}_{\quad\ \ \large =\,\color{#0a0}{(m,n)}\ {\rm by\ Bezout}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\! &=& {\color{#c00}u\,\bar v-\bar u\,\color{#c00} v}\phantom{I^{I^I}}\\ \\ \end{eqnarray}$

Hence we deduce that $\ \color{#c00}{(u,v)}\mid \color{#0a0}{D\ (m,n)}\ $ since $\,\color{#c00}{(u,v)}\,$ divides the $\rm\color{#c00}{rhs}$ above, so also the $\rm\color{#0a0}{lhs.}$ $ $ QED

Alternatively, this can also be deduced using Cramer's Rule, e.g. see this answer. Notice that the question is simply the special case when the determinant $\,D = \pm1.$

Showing $(m,n)=(u,v)$ given initial conditions

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