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I am studying the following problem, in which I don't see why one of the congruences has a single solution:

What is the number of solutions $(\bmod m n)$ for the equations $x=a(\bmod m)$ and $x=b(\bmod n)$ where $\operatorname{gcd}(m, n)=\operatorname{gcd}(a, m)=\operatorname{gcd}(b, n)=d ?$ Answer: $d$. There is a single solution for $y=a / d(\bmod m / d)$ and $y=b / d(\bmod n / d)$, and $x=$ $y d+i m n / d(\bmod m n)$ for $i \in\{0, \ldots, d-1\}$ satisfies $x=a(\bmod m)$ and $x=b(\bmod n)$ as $m n / d=0$ $(\bmod m)$ and $m n / d=0(\bmod n)$. Any $z=0 \bmod m$ and $z=0(\bmod n)$ implies that $z$ is a multiple of $m n / d$, these are the only solutions.

I am not seeing why there is a single solution for $y=a / d\pmod {m / d}$ and $y=b / d\pmod {n / d}$, as $m \over d$ and $n \over d$ don't seem necessarily coprime, as they could have other factors in common besides $d$, so eliminating $d$ doesn't seem to put us in a position where we could use CRT to conclude a unique solution here. What basis do we have to conclude that there exists a single solution here?

Princess Mia
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  • See the uniqueness part of the proof of General Easy CRT in the linked dupe. – Bill Dubuque Sep 30 '23 at 04:05
  • @billDubuque I read the uniqueness part which showed how the solution was unique in mod $mn \over d$; however I am actually not seeing how to apply it here as I believe $y$ is unique in mod $mn$ instead. How could I use this argument to show $y$ is unique here? I believe the difference between the post you showed and the one I have are the constraints on $a,b$ having gcd $d$ with their moduli here. – Princess Mia Sep 30 '23 at 08:41
  • @BillDubuque actually, is the argument that $m \over d$ and $n \over d$ are actually coprime, as $d$ is gcd so dividing through renders them coprime? So by CRT there is solution? – Princess Mia Sep 30 '23 at 18:33
  • Yes $,(m,n)=d\Rightarrow (m/d,n/d)=1,$ by the gcd distributive law. – Bill Dubuque Sep 30 '23 at 18:55
  • @BillDubuque actually, if we use CRT on the basis of $m \over d$ and $n \over d$ being coprime, then we could only say $y$ is unique mod $mn \over d^2$; which argument would we use to show $y$ is unique mod $mn$, which seems to be the intended modulus in which $y$ is said to be unique in the solution I presented? – Princess Mia Sep 30 '23 at 19:13
  • $x/d\equiv c\pmod{!mn/\color{#c00}{d^2}}!!!\overset{\times\ \color{#c00}d!}\iff x\equiv cd\pmod{!mn/{\color{#c00}d}} \ \ $ – Bill Dubuque Sep 30 '23 at 19:54
  • @BillDubuque would this not imply that there are $d^2$ solutions in mod $mn$ then? Because there are $d$ possibilities for what $yd$ could be in mod $mn$ by the statement you just commented, and $d$ possibilities for $i$ in $y d+i m n / d(\bmod m n)$? – Princess Mia Sep 30 '23 at 20:57
  • A residue $,a\bmod m,$ maps to $,\color{darkorange}d,$ values $!\bmod \color{darkorange}dm,,$ i.e.

    $$\quad\ \begin{align} x \equiv a!!!\pmod{! m}, &\equiv, {, a + k:!m}_{,\large 0\le k<\color{darkorange}d}!!\pmod{!dm}\[.3em] &\equiv, {a,,\ a! +! m,,\ldots,, a! +! (\color{darkorange}d!-!1)m} \end{align}\qquad$$

     – Bill Dubuque Sep 30 '23 at 21:39
  • which is true because $\ k:!m\bmod dm =, (\color{#c00}{k\bmod d}), m\ $ by MDL, and the RHS takes exactly $,d,$ values $!\bmod dm,,$ i.e. $,\color{#c00}0m,, \color{#c00}1m,, \color{#c00}2m, \ldots, (\color{#c00}{d!-!1})m.,$ Here MDL = mod Distributive Law. – Bill Dubuque Sep 30 '23 at 21:41
  • e.g.

    $$ \overbrace{3x\equiv 6!!!!\pmod{!12}}^{{\rm\large cancel}\ \ \Large (3,12),=,\color{darkorange}3}\iff x\equiv 2!!!!\pmod{!4},\equiv, !!!!!!!!!!!!!!!!!!\overbrace{{2,6,10}}^{\qquad\ \ \Large{ 2,+,4k}_{\ \Large 0\le k< \color{darkorange}3}}!!!!!!!!!!!!!!!!!!!!!!\pmod{!\color{darkorange}3\cdot 4}\qquad $$ See also here.

     – Bill Dubuque Sep 30 '23 at 21:41

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