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We want to find the solution $x$ to the congruence system

$$\begin{align} x &\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \ \gcd(m_1, m_2) = 1$$

These can be rewritten as

$$\begin{align} x &= r_1 + m_1 j\\ x &= r_2 - m_2k\end{align}$$

for unknown integers $j, k$. So we set those two equations as equal and rearrange them to

$$m_1j + m_2k = r_2 - r_1$$

Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to

$$m_1u + m_2v = \gcd(m_1, m_2)$$

Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.

Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) = r_2 - r_1 = m_1j + m_2k$

At this point I am lost because I don't think we can just assume $hm_1u = m_1j$ and extract $j = hu$.

How do I get the values of $j$ or $k$ so I can get the value of $x$?

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    $\begingroup$ Why on earth did someone downvote this? I showed my work and had a clear question here. $\endgroup$ – user399664 Dec 16 '16 at 2:02
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    $\begingroup$ I just upvoted to compensate. save yourself the emotional and intellectual energy of trying to understand the illogic of voting here. You could give your post a better title, though. "Can the Chinese Remainder Theorem be proved this way?" or "How to proceed in this proof of the Chinese Remainder Theorem, if possible?" $\endgroup$ – symplectomorphic Dec 16 '16 at 2:06
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Correct, if $x$ is a root of the congruences then $\, x= j\,m_1 + r_1 = -k\, m_2 + r_2\,$ has roots $\,j,k\in\Bbb Z$.

This argument reverses:$ $ if $\,j,k\,$ are integers with $\ j\, \color{#c00}{m_1}+ k\, \color{#0a0}{m_2} =\, \color{#0a0}{r_2} - \color{#c00}{r_1}\, $ then by $\rm\color{#c00}{re}\color{#0a0}{arranging}$ $\ x :=\ \color{#c00}{r_1} +\, j\ \color{#c00}{m_1}^{\phantom{|}}\ =\,\ \ \color{#0a0}{r_2}\: -\,\ k\ \color{#0a0}{m_2}\ $ is one solution of the given system of congruences because $\,x\equiv \color{#c00}{r_1}\!\!\pmod{\!\!\color{#c00}{m_1}}^{\phantom{|^|}}\!\!\!,$ $x\equiv \color{#0a0}{r_2}\!\pmod{\!\!\color{#0a0}{m_2}}.\,$ Since you've already discovered one such solution for $\,j,k,\,$ viz. $\,j=hu,\,k^{\phantom{|^|}}\!\! = hv,\,$ you need only substitute into the above rearranged CRT solution for $\,x.$

Remark $ $ Combining both directions above and adding a final gcd equivalence yields the following

Theorem $\ \ \left.\exists\, x\in\Bbb Z\!: \begin{align}x\equiv r_1\!\!\!\pmod{\!m_1}\\ x\equiv r_2\!\!\!\pmod{\!m_2}\end{align}\right\} \begin{array}{l}\!\iff \exists\,j,k\in\Bbb Z\!:\ j\,m_1\! + k\, m_2 =\, r_2\!-r_1 \\ \!\iff\, \gcd(m_1,\,m_2)\mid r_2 -r_1\end{array}$

Proof $ $ Clearly $\,d := \gcd(m_1,m_2)\mid r_2-r_1 \,$ is a necessary condition for the equation to have roots $\,j,k\in \Bbb Z,\,$ by $\,d\mid m_1,m_2\Rightarrow\, d^{\phantom{|}}_{\phantom{i}}\!\mid j m_1\! + km_2 = r_2 - r_1.\,$ Further this condition is also sufficient by Bezout (or, constructively, by the extended Euclidean algorithm), i.e. we can scale the Bezout equation $\, a m_1\! + b m_2 = d\,$ by $\, c = \large \frac{r_2\,-\,r_1^{\phantom{.}}}{d}\,$ to get $\,ca\,m_1\!+cb\,m_2 = r_2-r_1 \,$ so, as above, rearranging this yields a congruence system solution: $\ x\, :=\, r_1 + ca\,m_1 = r_2 - cb\,m_2$.

Thus the congruence system is solvable $\iff d=\gcd(m_1,m_2)\mid r_2-r_1, \,$ i.e. iff the pair of congruences is consistent mod their moduli gcd, and when true we can constructively read off a solution from the Bezout equation for the moduli by translating it into the equivalent system language as above, i.e. scale the Bezout equation to obtain the residue difference $\,r_1-r_2\,$ then rearrange it as above to obtain $\,x.\,$ Here is a worked example from this viewpoint. Thus we've the following simple Bezout-based CRT method for solving congruence systems

$\! \small \textbf{ scale the Bezout equation for the moduli gcd}\!$ $\small \textbf{ to get the residue difference, then }\rm\color{#c00}{re}\color{#0a0}{arrange}$

If you are familiar with ideals and cosets then the above can be expressed more succinctly as

$$ \bbox[9px,border:1px solid #c00]{r_1\! +\! m_1\Bbb Z\,\cap\, r_2\! +\! m_2\Bbb Z \neq \phi \iff r_1-r_2 \in m_1\Bbb Z+m_2\Bbb Z}\qquad\qquad $$

Generally a congruence system is solvable $\iff$ each pair of congruences is solvable as above, and we can solve the system by successively replacing a pair of congruences by the single congruence obtained from solving the pair of congruences. By induction we eventually obtain a single congruence, which is the solution of the entire congruence system.

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  • $\begingroup$ I am saying I can't say that $j = hu$ and $k = hv$ $\endgroup$ – user399664 Dec 16 '16 at 2:41
  • $\begingroup$ @user399664 But they are valid solutions, so what do you mean by you "can't say that"? $\endgroup$ – Bill Dubuque Dec 16 '16 at 2:43
  • $\begingroup$ As in all I can really say is $hm_1u + hm_2v = m_1j + m_2k$ $\endgroup$ – user399664 Dec 16 '16 at 2:44
  • $\begingroup$ @user399664 What we can say is that the values $\,j =hu\,$ and $\,k=hv\,$ are solutions of $\ jm_1+km_2 = r_2 - r_1\ $ (see the overbraced equation above).. That is all that we need. $\endgroup$ – Bill Dubuque Dec 16 '16 at 2:46
  • $\begingroup$ So it is correct to say that $x = r_1 + m_1hu$ and $x = r_2 - m_2hv$? Both of these $x$'s will always be the same? $\endgroup$ – user399664 Dec 16 '16 at 2:48

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