CRT solvablity criterion: congruence system solvable iff pairwise solvable

Question

Let a system of three linear congruence equations in integers be given;

\begin{cases}x\equiv b_1\mod c_1\\ x\equiv b_2\mod c_2\\ x\equiv b_3\mod c_3\\ \end{cases} with $c_1,c_2,c_3\in\mathbb Z_{+}$. Show that, if every pair of congruence equations admits solutions, then the entire system admits solutions.

Now, if every pairs admits solutions, then for example $b_1\equiv b_2\equiv a\mod \gcd(c_1,c_2)$ for some $a$ then;

$x-a\equiv b_1-a\mod c_1$

$x-a\equiv b_2-a\mod c_2$

replacing $x$ by $x':=\frac{x-y}{\gcd(c_1,c_2)}$, we obtain:

$x'\equiv b_1'\mod c_1'$

$x'\equiv b_2'\mod c_2'$

and since $\gcd(c_1,c_2)=1$ by Chinese Remainder Theorem we have a unique solution $\mod{c_1'c_2'}$, so $x\equiv (\gcd(c_1,c_2)x'+a)\mod{\operatorname{lcm(c_1',c_2')}}\quad(1)$

We can do it with another pair (take $2$nd and $3$rd, since we know that such a pair admits solutions;

$x\equiv(\gcd(c_2,c_3)x''+a')\mod{\operatorname{lcm(c_2',c_3')}}\quad(2)$

Now how can I show that for the pair $(1)$ and $(2)$ also solutions exist ?

Answer 1

$\newcommand{\lcm}{\mathrm{lcm}}$All pairs of equations have a solution iff $\gcd(c_{i}, c_{j}) \mid b_{i} - b_{j}$ for all $i, j$. So assume this holds.

Let $b$ be a solution of the first two congruences, so that $b \equiv b_{1} \pmod{c_{1}}$ and $b \equiv b_{2} \pmod{c_{2}}$. The system of three equations thus reduces to a system of two equations: $$ \begin{cases} x \equiv b \pmod{\lcm(c_{1}, c_{2})}\\ x \equiv b_{3} \pmod {c_{3}} \end{cases} $$ This has a solution iff $b - b_{3}$ is divisible by $$ \gcd(\lcm(c_{1}, c_{2}), c_{3}) = \lcm(\gcd(c_{1}, c_{3}), \gcd(c_{2}, c_{3})).. $$ It is enough to show that $\gcd(c_{1}, c_{3})$ divides $b - b_{3}$, the argument for $\gcd(c_{2}, c_{3})$ being exactly the same. Now $\gcd(c_{1}, c_{3})$ divides $b_{1} - b_{3}$, and $b \equiv b_{1} \pmod{c_{1}}$, so that $\gcd(c_{1}, c_{3})$ divides $b - b_{1}$, and thus $\gcd(c_{1}, c_{3})$ divides $b - b_{3} = (b - b_{1}) + (b_{1} - b_{3})$.

Answer 2

The inductive proof given below essentially verifies that we will always obtain a solution by using the common imethod of repeatedly replacing any pair of congruences by the single congruence specifying their solution. As we see from the proof, the key to the success of the inductive step is that gcd distributes over lcm, which allows us to lift up the pairwise solvability conditions to what we need. It turns out that such distributivity is exactly what is needed in any commutative ring, i.e. the proof immeidately generalizes to any ring whose lattice of ideals i distributive.

By induction there is a solution $\,x = \color{#c00}b\,$ to the first $\,n\!-\!1\,$ congruences, i.e.

$$\ \ \color{#c00}{b\equiv b_i}\!\!\!\pmod{c_i},\,\ i=1,\ldots n\!-\!1\ \quad (\color{#c00}\#)$$

Combining this solution with the final $n$'th congruence it suffices to solve

$$\begin{align} &x \equiv b\! \pmod {\!c},\,\ \ c = {\rm lcm}(c_1,\ldots c_{n-1})\\ &x\equiv b_n\!\!\!\! \pmod{\!c_n}\end{align}$$

That's solvable $\!\iff (c_n,c)\mid b_n-b,\,$ true by gcd distributes over lcm, i.e.

$\qquad\ \,\color{#90f}{(c_n,c)} = (c_n,{\rm lcm}(c_1,\ldots,c_{n-1}) = {\rm lcm}(\color{#0a0}{(c_n,c_1),\ldots,(c_n,c_{n-1})}) =: \ell $

$\begin{align}{\rm thus}\ \ \color{#90f}{(c_n,c)}\mid b_n-b \iff\:\! \ell\mid b_n-b\!\!\overset{\color{darkorange}{\rm L}\!\!\!}\iff \:&\,\color{#0a0}{(c_n,c_i)}\mid b_n-b,\ \forall\, i < n,\ \rm true\ by\\[.1em] \bmod\! &\,\color{#0a0}{(c_n,c_i)}\!:\,{\underbrace{b_n\equiv\, \color{#c00}{b_i}}_{\!\!\!\large \rm hypothesis}\!\!\underbrace{\ \color{#c00}{\equiv b}\phantom{_.}}_{\large\ \ (\color{#c00}\#)\!\!\!}}\end{align}$

where we used the lcm Universal Property $\ \color{darkorange}{\rm L}\!:\,\ {\rm lcm}(d_1,\ldots,d_k)\mid a\iff d_1,\ldots,d_k\mid a$.

Remark $ $ Domains that satisfy this Chinese Remainder Theorem (CRT) solvability criterion, i.e. where a system of congruences $\,x\equiv x_j\pmod{\!I_j}\,$ is solvable $\iff$ $\,x_j\equiv x_k \pmod{\! I_j + I_k},\,$ are known as Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations (cf. a couple dozen).