Solution to Simultaneous Linear Congruences

Theorem
Let:

$$ $$ $$ $$

be a system of simultaneous linear congruences.

This system has a simultaneous solution iff:
 * $$\forall i, j: 1 \le i, j \le r: \gcd \left\{{n_i, n_j}\right\}$$ divides $$b_j - b_i$$.

If a solution exists then it is unique modulo $$\operatorname{lcm} \left\{{n_1, n_2, \ldots, n_r}\right\}$$.

Proof
We take the case where $$r = 2$$.

Suppose $$x \in \Z$$ satisfies both:

$$ $$

That is, $$\exists r, s \in \Z$$ such that:

$$ $$

Eliminating $$x$$, we get:
 * $$b_2 - b_1 = n_1 r - n_2 s$$.

The RHS is an integer combination of $$n_1$$ and $$n_2$$ and so is a multiple of $$\gcd \left\{{n_1, n_2}\right\}$$.

Thus $$\gcd \left\{{n_1, n_2}\right\}$$ divides $$b_2 - b_1$$, so this is a necessary condition for the system to have a solution.

To show sufficiency, we reverse the argument.

Suppose $$\exists k \in \Z: b_2 - b_1 = k \gcd \left\{{n_1, n_2}\right\}$$.

We know that $$\exists u, v \in \Z: \gcd \left\{{n_1, n_2}\right\} = u n_1 + v n_2$$ from Bézout's Identity.

Eliminating $$\gcd \left\{{n_1, n_2}\right\}$$, we have:
 * $$b_1 + k u n_1 = b_2 - k v n_2$$.

Then:
 * $$b_1 + k u n_1 = b_1 + \left({k u}\right) n_1 \equiv b_1 \pmod {n_1}$$
 * $$b_1 + k u n_1 = b_2 + \left({k v}\right) n_2 \equiv b_2 \pmod {n_2}$$

So $$b_1 + k u n_1$$ satisfies both congruences and so simultaneous solutions do exist.

Now to show uniqueness.

Suppose $$x_1$$ and $$x_2$$ are both solutions.

That is:
 * $$x_1 \equiv x_2 \equiv b_1 \pmod n_1$$
 * $$x_1 \equiv x_2 \equiv b_2 \pmod n_2$$

Then from Intersection of Congruence Classes Modulo m the result follows.

The result for $$r > 2$$ follows by a tedious induction proof.