Solution of Linear Congruence

Theorem
Let $$a x \equiv b \pmod n$$ be a linear congruence.

The following results hold:
 * It has solutions iff $$\gcd \left\{{a, n}\right\} \backslash b$$;
 * If $$\gcd \left\{{a, n}\right\} = 1$$, the congruence has a unique solution.
 * If $$\gcd \left\{{a, n}\right\} = d$$, the congruence has $$d$$ solutions which are given by the unique solution modulo $$\frac n d$$ of the congruence $$\frac a d x \equiv \frac b d \pmod {\frac n d}$$.

Proof
Consider the linear congruence $$a x \equiv b \pmod n$$.

Suppose $$\exists x_0 \in \Z: a x_0 \equiv b \pmod n$$.

Then $$\exists y_0 \in Z: a x_0 - b = n y_0$$ by definition of congruence.

Thus $$x = x_0, y = y_0$$ is a solution to the linear Diophantine equation $$ax - ny = b$$.

On the other hand, if $$x = x_0, y = y_0$$ is a solution to the linear Diophantine equation $$ax - ny = b$$, then it follows that $$a x \equiv b \pmod n$$.

Hence the problem of finding all integers satisfying the linear congruence $$a x \equiv b \pmod n$$ is the same problem as finding all the $$x$$ values in the linear Diophantine equation $$ax - ny = b$$.

Hence the following:


 * It has solutions iff $$\gcd \left\{{a, n}\right\} \backslash b$$:

This follows directly from Solution of Linear Diophantine Equation: the linear Diophantine equation $$ax - ny = b$$ has solutions iff $$\gcd \left\{{a, n}\right\} \backslash b$$.


 * If $$\gcd \left\{{a, n}\right\} = 1$$, the congruence has a unique solution:

Suppose then that $$\gcd \left\{{a, n}\right\} = 1$$.

From Solution of Linear Diophantine Equation, if $$x = x_0, y = y_0$$ is one solution to the linear Diophantine equation $$ax - ny = b$$, the general solution is:
 * $$\forall k \in \Z: x = x_0 + n k, y = y_0 + a k$$

But $$\forall k \in \Z: x_0 + n k equiv x_0 \pmod n$$.

Hence $$x \equiv x_0 \pmod n$$ is the only solution of $$a x \equiv b \pmod n$$.


 * If $$\gcd \left\{{a, n}\right\} = d$$, the congruence has $$d$$ solutions which are given by the unique solution modulo $$\frac n d$$ of the congruence $$\frac a d x \equiv \frac b d \pmod {\frac n d}$$:

But $$\gcd \left\{{\frac a d, \frac n d}\right\} = 1$$ from Divide by GCD for Coprime Integers.

So the RHS has a unique solution modulo $$\frac n d$$, say:
 * $$x \equiv x_1 \pmod {\frac n d}$$.

So the integers $$x$$ which satisfy $$a x \equiv b \pmod n$$ are exactly those of the form $$x = x_1 + k \frac n d$$ for some $$k \in \Z$$.

Consider the set of integers $$\left\{{x_1, x_1 + \frac n d, x_1 + 2 \frac n d, \ldots, x_1 + \left({d-1}\right)\frac n d}\right\}$$.

None of these are congruent modulo $$n$$ and none differ by as much as $$n$$.

Further, for any $$k \in Z$$, we have that $$x_1 + k \frac n d$$ is congruent modulo $$n$$ to one of them.

To see this, write $$k = d q + r$$ where $$0 \le r < d$$ from the Division Theorem.

Then:

$$ $$ $$

So these are the $$d$$ solutions of $$a x \equiv b \pmod n$$.