Prove that if $b = dk$ where $gcd(m, a) = d$, there are $d$ solutions to $ax \equiv b \mod m$

Question

I know that there exists at least one solution, because $ax \equiv b \mod m \implies ax - b = qm$

Rearranging terms gives $ax - qm = b$, and since $d | a$ and $d | m$, so $d | b$ it can be said that $b \equiv 0 \mod d$.

I implemented Euclid's extended algorithm and have been able to experimentally verify that the claim holds, but I do not see a pattern or how I can begin to show there are exactly $d$ solutions.

def euclid_extended(x, y, i=0): if y == 0: print("iteration {0} returns ({1}, {2}, {3}) - base case hit with ({4}, {5})".format(i, x, 1, 0, x, y)) return (x, 1, 0) else: (d, a, b) = euclid_extended(y, x % y, i + 1) print("iteration {0}. ({4}, {5}) => ({1}, {2}, {3})".format(i, d, b, a - b * (x//y), x, y)) return (d, b, a - b * (x//y))

Answer 1

Case 1: $\gcd (m,a) = 1; b = 1$ then

$ax \equiv 1 \mod m$ has exactly one solution modulo $m$.

If $ax \equiv ay \equiv 1 mod m$ then $m$ divides $a(x-y)$ and as $m$ and $a$ are relatively prime and have no factors in common, $m$ divides $x-y$ and $x \equiv y \mod m$. So $x \equiv y \mod m$ and $x$ is unique solution. If it exists.

As $\gcd(m,a) =1$ there are $ax + ym = 1$ and ... $ax \equiv 1 \mod m$.

Case 2: $\gcd(m,a) = 1$ $b = k$.

Solve for $ay \equiv 1 \mod m$. Let $x = ky$. That is clearly a solution.

If $aw \equiv k \mod m$ is another solution then $aw \equiv ax \mod m$ and as $\gcd(m,a)=1$ and $a$ and $m$ have no factors in common then $w \equiv x \mod m$.

Case 3: $\gcd(m,a) = d$ $b = dk$.

Let $m' = m/d$ and $a' = a/d$. Note: $\gcd(a',m') = 1$ (else that'd be a common divisor of $m$ and $d$ and $d*\gcd(a',m')$ would be a common divisor larger than the greatest common divisor.)

Then $ax \equiv b \mod m \implies ax - b = a'dx - dk$ is divisible by $m=m'd$. So $a'x - k$ is divisible by $m'$ or $a'x \equiv k \mod m'$ which has exactly one solution.

Call that solution $g$. Any solution $x$ must be so that $x \equiv g \mod m'$.

There are precisely $d$ such $x$, namely $x = g + jm'; 0 \le j < d$.

Answer 2

The linear congruence $ax \equiv b \mod m$ has a solution $x'$ since $(a,m) = d | b$ and there exists $p,q$ such that $ap -mq = d$.

Note that there exists $j$ such that $b = jd$ and $a(pj)-m(qj) = jd = b$, implying that the solution is $x' = pj$.

It is easy to show all solutions must be of the form

$$x = x' + t \frac{m}{d}$$

with $t = 0, \pm1, \pm 2, \ldots.$

The $d$ solutions $x', x' + \frac{m}{d}, x' + 2\frac{m}{d}, \ldots, x' + (d-1)\frac{m}{d}, $ are mutually incongruent modulo $m$ since the difference between any two is less than $m$. If $x'' = x' + p \frac{m}{d}$ is any other solution then Euclid's division lemma gives $p = qd + r$ with $0 \leqslant r < d$ and $x'' = x' + (qd +r)\frac{m}{d} \equiv x' + r \frac{m}{d}.$ Therefore, there are $d$ mutually incongruent solutions modulo $m$.