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The Theorem state:

If $(k,m)=d,$ then the congruence $$(1)\ kx≡l({\rm mod}\ m)$$ is soluble if and only if $d|l.$ It has then just d solutions. In particular, if $(k,m)=1,$ the congruence has always just one solution.

Here is one part of the proof:

If $d＞1,$ the congruence (1) is clearly insoluble unless $d|l.$ If $d|l,$ then $$m=dm',\ k=dk',\ l=dl',$$ and the congruence is equivalent to $$(2)\ k'x=l'({\rm mod}\ m').$$ Since $(k',m')=1,$ (2)has just one solution. If this solution is $$x≡t({\rm mod}\ m'),$$ then $$x=t+ym',$$ and the complete set of solutions of (1) is found by giving $y$ all values which lead to values of $t+ym'$ incongruent to modulus $m.$ Since $$t+ym'≡t+zm'({\rm mod}\ m) = m|m'(y-z) = d|(y-z)$$ there are just d solutions, represented by $$t,\ t+m',\ t+2m',…,\ t+(d-1)m'$$ This proves the theorem.

I don't know why the complete set of solutions of (1) is given by all $y$ values lead to values of $t+ym'$ incongruent to modulus m and why does $d|(y-z)$ shows there are just $d$ solutions.

HeroZhang001
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John He
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3 Answers3

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$x\equiv t\bmod{m'}$ is, by hypothesis, a solution.

It follows that for all $y$, $x=t+ym'$ is a solution. But it may be that two different values of $y$ give the same solution, modulo $m$ – which, since we are solving a congruence to the modulus $m$, means those two values of $y$ give the same solution. So we only want values of $y$ that give solutions that are incongruent, modulo $m$. The condition for two values of $y$, call them $y$ and $z$, to give the same solution modulo $m$ is obtained in your second-last display; it turns out to be $d\mid(y-z)$. So we get solutions incongruent modulo $m$ by taking $y=0,1,\dots,d-1$ (you seem to have written $d-l$ where what's wanted is $d-1$), since no two of these $d$ values of $y$ differ by a multiple of $d$, but any other value of $y$ will differ from one of these values of $y$ by a multiple of $d$, so must be excluded. So, there are $d$ values of $y$, hence, $d$ solutions.

Gerry Myerson
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1

Since $\frac {k}{d}$ and $\frac {m}{d}$ belong to $\Bbb Z,$ the necessity of $d|l$ for a solution $x\in \Bbb Z$ of the congruence to exist can be shown by $$kx\equiv l \pmod m\implies \frac {kx-l}{m}\in \Bbb Z\implies$$ $$\implies \frac {kx-l}{m}\cdot\frac {m}{d}\in \Bbb Z\implies$$ $$\implies \frac {kx-l}{d}\in \Bbb Z\implies$$ $$\implies \frac {k}{d}\cdot x-\frac {l}{d}\in \Bbb Z\implies$$ $$\implies -\frac {l}{d}\in \Bbb Z\implies d|l. $$

DanielWainfleet
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The essence is simple: $\ \overbrace{ym'\bmod dm'}^{\rm LHS} =\, \overbrace{(\color{#c00}{y\bmod d})\, m'}^{\rm RHS}\ $ by the mod Distributive Law. The $\small\rm RHS$ has $\:\!\color{#c00}d\:\!$ values $\!\bmod dm',\,$ i.e. $\,\color{#c00}0m',\, \color{#c00}1m',\, \color{#c00}2m', \ldots, (\color{#c00}{d\!-\!1})m',\, $ so ditto for $\,t+\small{\rm LHS} = x$.

So, in summary, the argument is as follows, using $\color{darkorange}{1/k'}\bmod m'$ exists by $\gcd(k',m')=1$

$$\begin{align} kx&\equiv l\:\!\pmod{\!m}\\[.2em] \iff k'x&\equiv l'\!\pmod{\!m'}\ \ \text{by cancelling $\,d\,$ from above}\\[.2em] \iff\ \ \ x&\equiv t\pmod{\!m'},\ {\rm for}\,\ t\equiv l'/\color{darkorange}{k'}\!\!\!\pmod{\!m'}\\[.2em] \iff\ \ \ x&\equiv t\!+\!ym'\!\!\!\pmod{\!dm'},\ 0\le y < d,\,\ \text{by proof above} \end{align}\qquad\qquad$$

Bill Dubuque
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