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Here on this pdf on page no $5$ it is stated that

Let $g = (a, m)$. Then there is a solution to $ax \equiv b \mod m$ if and only if $g|b$. If it has solutions, then it has exactly $g$ solutions $\mod m$.

I do not understand the proof that it has exactly $g$ solutions. How we conclude the second last and last sentence on the pdf?

Mula Ko Saag
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1 Answers1

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If $x_0$ is one solution, then $x_1 = x_0 + m/g$ is another. So is $x_2 = x_0 + 2m/g$, and so on. Modulo $m$ there are exactly $g$ different congruence classes hit by the different $x_n$. Therefore there are at least $g$ solutions.

As for why there cannot be more than $g$ solutions, try solving $ay \equiv 0 \pmod m$ first. Then notice that any two solutions to $ax \equiv b$ must have a difference equal to one of those $y$.

This question is more or less identical. See if you can understand any of that.

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Arthur
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