For which positive integer values of $n$ does $a|mn-b?$

Question

I am taking a number theory class, and I was not able to solve this homework problem.

Let $a,$ $b,$ and $m$ be constant positive integers. For which positive integer values of $n$ does $a|mn-b?$

I tried this out for a couple of values, and the values of $n$ satisfying this expression seem to occur in a cycle. (For example, let $m=2019,$ $a=20,$ $b=19.$ The values of $n$ that work are $1,21,41,61,....$) However, I am not able to show that this always happens.

Welcome to Math Stack Exchange. Cf. Bezout's identity $mn+ka=b$ — J. W. Tanner, Nov 10 '19 at 16:53
let S be the set of integer values n such that : $a|mn-b$ and let D be the following set : $$D={0\leq r\leq a-1 ; / a|mr-b }$$ you can prove that : $$S={aq+r ; / q\in\mathbb{N} ; r\in D}$$ — Laassila souhayl, Nov 10 '19 at 17:12

Bill Dubuque · Accepted Answer · 2021-01-17T19:44:57.690

1

Let $\,d = \gcd(a,m).\,$ $\ d\mid m\ $ & $\ d\mid a\mid nm\!-\!b\,\Rightarrow\,\color{#c00}{d\mid b}\,$ so $\:\!\color{#90f}{{\rm cancelling}\ d}\,$ from the divisibility

$$ a\mid nm-b\!\!\overset{\large\color{#90f}{\div\ d_{\phantom{.}}}\!\!}\iff a/d\mid n(m/d) - b/d\iff n\equiv \dfrac{\color{#c00}{b/d}}{\color{#0a0}{m/d}}\!\!\!\pmod{a/d}\qquad$$

i.e. to evaluate a (multivalued) modular fraction $\, n\equiv b/m\pmod{\!a},\,$ cancel $\,d = \gcd(a,m)\,$ everywhere, i.e. from the top $\,b\,$ & bottom $\,m\,$ & modulus $\,a\,$ (if $\,\color{#c00}{d\nmid b}\,$ then it does not exist, by above). Notice that $\,\gcd(a/d,m/d) = 1\,$ so $\,\color{#0a0}{(m/d)^{-1}}$ exists $\!\pmod{\!a/d}\ $ (e.g. by Bezout).

See here for much further discussion (including how to use such multivalued modular fractions in the extended Euclidean algorithm)

edited Jan 17 '21 at 19:44

answered Nov 10 '19 at 18:42

Bill Dubuque

272,048

why does $d|b?$ – Naleba Nov 10 '19 at 19:51
@Naleba $\ d,$ divides $mn$ and $,mn-b,$ so also their difference $= b\ \ $ – Bill Dubuque Nov 10 '19 at 19:57
So, the final result was that $n\equiv b/m \pmod a?$ – Naleba Nov 10 '19 at 20:06
See the displayed congruence. The final result is the RHS. There are examples in the linked answer (and questions linked to that). – Bill Dubuque Nov 10 '19 at 20:10
The linked answer states that $x \equiv ad/bd \pmod{md}$ is the same as $x\equiv a/b \pmod{m},$ which would mean in this case that $n \equiv b/m \pmod{a},$ right? – Naleba Nov 10 '19 at 20:24
@Naleba Yes, if we interpret that "fractional" notation in the multivalued manner I describe there. – Bill Dubuque Nov 10 '19 at 20:34

For which positive integer values of $n$ does $a|mn-b?$

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