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It's easy to show that if the divisor has a modular multiplicative inverse under m then dividing by it would be possible. I'd simply multiply both sides of the antecedent below by $b$ and let $x=yb$ be the witness.

$$there\:is\:a\:unique\:y\:such\: that\: ay=1\:\left(mod\:m\right)\rightarrow there\:is\:a\:unique\:x\:such\: that\: ax=b\:\left(mod\:m\right) where\:b\in \left\{0,1,\:...,\:m-1\right\} $$However, it doesn't seem as easy when I try to go the other way. Any suggestions?

Bill Dubuque
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Malzahar
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    The opposite direction is not true. As explained in the linked dupe (and its links, esp. here) we have $ax\equiv b\pmod{m}$ is solvable $\iff d:=\gcd(a,m)\mid b;,$ if so there are $,d,$ solutions $\bmod m\ \ $ – Bill Dubuque Oct 28 '21 at 05:28
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    To make it true you can change it to $,ax\equiv b\pmod{m},$ has a unique solution, i.e. the modular fraction $b/a$ uniquely exists. The prior linked post explains this at length from a fractional viewpoint. – Bill Dubuque Oct 28 '21 at 05:35
  • So both ways are true now? – Malzahar Oct 28 '21 at 05:56
  • Yes, after incorporating the above fix it is now true, since, like the first, the second congruence has a unique root $\iff d := \gcd(a,m) = 1\ \ $ – Bill Dubuque Oct 28 '21 at 06:02
  • See also here for an introduction to modular fractions. – Bill Dubuque Oct 28 '21 at 06:06
  • I can't see that the answers in any of the links you provided exactly answer my question. I'm trying to prove that if we can perform modular division for any $b$ in ${0,1,..,m-1}$ with $a$ being the divisor under $m$ then the modular multiplicative inverse of $a$ under $m$ must exist (and vice versa.) – Malzahar Oct 28 '21 at 06:57
  • They do - see my prior comment. – Bill Dubuque Oct 28 '21 at 13:39
  • Why do I have a to understand a different question and read its lengthy answer just so I can stumble upon an answer to my straightforward question? – Malzahar Oct 28 '21 at 14:07
  • You only need to read paragraphs 2 & 3 in said answer, which shows that when $d>1$ the solution is not unique (or does not exist). – Bill Dubuque Oct 28 '21 at 14:37
  • If you seek something more direct then note that if $d>1$ and $,x_0,$ is a solution of $,ax\equiv b\pmod m,$ then $,x_1 := x_0 + m/d,$ is a different solution $x_1\not\equiv x_0\pmod m,,$ because $,m/d \not \equiv 0\pmod m$ are different solutions of the associated homogeneous equation $,ax\equiv 0\pmod m,$ as explained here. – Bill Dubuque Oct 28 '21 at 14:53

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