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Everything I've read about modular arithmetic generally concerns doing things in some "mod m" world where "m" is some constant. But I'm perplexed how to tackle modular arithmetic problems where the aim is to find the modulus which produces some result.

More specifically: "Given positive integers $a,b,c$, how can I go about efficiently finding $x$ such that $ax \equiv 1 \pmod{bx+c}$?"

For example, with $a=2021$, $b=89$, $c=1932$, I'm looking for $x$ such that $2021x \equiv 1 \pmod{89x+1932}$. By brute force search, I happen to know that this is solved by $x=83$, but surely there's a better way...

For what it's worth, in the problems I'm concerned with $c=a-b$ (as hinted by the example numbers given above), but I'm not sure how useful that is if there's a more general technique anyway.

Bill Dubuque
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timday
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2 Answers2

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Hint $ $ Completing the product (rectangle) yields $$\begin{align} ax - 1 &= (bx+c)y\\[.2em] \iff (bx+c)(a-by) &= ac+b\end{align}\qquad\qquad$$

Now find factors $\,d\,$ of $\,ac+b\,$ of form $\,bx+c,\,$ i.e. $\,d\equiv c\pmod{\!b}$.

OP has $\, ac+b = 419\cdot 9319 = $ product of primes, so only a few factors to test.

Bill Dubuque
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Hint: Since $a,b,c,x \in \Bbb{N}$ $$ax \equiv 1 \mod bx+c\Rightarrow ax=(bx+c)y+1\Rightarrow x=\frac{cy+1}{a-by}\in \Bbb{N} \Rightarrow a-by \gt 0 \\ \Rightarrow 0 \le y \le \lfloor\frac{a}{b}\rfloor$$ so you're searching for $y$ in a bounded set...With no further information about $a,b,c$ you have to make the computer search for $y$ in this set, as brute force would be the only possibility..

sirfoga
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    Such a brute force search will be very slow if $,a/b,$ is large. Then it will generally be much quicker to exploit the innate arithmetical structure and employ factorization - see my answer. – Bill Dubuque Apr 26 '14 at 14:23
  • @BillDubuque And if big prime factors are not involved?? Moreover if there are big numbers involved, it's not quite easy to be sure they're primes..first you have to check every number smaller than $\sqrt{n}$... – sirfoga Apr 26 '14 at 17:04
  • Generally solving a diophantine equation of this form is equivalent to factoring integers, as in my answer. So generally one cannot do any better than to employ a fast integer factorization algorithm (which generally will be much faster than a brute-force search). – Bill Dubuque Apr 26 '14 at 17:16
  • Maybe, but your method needs a composition of multiple algorithms (prime finder, diophantine equation solver ...) – sirfoga Apr 26 '14 at 17:20
  • Not maybe, but definitely. Brute-force (trial division) integer factorizations algorithms become impractical many orders of magnitude sooner than do the best known factorization algorithms. – Bill Dubuque Apr 26 '14 at 17:24