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Inspired by Bill Dubuque's Gauss Algorithm I have been interested in finding such optimal ways of solving simple linear congruences and finding multiplicative inverses.

Bill presented the algorithm for finding a multiplicative inverse modulo $p$ prime. The context is present in the linked answer. I am thinking why is this prime modulus required for such computations. The only requirement is that the number with the coefficient $x$ is coprime to the modulus. Let me give an example:

Suppose $p$ is not prime and we are trying to find a multiplicative inverse of $5$ modulo $8$

$$5x \equiv 1 \pmod 8 $$

What we can do very similarly to Bill's algorithm is multiply both sides by a number coprime to the modulus. Number must be coprime so that the resulting congruence will keep the coefficient of $x$ coprime to the modulus and congruence will have a solution.

We can multiply this by $5$, since $\gcd(5, 8) = 1$.

$$25x \equiv 5 \pmod 8$$

Now we can reduce $25$ modulo $8$ because of the congruence property like so:

$$x \equiv 5 \pmod 8$$

And we have our inverse $x \equiv 5$.

This is a simple example, but I tried doing some other congruences and it worked. Important is to make sure that we multiply by a number coprime to modulus and that after reducing the resulting product modulo we get a lesser number than we started with.

I am new to number theory and this might just be something very obvious, but I wanted to present it here just in case. Can anyone provide some kind of feedback on this?

Bernard
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Michael Munta
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  • The answer you linked says that modular fraction arithmetic is valid for fractions with denominator coprime to the modulus. – J. W. Tanner Jul 23 '19 at 13:36
  • Oh yes, in fact it is the same now. I recall that he always referred to this only when modulus was prime. Like here https://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency/3230#3230 – Michael Munta Jul 23 '19 at 13:56
  • e.g. try computing $,7^{-1}\bmod 30\ $ by Gauss's algorithm to see the problems that arise. – Bill Dubuque Jul 23 '19 at 14:06
  • https://math.stackexchange.com/questions/257127/how-to-convert-a-diophantine-equation-into-parametric-form – lab bhattacharjee Jul 23 '19 at 14:35
  • @lab How is the link you gave supposed to be related to the specific algorithm being discussed here? It seems there is no relation at all (other than being a different method to compute modular inverses). – Bill Dubuque Jul 23 '19 at 16:52
  • @Bill well it is still possible, it just makes it not as efficient. Multiply by $17$, reduce, multiply by $-31$, reduce. Inverse is $-3689 \equiv 13$ (mod $30$) – Michael Munta Jul 23 '19 at 19:09
  • Since an inverse exists it is always possible to find it (e.g. by brute force search). But the point is that the efficient descent process used in Gauss's algorithm doesn't work in general for composite moduli. Instead we can employ more general efficient algorithms, e.g. the fractional extended Euclidean algorithm, but that's a bit more complex (using multi-valued modular fractions). If you are first learning these ideas then it's better to work with the equations vs. such generalized fractions (compare the two in the prior-linked post) – Bill Dubuque Jul 23 '19 at 19:17

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