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I have found two different methods for finding the multiplicative inverse of a number modulo $n$ in $\mathbb{Z}/n\mathbb{Z}$.

One is using the Extended Euclidean Algorithm, which produces a different result to the much faster approach below.

Say I am trying to find the multiplicative inverse of the value $5$ in $\mathbb{Z}/8\mathbb{Z}$.

$$\begin{split} 5x &= 1+8n\\ x &= \frac{1 + 8n}{5}\\ \end{split}$$

Then find the smallest value of $n$ that makes $x$ an integer. In this case, $n=3$ which produces an inverse value of $5$.

Is this a correct method?

Christian E. Ramirez
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Benjamin McDowell
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  • You should get the same result if you do it right, since there's only one inverse. You can always check for correctness: $x \cdot x^{-1} \equiv 1 \mod n$. – Robert Israel Jan 24 '23 at 21:13
  • How is this less work-intensive that multiplying $5$ by $1$, $2$, $3$, etc until you get one that is one more than a multiple of $8$? You are multiplying $8$ by $1$, $2$, $3$, etc. until you hit one that is one less than a multiple of $5$. Granted, detecting multiples of $5$ is straightforward, but how much work would you have to do to find the inverse of $17$ modulo $215$? – Arturo Magidin Jan 24 '23 at 21:16
  • But 1 mod any number is always 1...@RobertIsrael – Benjamin McDowell Jan 24 '23 at 21:35
  • Yes, and we can finish via $\bmod 5!:\ \color{#c00}n\equiv \dfrac{1}{-8}\equiv \dfrac{6}2\equiv \color{#c00}3,,$ hence $\bmod 8!:\ \dfrac{1}5\equiv \dfrac{1+8(\color{#c00}3)}5\equiv 5.\ $ This is the inverse reciprocity method in the linked dupe (essentially an optimization of the Extended Euclidean Algorithm when it terminates in two steps). See there and its links for many more methods. $\ \ $ – Bill Dubuque Jan 24 '23 at 21:48
  • @Arturo It ends up being less work since it reduces the inversion modulus (from $\color{#0a0}8$ to $\color{#c00}5),,$ i.e. we reduce from computing $,\color{#c00}5^{-1}\pmod{!\color{#0a0} 8},$ to $,\color{#0a0}8^{-1}\pmod{!\color{#c00}{!5}},,$ hence the name $\rm inverse\ \color{#0a0}{recip}\color{#c00}{rocity}$. – Bill Dubuque Jan 24 '23 at 22:18
  • That's not a huge optimization in this instance - but it is when the modulus is much larger than $\color{#0a0} 8,,$ e.g. to compute the inverse of $5$ mod $n = 1000000001$ we reduce to computing the inverse of $n\equiv 1\pmod{!5}\ \ $ – Bill Dubuque Jan 25 '23 at 00:08

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Yes, that works. However, for large values of $8$ it will be MUCH slower than the Euclidean algorithm.

Igor Rivin
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