Using the Euclidean Algorithm how to find the inverse of 41 in Z(131)

Question

I need to find the inverse of 41 in the integers of Z131 and am confused as to how to go about it.

Do I use the Euclidean Algorithm as 41 mod 131?

Answer 1

To find an inverse we need to find a solution to the following equation (since if you take $\pmod{131}$ this equation will give you $41X \equiv 1$ for modulo $131$):

$$41X + 131Y = 1$$

We will find some pair $(X, Y)$ that satisfies this linear Diophantine equation by extended euclidean algorithm. First, we calculate the GCD of the pair $(131, 41)$.

$$(131, 41) = (41, 8) = (8, 1) = 1$$

This means we have a solution since GCD divides the result of Diophantine equation. Now we go the other way around. We have:

$$131 = 3 \cdot 41 + 8,$$ $$41 = 5 \cdot 8 + 1,$$ $$8 = 3 \cdot 1 + 0.$$

We will rewrite them in the following form:

$$131 - 3 \cdot 41 = 8,$$ $$41 - 5 \cdot 8 = 1.$$

We are going to substitute $131 - 3 \cdot 41$ for $8$:

$$41 - 5 \cdot (131 - 3 \cdot 41) = 1$$ $$16 \cdot 41 - 5 \cdot 131 = 1$$

Hence our pair of solution is $(16, -5) = (x, y)$. Yet don't forget there are infinitely many other solutions in this case which you can derive by one initial solution. But because they are all in the congruence class $[16]_{131}$ (AKA the set of numbers which leave the remainder $16$ when divided by $131$), they aren't different at all for the modular arithmetic.

$X \equiv 16 \pmod{131}$ will be our inverse.

Answer 2

In this case, the extended Euclidean algorithm is particularly fast:

\begin{array}{rrrr} r_i&u_i&v_i&q_i\\ \hline 131 & 0 & 1\\ 41 & 1 & 0 & 3\\ \hline 8 & -3 & 1 & 5\\ 1 & 16 & -5\\ \hline \end{array}

so a Bézout's relation is $$16\cdot 41-5\cdot131 = 1$$ and $\;41^{-1}\equiv 16\bmod 131$.

Answer 3

$131 = 3*41 +8$ so $8 = 131 - 3*41$.

$41 = 5*8 + 1$ so $1 = 41- 5*8 = 41 - 5(131 - 3*41)= 16*41 - 5*131$

So $16*41 - 5*131 =1$.

Or $16*41 = 1 + 5*131$

Or $16*41 \equiv 1 \pmod {131}$.

So $16$ is the multiplicative inverse $\mod 131$ of $41$.

Answer 4

Below is a calculateion of $\ \color{#0a0}{x \equiv 41^{-1}}\pmod{\!131}\,$ by the forward extended Euclidean Algorithm.

$\ \ \ \ \begin{array}{rr} [\![1]\!] &131\, x\equiv\, 0 \\ [\![2]\!] & \color{#0a0}{41\,x \equiv\, 1}\\ [\![1]\!]-3\,[\![2]\!] \rightarrow [\![3]\!] & 8\,x \equiv -3\!\!\! \\ [\![2]\!]-5\,[\![3]\!] \rightarrow [\![4]\!] & \bbox[5px,border:1px solid #c00]{x \equiv 16}\!\!\! \end{array}$

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Or $\bmod 131\!:\,\ \dfrac{1}{41}\equiv \dfrac{3}{123}\equiv \dfrac{3}{-8}\equiv \dfrac{-128}{-8}\equiv\, \bbox[5px,border:1px solid #c00]{16}\ $ by Gauss's algorithm

Beware $ $ Modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus. See here for further discussion.

Answer 5

I use scaling method. Since 8*41 > 131, go straight for final inverse.

1
8
41 → $(-1)^2\lfloor 1/8*131 \rfloor = 16 \equiv 41^{-1} \text{ (mod 131)} $
131

Example, it takes only 4 scalings so to calculate:

1/1223334444 ≡ 3171632349 (mod 9988776655)