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Having some trouble working my way back up the Extended Euclidean Algorithm. I'm trying to find the multiplicative inverse of $497^{-1} (mod 899)$. So I started working my way down first finding the gcd:

\begin{align} 899&=497\cdot1 + 402\\ 497&=402\cdot1 + 95\\ 402&=95\cdot4 + 22\\ 95&=22\cdot4 + 7\\ 22&=7\cdot3 + 1 \end{align} Now I work my way back up using the extended algorithm and substituting: \begin{align} 1&=22-(7\cdot3)\\ 1&=22-(95-(22\cdot4))\cdot3\\ 1&=22-(95-(402-(95\cdot4)\cdot4))\cdot3\\ 1&=22-(95-(402-((497-402)\cdot4)\cdot4))\cdot3\\ 1&=22-(95-(402-((497-(899-497))\cdot4)\cdot4))\cdot3\\ \end{align}

Am I going about this right? Do I just keep substituting up the chain? It gets difficult to follow for me. And Here's what the terms equal going up:

\begin{align} 7&=95-(22\cdot4)\\ 22&=402- ( 95\cdot4)\\ 95&=497- 402\\ 402&=899- 497\\ \end{align}

pfinferno
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1 Answers1

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For an (iterative) implementation it is easier to compute the inverse resp. the Bezout coefficients while going down.

You start with $0·497 \equiv r_0=899\mod899$ and $1·497 \equiv r_1=497\mod899$ and apply the same sequence of computations as to the remainder to the quotient sequence starting with $u_0=0, u_1=1$. \begin{align} r_2&=r_0-1·r_1&\implies u_2&=u_0-1·u_1=-1 &:&&-1·497 &\equiv r_2=402\mod899 \\ r_3&=r_1-1·r_2&\implies u_3&=u_1-1·u_2=2 &:&&2·497 &\equiv r_3=95\mod899 \\ r_4&=r_2-4·r_3&\implies u_4&=u_2-4·u_3=-9 &:&&-9·497 &\equiv r_4=22\mod899 \\ r_5&=r_3-4·r_4&\implies u_5&=u_3-4·u_4=38 &:&&38·497 &\equiv r_5=7\mod899 \\ r_6&=r_4-4·r_5&\implies u_6&=u_4-3·u_5=-123 &:&&-123·497 &\equiv r_6=1\mod899 \end{align} Thus the inverse is $-123$ or in the same equivalence class $899-123=776$

Lutz Lehmann
  • 126,666