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Given the equation: $42x\equiv 1\pmod 5$ I have determined the class $[-2]_{5}$ as $x$ solution of the given equation. Now I have to find the inverse of $x$ (i.e. $x^{-1}=[-2]_5^{-1}$ ). As far as I know, the $x=[-2]_5$ first found is just $[42]_5^{-1}$, and so the inverse of inverse (i.e. $([42]_5^{-1})^{-1}$) is just $42$ again, so $[42]_5=[-2]_5=[x]_5^{-1}$

Am I wrong=

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    @MichaelHardy: Noted!2012-03-08

3 Answers 3

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You wrong in saying that $[42]_5=[-2]_5$.

You found $x$ correctly, since $[42]_5[-2]_5=[2]_5[3]_5=[6]_5=[1]_5$ , although since we are working mod 5 it is traditional to choose the representative between $0$ and $4$ so instead of $[-2]_5$ you might write $[3]_5$. Thus $[42]_5$ (i.e. $[2]_5$) and $[3]_5$ are inverses of each other in $\mathbb{Z}/5$

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Write down the powers of $2$ modulo $5$. They are $2^0=1$, $2^1=2$, $2^2=4\equiv-1$, $2^3=8\equiv3$, and $2^4=16\equiv1$. Now all is clear.

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Using Gauss Method and taking modulo 5: 1/42 = 1/2 = 3/6 = 3