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Jeremy Ting edited p7.tex
about 10 years ago
Commit id: cc544606dcc9a3d206a2e422621c885b3e8b3d0b
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\section{Problem 7}
\begin{itemize}
\item
If a number is it's own inverse modulo $x*x^{-1} = 1modn$, then we know that the number is itself:
$x^2 = 1modn$
$x^2 = 1modn$ \Rightarrow \frac{x^2-a}{n} \Rightarrow $n | (x-1)(x+1$
So (x-1) = np or (x+1) = np; \Rightarrow x = np\pm1
\end{itemize}