Jeremy Ting edited p7.tex  about 10 years ago

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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}