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Let $n$ be a positive integer greater than or equal to $3$. Let $a,b$ be integers such that $ab$ is invertible modulo $n$ and $(ab)^{-1} \equiv 2$ (mod $n$). Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by $n$?

 Jun 16, 2021
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Multiply \({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\) by ab

 

\({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\)(ab) =\({(a+b)}^{-1}({a}^{-1}(ab)+{b}^{-1}(ab))\)

 

                                             \(={(a+b)}^{-1}(b+a)\)

 

                                            = 1 (mod n)

 

This shows that \({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\) = \({(ab)}^{-1}\)= 2 (mod n)

 Jun 17, 2021

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