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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 $(ab)^{-1}\equiv 2\pmod n$. Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by n?

 Jan 25, 2021
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This is my take:

 

\((ab)^{-1}=2\mod n\\ \text{What does this even mean, well}\\ \text{the inverse of 7 is 1/7 because } 7*(1/7)=1\\ so\\ (ab)^{-1}*(ab)=1\mod n\\ 2ab=1\mod n\\~\\ (a+b)^{-1}(a^{-1}+b^{-1})\\ =\frac{1}{a+b}*(\frac{b+a}{ab})\\ =\frac{1}{ab}\\ \equiv 2\mod n \)

 

 

Coding:

(ab)^{-1}=2\mod n\\
\text{What does this even mean, well}\\
\text{the inverse of 7 is  1/7  because } 7*(1/7)=1\\
so\\
(ab)^{-1}*(ab)=1\mod n\\
2ab=1\mod n\\~\\

(a+b)^{-1}(a^{-1}+b^{-1})\\
=\frac{1}{a+b}*(\frac{b+a}{ab})\\
=\frac{1}{ab}\\
\equiv 2\mod n

 Jan 25, 2021

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