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 $n$?
+26387 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 \(n\)?
Answer here: https://web2.0calc.com/questions/i-do-not-like-this-question
+287 For brevity, all congruences below are taken modulo $n$. $(ab)^{-1}\equiv 2$ means $2(ab)\equiv1$. So we have $(2a)b\equiv a(2b)\equiv1$, meaning both $a$ and $b$ are invertible with $a^{-1}= 2b\bmod n$ and $b^{-1}=2a\bmod n$. Thus,
\begin{eqnarray*}
(a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\
&\equiv& ((a+b)^{-1}(a+b))2 \\
&\equiv& 1\cdot 2 \\
&\equiv& 2.
\end{eqnarray*}
+26387 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
\((a+b)^{-1} \left(a^{-1}+b^{-1} \right)\) is divided by \(n\)?
\(\begin{array}{|lrclcc|} \hline & && & \quad I. & \quad II. \\ & (ab)^{-1}=\dfrac{1}{ab} &\equiv& 2 \pmod{n} &| *a &| *b \\\\ \hline I. & \dfrac{a}{ab}&\equiv& 2a \pmod{n} \\ & \dfrac{1}{b} &\equiv& 2a \pmod{n} \\ & \mathbf{b^{-1}} &\equiv& \mathbf{2a \pmod{n}} \\\\ II. & \dfrac{b}{ab}&\equiv& 2b \pmod{n} \\ & \dfrac{1}{a} &\equiv& 2b \pmod{n} \\ & \mathbf{a^{-1}} &\equiv& \mathbf{2b \pmod{n}} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\ &\equiv& 2(a+b)^{-1}(b+a) \\ &\equiv& 2*\left(\dfrac{a+b}{a+b}\right) \\ &\equiv& 2*1 \\ \mathbf{ (a+b)^{-1}(a^{-1}+b^{-1}) } &\equiv& \mathbf{ 2 \pmod{n} }\\ \hline \end{array}\)
