question

Let n be a positive integer greater than or equal to 3.
Let a,b be integers such that 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?

\(\begin{array}{|rcll|} \hline (ab)^{-1} &\equiv& 2\pmod n \\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot a \\ \mathbf{b^{-1}} &\equiv&\mathbf{ 2a\pmod n} \\\\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot b \\ \mathbf{a^{-1}} &\equiv& \mathbf{2b\pmod n} \\\\ (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& x \pmod n \\ (a+b)^{-1}(2b+2a) &\equiv& x \pmod n \\ 2(a+b)^{-1}(ab) &\equiv& x \pmod n \\ 2 &\equiv& x \pmod n \\ x &\equiv& 2 \pmod n \\ \hline \end{array}\)