I've been stuck on this one for a while, I just don't know what to do with all of this info :(

Given \(m\geq 2\), denote by \(b^{-1}\) the inverse of \(b\pmod{m}\). That is, \(b^{-1}\) is the residue for which \(bb^{-1}\equiv 1\pmod{m}\). Sadie wonders if \((a+b)^{-1}\) is always congruent to \(a^{-1}+b^{-1}\) (modulo m). She tries the example a=2, b=3, and m=7. Let L be the residue of \((2+3)^{-1}\pmod{7}\), and let R be the residue of \(2^{-1}+3^{-1}\pmod{7}\), where L and R are integers from 0 to 6 (inclusive). Find L-R.

Guest Dec 8, 2019