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 ). She tries the example $a=2$, $b=3$ and $m=17$. Let be the residue of $(2+3)^{-1}\pmod{17}$, and let $R$ be the residue of $2^{-1}+3^{-1}\pmod{17}$, where $L$ and $R$ are integers from $0$ to $16$ (inclusive). Find $L-R$
1 answer
Since $a \equiv a \cdot 1 \equiv a \cdot (b \cdot b^{-1}) \equiv (a \cdot b) \cdot b^{-1}$, we see that $\boxed{L = R}$. Therefore, the answer is zero.