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 $L-R$
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.
To find $L$, the residue of $(2+3)^{-1} \pmod{17}$, we first need to find the inverse of $2+3$ modulo $17$.
To do this, we need to find a number $x$ such that $(2+3)x \equiv 1 \pmod{17}$.
We can simplify $(2+3)x \equiv 1 \pmod{17}$ to $5x \equiv 1 \pmod{17}$.
Now we need to find a number $x$ such that $5x \equiv 1 \pmod{17}$.
We can observe that $5 \cdot 7 \equiv 1 \pmod{17}$.
Therefore, $x \equiv 7 \pmod{17}$.
This means that $L = 7$.
Now, let's find $R$, the residue of $2^{-1}+3^{-1} \pmod{17}$.
Using the same process, we need to find $x$ such that $2x \equiv 1 \pmod{17}$.
We observe that $2 \cdot 9 \equiv 1 \pmod{17}$.
Therefore, $x \equiv 9 \pmod{17}$.
Similarly, we need to find $y$ such that $3y \equiv 1 \pmod{17}$.
We observe that $3 \cdot 6 \equiv 1 \pmod{17}$.
Therefore, $y \equiv 6 \pmod{17}$.
Thus, $2^{-1}+3^{-1} \equiv 9+6 \equiv 15 \pmod{17}$.
This means that $R = 15$.
Finally, we can find $L-R$:
$L-R = 7 - 15 = -8$.
Therefore, $L-R = \boxed{-8}$.