Given that \(p\ge 7\) is a prime number, evaluate \(1^{-1} \cdot 2^{-1} + 2^{-1} \cdot 3^{-1} + 3^{-1} \cdot 4^{-1} + \cdots + (p-2)^{-1} \cdot (p-1)^{-1} \pmod{p}.\)
\(1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}\)
So our work is to evaluate
\(\dfrac{p-3}{p-2}\pmod p\).
Don't know how to do that, if I were you, I would try some primes.
So let's try it.
When p = 7,
\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{4}{5}\mod 7\\ =\dfrac{12}{15}\mod 7\\ =5\)
When p = 11,
\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{8}{9}\pmod {11}\\ =\dfrac{40}{45}\pmod {11}\\ =7\)
When p = 13,
\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{10}{11}\pmod {13}\\ =\dfrac{60}{66}\pmod {13}\\ =8\)
When p = 17,
\(\dfrac{14}{15}\pmod{17}\\ =\dfrac{112}{120}\pmod {17}\\ =10\)
Welp can't see a pattern :(
a mod b- the remainder of the division a/b.
(P-3)/(P-2) mod P- the remainder of the division ((P-3)/(P-2))/P. P>(P-3)/(P-2),
therefore ((P-3)/(P-2))/P<1, therefore (P-3)/(P-2) mod P=(P-3)/(P-2)-P*0=(P-3)/(P-2)
Guest,
We can still get a 'proper' answer with a fraction.
Example:
\(\dfrac{41}{7}\pmod{13}\\ \equiv\dfrac{2}{7}\pmod{13}\text{ because 41 mod 13 = 2}\\ \equiv\dfrac{4}{14}\pmod{13}\\ \equiv \dfrac{4}{1}\pmod {13}\text{ similarly, because 14 mod 13 = 1}\\ \equiv 4 \pmod {13}\\ \equiv 4\)
Or
\(\dfrac{a}{b}\pmod c\\ \text{We first find numbers m and n that } bm = cn+1\\ =\dfrac{am}{bm}\pmod c\\ =\dfrac{am}{cn + 1}\pmod c\\ =am\pmod c\text{ (because (cn + 1) mod c = 1)}\\ \text{So we find the answer.}\)