Given that ​ is a prime number, evaluate

$1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}$

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)

Hi Max

Your partial fractions, line 3, are the wrong way round, should be  $\displaystyle \frac{1}{p-2}-\frac{1}{p-1}$, it works out nicely after that.

Tiggsy