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Given that \(p\ge 7\) is a prime number, evaluate \(1^{-1} \cdot 2^{-1} + 2^{-1} \cdot 3^{-1} + 3^{-1} \cdot 4^{-1} + \cdots + (p-2)^{-1} \cdot (p-1)^{-1} \pmod{p}.\)

Guest Jul 8, 2017
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 #1
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\(1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}\)

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 :(

MaxWong  Jul 11, 2017
 #2
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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 Jul 11, 2017
edited by Guest  Jul 11, 2017
edited by Guest  Jul 11, 2017
 #3
avatar+6705 
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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

\(\dfrac{a}{b}\pmod c\\ \text{We first find numbers m and n that } bm = cn+1\\ =\dfrac{am}{bm}\pmod c\\ =\dfrac{am}{cn + 1}\pmod c\\ =am\pmod c\text{ (because (cn + 1) mod c = 1)}\\ \text{So we find the answer.}\)

MaxWong  Jul 12, 2017
 #4
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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

Guest Jul 12, 2017

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