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

 
 Jun 11, 2024
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We can rewrite this as \(\frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + ... + \frac{1}{(p-2)\times (p-1)}\). All of these terms are in the form of \(\frac{1}{n(n+1)}\), so let's first try to simplify that.

\(\frac{1}{n(n+1)}\)

\(\frac{1+n-n}{n(n+1)}\)

\(\frac{1+n}{n(n+1)} - \frac{n}{n(n+1)}\)

\(\frac{1}{n} - \frac{1}{n+1}\)

 

Now, we can rewrite all of out fractions as \((\frac{1}{1} - \frac{1}{2})+(\frac{1}{2} - \frac{1}{3})+(\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{p-2} - \frac{1}{p-1})\)

We can finally simplify this to \(1 - \frac{1}{p-1}\) or \(\mathbf{\frac{p-2}{p-1}}\), which is the final answer.

 

 

You could also solve this by induction, because you could notice that \(\frac{1}{2}+\frac{1}{2\times3}=2/3\) and that  \(\frac{1}{1\times2}+\frac{1}{2\times3} + \frac{1}{3\times4}=3/4\).

 Jun 11, 2024

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