- #1

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## Homework Statement

Prove that

[tex] \frac{1}{p} c(p,n) = (-1)^{n-1}/n (mod p) [/tex]

I expanded that combination in every way I could think and I tried to use Wilson's Theorem and I couldn't get :(

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- Thread starter ehrenfest
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- #1

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Prove that

[tex] \frac{1}{p} c(p,n) = (-1)^{n-1}/n (mod p) [/tex]

I expanded that combination in every way I could think and I tried to use Wilson's Theorem and I couldn't get :(

- #2

NateTG

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Try writing the LHS out as a fraction with the stuff in the numerator as negative representatives. It should nicely cancel to give the result.

- #3

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What do you mean "negative representatives"?

- #4

Gokul43201

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-(p-1)/2, -(p-2)/2,..., -1 for odd p

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- #5

matt grime

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What is 1/n, or -1/n mod p supposed to mean?

- #6

NateTG

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What is 1/n, or -1/n mod p supposed to mean?

Usually those would be the multiplicative inverses of n and -n respectively.

- #7

matt grime

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Usually? I beg to differ. Writing 1/n would indicate that the OP hasn't grasped what's going on. As would the fact there is an equals sign. I can't think of anyone who writes 1/2 mod 3 and not -1 0 it is incredibly bad notation. There is a difference from what I infer and what the OP ought to have written.

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- #8

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I just realized that my book my book defines congruence as

[tex] x \equiv y \mod p [/tex]

when x-y is a rational number whose numerator, in reduced form, is divisible by p.

So, it is like a generalized congruence or something...

Are there different rules for these generalized congruences?

I am not sure why what Gokul43201 wrote cancels nicely?

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