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Math Help - Inverse product of residues of a prime

  1. #1
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    Inverse product of residues of a prime

    Hi

    I'm trying to prove that \prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv -1 \ (mod \ p)

    The way I see it, the inverse of 1 is itself, and from 2 to p-2, the inverses are paired up, so we are left with the inverse of p-1 which is itself.

    So, we have that \prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv p-1 \equiv -1 \ (mod \ p)

    My problem (with this proof) is in showing that the inverses are all paired up from 2 to p-2.
    a) Is there a way to prove/show this?

    or
    b) Is there another way to prove \prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv -1 \ (mod \ p)?
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  2. #2
    Super Member PaulRS's Avatar
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    That is Wilson's Theorem actually. - see here -

    To show the inverses are 'paired' show that f:\left\{ {2,...,p - 2} \right\} \to \left\{ {2,...,p - 2} \right\} defined by: <br />
f\left( x \right) = \left( {x^{ - 1} \bmod .p} \right)<br />
is injective and that f(x)=x can't happen. Then for each number 2 ... p-2 you will have a unique inverse there - different from itself-, which will be the inverse of that number and no other.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Here is one other possible proof. (\mathbb{Z}/p\mathbb{Z})^\times is cyclic of order p-1; let \alpha be a generator. Then

    \prod_{j=1}^{p-1}j^{-1}=\alpha^{-(1+2+...+(p-1))}=\alpha^{-p(p-1)/2}=\alpha^{-(p-1)/2}=-1

    because \alpha^{\pm (p-1)/2} is just the Legendre symbol, and a primitive root is obviously not a square (or else all elements of \mathbb{Z}/p\mathbb{Z}) would be squares).
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  4. #4
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    Yes, it is indeed Wilson's theorem, written a little differently . I didn't see that before, hehehe, my bad

    \prod_{i=1}^{p-1}\displaystyle\frac{1}{i} \equiv \prod_{j=1}^{p-1}j \equiv (p-1)! \equiv -1 \ (mod \ p)

    Hehehe ... I'm learning lots of interesting things, and I'm just in awe of how these things work out , like how there's different ways of showing the same thing
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