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Math Help - Prove that if p is an odd prime, 1^2 * 3^2 ...(p-2)^2 = (-1)^(p+1)/2 (mod p)

  1. #1
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    Prove that if p is an odd prime, 1^2 * 3^2 ...(p-2)^2 = (-1)^(p+1)/2 (mod p)

    If p is an odd prime, prove that 1^2 * 3^2 * 5^2 * ...(p-2)^2 = (-1)^(p+1)/2 (mod p)
    and
    2^2 * 4^2 * 6^2 * ... (p-1)^2 = (-1)^(p+1)/2 (mod p)

    I know that (p-1)! = -1 mod p using Wilson's theorem, but then I am stuck. Help?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    For the first one : first, notice that 1^2,3^2,...,(p-2)^2 are all the squares in \mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}. The squares of \mathbb{F}_p, by Euler's criterion, are the roots of the polynomial f(x)=x^{(p-1)/2}-1. Therefore, since \mathbb{F}_p[x] is a unique factorization domain, we can factor this polynomial as :

    x^{(p-1)/2}-1 = (x-1^2)(x-3^2)...(x-(p-2)^2)

    Equating the constant terms, you get the theorem.

    The second one is exactly the same.
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  3. #3
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    Very interesting information. The problem is that I haven't learned Euler's criterion in class yet, so I don't really understand your solution. I do know how to manipulate mods, use Wilson's Theorem, and Euler's generalization of Fermat's little theorem. Is there another way using only basic modular arithmetic and congruence to prove that 1^2 * 3^2 * 5^2 * ...(p-2)^2 = (-1)^(p+1)/2 (mod p) and 2^2 * 4^2 * 6^2 * ... (p-1)^2 = (-1)^(p+1)/2 (mod p) are both true? Thanks to all.
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  4. #4
    Super Member PaulRS's Avatar
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    Also, you can take a primitive root module p, call it g.

    Since you have all the quadratic residues (*) there -in both cases-, and they are congruent to: g^2,g^4, ..., g^{p-1} we find that both products are congruent to g^2\cdot g^4 \cdot ...\cdot g^{p-1}=g^{2\cdot \sum_{k=1}^{\frac{p-1}{2}}{k}}=g^{\tfrac{p^2-1}{4}}

    But g^{\tfrac{p-1}{2}}\equiv{-1}(\bmod.p) -since g^{p-1}\equiv{1}(\bmod.p) and it's primitive- and so we find that the products are congruent to: (g^{\tfrac{p-1}{2}})^{\tfrac{p+1}{2}}\equiv{(-1)^{\tfrac{p+1}{2}}}(\bmod.p)

    (*) x is a quadratic residue module p if and only if there exist an integer y such that y^2\equiv{x}(\bmod.p)

    EDIT: What you have there is congruent to 1^2\cdot 2^2\cdot ... \cdot \left(\tfrac{p-1}{2}\right)^2 now note that 1\equiv -(p-1)(\bmod.p) ... \left(\tfrac{p-1}{2}\right)\equiv -(p-\left(\tfrac{p-1}{2}\right))(\bmod.p) but then we have: 1^2\cdot 2^2\cdot ... \cdot \left(\tfrac{p-1}{2}\right)^2\equiv 1\cdot(-(p-1))\cdot ... \cdot \left(\tfrac{p-1}{2}\right)\cdot  \left(p-\tfrac{p-1}{2}\right)\equiv (-1)^{\tfrac{p-1}{2}}\cdot (p-1)! \equiv{(-1)^{\tfrac{p+1}{2}}}(\bmod.p) -just by applying Wilson's Theorem-
    Last edited by PaulRS; October 3rd 2009 at 06:30 AM.
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