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Math Help - number theory

  1. #1
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    number theory

    Given p≡1(mod 4), prove [((p-1)/2)!]2≡-1 (mod p)
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  2. #2
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    Re: number theory

    Use induction? p = 4k+1 for some integer k. So, \dfrac{p-1}{2} = 2k. Obviously, [(0)!]^2 = 1 \equiv -1 \pmod{1} is true, so assume it is true for all nonnegative integers less than or equal to k. Then p = 4(k+1)+1 = 4k+5 implies \dfrac{p-1}{2} = 2k+2. So, [(2k+2)!]^2 = (2k+2)^2(2k+1)^2[(2k)!]^2. By the induction assumption, you have [(2k)!]^2 \equiv -1 \pmod{p-4}, so there exists an integer q with [(2k)!]^2 = -1+q(p-4). Plug that in and see if you can get -1\pmod{p}.
    Thanks from virebbala90
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  3. #3
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    Re: number theory

    i'm sorry , i could'nt get the final answer after substitution
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  4. #4
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    Re: number theory

    There is a good reason you are not finding it easy to prove. 9 \equiv 1 \pmod{4}. \left[\left(\dfrac{9-1}{2}\right)!\right]^2 = 24^2 = 576 = 64\cdot 9 \equiv 0 \pmod{9}. Since this is a counterexample, the proposition is disproven.

    Edit: If the problem added the constraint that p is prime, it will be true.

    Assuming p is prime and p \equiv 1 \pmod{4}, then
    \begin{align*}\left[\left(\dfrac{p-1}{2}\right)!\right]^2 & \equiv [(-1)1(p-1)][(-1)2(p-2)][(-1)3(p-3)]\cdots \left[(-1)\left(\dfrac{p-1}{2}\right) \left(p-\dfrac{p-1}{2}\right)\right] \pmod{p} \\ & \equiv (-1)^{(p-1)/2}(p-1)! \pmod{p} \end{align*}

    Since p \equiv 1 \pmod{4}, (-1)^{(p-1)/2} = 1. So, you are just trying to show that (p-1)! \equiv -1 \pmod{p} where p is a prime. Since \mathbb{Z} / p\mathbb{Z}\setminus \{[0]\} is a group under multiplication, each nonzero element has a unique multiplicative inverse. Since given any prime p, if k^2 \equiv 1 \pmod{p}, then k \equiv \pm 1 \pmod{p}, the product (p-2)! \equiv 1 \pmod{p}, so (p-1)! \equiv -1 \pmod{p}.
    Last edited by SlipEternal; October 26th 2013 at 05:54 PM.
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