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Thread: 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? $\displaystyle p = 4k+1$ for some integer $\displaystyle k$. So, $\displaystyle \dfrac{p-1}{2} = 2k$. Obviously, $\displaystyle [(0)!]^2 = 1 \equiv -1 \pmod{1}$ is true, so assume it is true for all nonnegative integers less than or equal to k. Then $\displaystyle p = 4(k+1)+1 = 4k+5$ implies $\displaystyle \dfrac{p-1}{2} = 2k+2$. So, $\displaystyle [(2k+2)!]^2 = (2k+2)^2(2k+1)^2[(2k)!]^2$. By the induction assumption, you have $\displaystyle [(2k)!]^2 \equiv -1 \pmod{p-4}$, so there exists an integer $\displaystyle q$ with $\displaystyle [(2k)!]^2 = -1+q(p-4)$. Plug that in and see if you can get $\displaystyle -1\pmod{p}$.
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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. $\displaystyle 9 \equiv 1 \pmod{4}$. $\displaystyle \left[\left(\dfrac{9-1}{2}\right)!\right]^2 = 24^2 = 576 = 64\cdot 9 \equiv 0 \pmod{9}$. Since this is a counterexample, the proposition is disproven.

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

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

    Since $\displaystyle p \equiv 1 \pmod{4}$, $\displaystyle (-1)^{(p-1)/2} = 1$. So, you are just trying to show that $\displaystyle (p-1)! \equiv -1 \pmod{p}$ where $\displaystyle p$ is a prime. Since $\displaystyle \mathbb{Z} / p\mathbb{Z}\setminus \{[0]\}$ is a group under multiplication, each nonzero element has a unique multiplicative inverse. Since given any prime $\displaystyle p$, if $\displaystyle k^2 \equiv 1 \pmod{p}$, then $\displaystyle k \equiv \pm 1 \pmod{p}$, the product $\displaystyle (p-2)! \equiv 1 \pmod{p}$, so $\displaystyle (p-1)! \equiv -1 \pmod{p}$.
    Last edited by SlipEternal; Oct 26th 2013 at 04:54 PM.
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