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I also have to prove this for $$2^2 \times 4^2 \times 6^2 \cdots \times (p-1)^2 \equiv (-1)^{(p+1)/2} \pmod{p}$$ I made some progress so far and got stuck. I said that since p is odd, $(p+1)/2$ is even. Then we can say that $(-1)^{\mathrm{even}}= 1$, so $(-1)^{(p+1)/2}\pmod{p}$ can be written as $1\pmod{p}$. I know that the product of odds is odd and the product of evens is even, but I cant prove that the left side of this equation in either case is congruent to $1 \pmod{p}$. Any help here would be greatly appreciated.

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    Homework I presume?2011-02-16
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    You would probably use Wilson's Theorem.2011-02-16
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    Hint: if $z$ is even then $p-z$ is odd and vice versa.2011-02-16
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    Also, $(p+1)/2$ is not always even: try $p=5$.2011-02-16

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