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.
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$
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1Also, $(p+1)/2$ is not always even: try $p=5$. – 2011-02-16
2 Answers
You can start with Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}.$ Rearrange the factorial thus, with $p=2m+1,$ $-1 \equiv 1(p-1)2(p-2) \ldots m(p-m) \equiv 1(-1)2(-2) \ldots m(-m) \equiv 1^22^2 \ldots m^2(-1)^m \pmod{p}.$
So we have shown
$1^22^2 \ldots m^2 \equiv (-1)^{m+1} $
or, noting that $m=(p-1)/2,$ in terms of $p$
$1^12^2 \ldots \left( \frac{p-1}{2} \right)^2 \equiv (-1)^{(p+1)/2}.$
Now we construct the product on the LHS, so consider first $2^24^2 \ldots (p-1)^2$ and factor out $2^{2p-2}$ then use Fermat, $2^{p-1} \equiv 1 \pmod{p}.$
You should now have the result for the product with the even numbers, use that along with Wilson's theorem again to prove the desired result for the product with the odd numbers.
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0Hi Derek, I was just wondering how to factor out $2^{2p-2}$ as this factor might not exist - take $p=7$ for example. The even numbers are $2^2 \cdot 4^2 \cdot 6^2$ but this has the common factor $2^8$. Please advise. – 2013-02-09
$p=2m+1$,
we can write $1\equiv -(p-1)\pmod p$,
Now write $1^2 \times 3^2 \dots(p-2)^2$ as $1\times(-(p-1))\times 3\times(-(p-3)\dots(p-2)(-(P-(p-2)))$, $\implies 1^2 \times 3^2 \dots(p-2)^2 = 1\times(-(p-1))\times3\times(-(p-3)\dots(p-2)(-(P-(p-2)))\\\equiv-1^m (p-1)! \equiv-1^{m+1}\pmod p$ wilson theorem, $1^2 \times 3^2 \dots(p-2)^2\equiv-1^{(p+1)/2} \pmod p$ (as taken $p=2m+1$)
similar for another case
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0Your post is hardly legible (please use LateX) and the quality of the explanations needs to be improved. (-1) – 2016-02-19