<s>[INST] If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots  \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$  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. 
 [/INST] $p=2m+1$,
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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