Hey, I was just wondering gif someone could help me prove the following.

((p-1)/2)!)^ 2=(+ or -) 1 mod (p), where p is prime.

THanks

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- Dec 10th 2010, 10:07 PMHelloWorld2Proof involving Wilson's theorem.
Hey, I was just wondering gif someone could help me prove the following.

((p-1)/2)!)^ 2=(+ or -) 1 mod (p), where p is prime.

THanks - Dec 11th 2010, 02:39 AMtonio

For p an odd prime, we have:

$\displaystyle \displaystyle{(p-1)! = 1\cdot 2\cdot\ldots\cdot(p-1)=1\cdot\ldots\cdot \frac{p-1}{2}\cdot \frac{p+1}{2}\cdot\ldots\cdot (p-1)!=(-1)^{\frac{p-1}{2}}\left(1\cdot\ldots\cdot \frac{p-1}{2}\right)^2}$ (why?),

so...

Tonio

Pd. If you in fact understand the above, then you must be able even to tell whenwe'll__exactly__

get 1 and when -1 modulo p. - Dec 11th 2010, 09:04 AMHelloWorld2
well, we'll get 1 when p=1mod(4), and -1 when p=-1mod(4). I realy don't get how you got that equaility.

- Dec 11th 2010, 10:55 AMtonio
- Dec 11th 2010, 02:53 PMmelese
But, for example, $\displaystyle 3\cdot4\equiv1\pmod{11}$, and $\displaystyle \displaystyle3,4\leq\frac{11-1}{2}$ (If I understand what you meant.).

The observation here is that for $\displaystyle \displaystyle1\leq x\leq\frac{p-1}{2}$, we have $\displaystyle \displaystyle\frac{p+1}{2}\leq p-x\leq p-1$.

Then noting that $\displaystyle p-x\equiv-x\pmod p$ shows that $\displaystyle \displaystyle\frac{p+1}{2}\cdots(p-2)(p-1)\equiv(-\frac{p-1}{2})\cdots(-2)(-1)\pmod p$. - Dec 11th 2010, 06:30 PMtonio