Thread: Congruence Problems

a) Let p be an odd prime and let
a = $\displaystyle \prod$ (2j-1) = (1)(3)(5)....(p-2)
Prove that $\displaystyle a^2$ = $\displaystyle (-1)^{(p+1)/2}$ mod p

(Product is from j=1 to (p-1)/2)

b) Let p be a prime. Prove that
$\displaystyle \left(\begin{array}{cc}2p\\p\end{array}\right)$ = 2 mod p

$\displaystyle \left(\begin{array}{cc}2p\\p\end{array}\right)$ is 2p choose p, as in statistics. so $\displaystyle (2p)!/p!(2p-p)!$

Hint for a:

Wilson theorem.

Originally Posted by Also sprach Zarathustra

Hint for a:

Wilson theorem.

Wilson's crossed my mind but I don't see how to work (p-1)! into the proof

Another hint(or solution)...

$\displaystyle t\equiv -(p-t)(\mod p)$, so

$\displaystyle 2\cdot4\cdot6 ... (p-1)\equiv (-1)^{\frac{p-1}{2}}1\cdot3\cdot5...(p-2)(\mod p)$

...