# Math Help - h+k=p-1 ==> h!k!+[(-1)^h]=0 (mod p)

1. ## h+k=p-1 => h!k!+[(-1)^h]=0 (mod p)

$p$, prime odd number

$h \geq 0 ,k \geq 0$

Show that:

$h+k=p-1 \Rightarrow h!k!+(-1)^{h}=0 (mod p)$

2. Originally Posted by demdirbu
$p$, prime odd number

$h \geq 0 ,k \geq 0$

Show that:

$h+k=p-1 \Rightarrow h!k!+(-1)^{h}=0 \pmod p$
This is a lightly disguised form of Fermat's theorem. Notice that $(-1)^hh! = (-1)(-2)\cdots(-h)$. Modulo p, this is equivalent to $(p-1)(p-2)\cdots(p-h) = (p-1)(p-2)\cdots(k+1)$. When you multiply this by k! you get (p-1)!.

>, 1h0, kp1, mod