Let p be odd. Then 2(p-3)!\equiv -1 (mod p)

**dwsmith** · July 12th 2010, 02:09 PM

Let $p$ be odd. Then $2(p-3)!\equiv -1 \ \mbox{(mod p)}$

Don't know how to do this one.

**chiph588@** · July 12th 2010, 02:18 PM

Originally Posted by dwsmith

Let $p$ be odd. Then $2(p-3)!\equiv -1 \ \mbox{(mod p)}$

Don't know how to do this one.

$2(p-3)!=(-1)(-2)(p-3)!\equiv(p-1)!\equiv-1\bmod{p}$

**dwsmith** · July 12th 2010, 02:26 PM

Originally Posted by chiph588@

$(-1)(-2)(p-3)!\equiv(p-1)!$

How do you show those two are congruent though?

**chiph588@** · July 12th 2010, 02:29 PM

Originally Posted by dwsmith

How do you show those two are congruent though?

$p-a\equiv-a\bmod{p}$

**dwsmith** · July 12th 2010, 02:51 PM

Originally Posted by chiph588@

$p-a\equiv-a\bmod{p}$

I don't see the connection.

**chiph588@** · July 12th 2010, 03:42 PM

Originally Posted by dwsmith

I don't see the connection.

What are you having trouble with?

Thread: Let p be odd. Then 2(p-3)!\equiv -1 (mod p)