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

• Jul 12th 2010, 02:09 PM
dwsmith
Let p be odd. Then 2(p-3)!\equiv -1 (mod p)
Let $\displaystyle p$ be odd. Then $\displaystyle 2(p-3)!\equiv -1 \ \mbox{(mod p)}$

Don't know how to do this one.
• Jul 12th 2010, 02:18 PM
chiph588@
Quote:

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

Don't know how to do this one.

$\displaystyle 2(p-3)!=(-1)(-2)(p-3)!\equiv(p-1)!\equiv-1\bmod{p}$
• Jul 12th 2010, 02:26 PM
dwsmith
Quote:

Originally Posted by chiph588@
$\displaystyle (-1)(-2)(p-3)!\equiv(p-1)!$

How do you show those two are congruent though?
• Jul 12th 2010, 02:29 PM
chiph588@
Quote:

Originally Posted by dwsmith
How do you show those two are congruent though?

$\displaystyle p-a\equiv-a\bmod{p}$
• Jul 12th 2010, 02:51 PM
dwsmith
Quote:

Originally Posted by chiph588@
$\displaystyle p-a\equiv-a\bmod{p}$

I don't see the connection.
• Jul 12th 2010, 03:42 PM
chiph588@
Quote:

Originally Posted by dwsmith
I don't see the connection.

What are you having trouble with?