Let $\displaystyle p$ be odd. Then $\displaystyle 2(p-3)!\equiv -1 \ \mbox{(mod p)}$ Don't know how to do this one.
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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} $
Originally Posted by chiph588@ $\displaystyle (-1)(-2)(p-3)!\equiv(p-1)! $ How do you show those two are congruent though?
Originally Posted by dwsmith How do you show those two are congruent though? $\displaystyle p-a\equiv-a\bmod{p} $
Originally Posted by chiph588@ $\displaystyle p-a\equiv-a\bmod{p} $ I don't see the connection.
Originally Posted by dwsmith I don't see the connection. What are you having trouble with?