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$p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{\left(\frac{p-1}{2}\right)}\cdot(p-1)! \equiv 1 \pmod p$

From Wilson's thm: $(p-1)!= -1 \pmod p$.

hence, need to show that $2^{\left(\frac{p-1}{2}\right)} \equiv -1 \pmod p. $

we know that $2^{p-1} \equiv 1 \pmod p.$

Hence: $2^{\left(\frac{p-1}{2}\right)} \equiv \pm 1 \pmod p. $

How do I show that this must be the negative option?

  • 2
    Notice the Euler criterion and that 2 is not a quadratic residue modulo $p$.2012-11-21
  • 0
    www4.ncsu.edu/~singer/437/proj32.pdf --Theorem 11.6 or http://www.mathreference.com/num-mod,qr12.html2012-11-21

2 Answers 2