0
$\begingroup$

Let $p$ be and odd prime. Use Wilson's Theorem to show that:

$[(\frac{p -1}{2}) !]^2$ $\equiv$ $(-1)^{(p+1)/2}$ mod $p$

  • My understanding is that this should be as simple as picking an odd prime and simplifying, thus far I have:

$[(\frac{7 - 1}{2}) !]^2$ $\equiv$ $(-1)^{(7+1)/2}$ mod $7$

$[(3) !]^2$ $\equiv$ $(-1)^{4}$ mod $7$

$[6]^2$ $\equiv$ $(-1)^{4}$ mod $7$

$36$ $\equiv$ $1$ mod $7$

Is this the correct approach and if so what would be the following steps (if any), a bit confused as to how i am supposed to "show" this. Any help is greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ You need to prove it for general $p$, not just a few special cases. $\endgroup$ – M10687 Apr 13 '16 at 3:08
  • $\begingroup$ Via induction I would assume? $\endgroup$ – Nick Powers Apr 13 '16 at 3:09
  • $\begingroup$ @NickPowers No, not by induction. (BTW, Wilson's theorem isn't proved by induction either. Have you studied the proof?) $\endgroup$ – Erick Wong Apr 13 '16 at 3:46
1
$\begingroup$

Wilson's theorem gives $(p-1)!\equiv -1\pmod{p}$, but: $$(p-1)! = 1\cdot 2\cdot\ldots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\ldots\cdot(p-2)\cdot(p-1)$$ that $\pmod{p}$ is the same as: $$ \left(\frac{p-1}{2}\right)!\cdot \left(-\frac{p-1}{2}\right)\cdot\ldots\cdot(-1) = (-1)^{\frac{p-1}{2}}\left(\frac{p-1}{2}\right)!^2.$$ It follows that: $$\left(\frac{p-1}{2}\right)!^2\equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.