2
$\begingroup$

Prove or disprove:

Suppose that $p$ is prime and $≡ 3\bmod4$ then $((p-1)/2)!≡-1\bmod p$ or $((p-1)/2)!≡1\bmod p$

After I checked it I see it is true statement

so by Wilson's theorem we have $(p-1)!≡-1\bmod p$

so $1\cdot2\cdot3\cdots((p-1)/2)((p+1)/2)\cdots(p-1) ≡ -1\bmod p$

so $(p-1)/2)!((p-1)/2)! -2) ≡ -1\bmod p$

then $((p-1)/2)! ≡ -1\bmod p$ or $((p-1)/2)! ≡ 1\bmod p$

is which I did right?

$\endgroup$
  • $\begingroup$ When you write $p-1/2$, it means $p-0.5$. Obviously that's not what you mean because it doesn't make sense. You really mean $(p-1)/2$, or better yet $\frac {p-1}2$. $\endgroup$ – Matt Samuel Dec 7 '17 at 4:58
2
$\begingroup$

Your idea of using Wilson's theorem is correct, but when you get to $$1\cdot2\cdot3\cdots\frac{p-1}2\cdot\frac{p+1}2\cdots(p-1) ≡ -1\bmod p$$ you need to take a different approach. Rewrite $p-1$ as $-1$, $p-2$ as $-2$ and so on until you get $$1\cdot2\cdot3\cdots\frac{p-1}2\cdot\left(-\frac{p-1}2\right)\cdots(-1)≡-1\bmod p$$ Because $p\equiv3\bmod4$, the number of terms is singly even, so there are an odd number of terms that have become "negative". The above is thus equivalent to $$-\left(1\cdot2\cdot3\cdots\frac{p-1}2\right)^2≡-1\bmod p$$ $$\left(\frac{p-1}2!\right)^2≡1\bmod p$$ $$\frac{p-1}2!≡\pm1\bmod p$$

$\endgroup$

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.