Find $$(1008)!\equiv ? \pmod {2017}$$
From the Wilson theorem I can $2016!\equiv -1\pmod{2017}$
Considering the general： $$\left(\dfrac{p-1}{2}\right)!\equiv ?\pmod p$$ where $p$ is prime number
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2 
- 1$\begingroup$ See this duplicate, but with $p\equiv 1\bmod 4$. $\endgroup$Dietrich Burde– Dietrich Burde2017-10-20 11:25:58 +00:00Commented Oct 20, 2017 at 11:25
- $\begingroup$ How to Determine the number of positive integers less than $\frac{p}{2}$ that are quadratic nonresidues of p $\endgroup$math110– math1102017-10-20 11:30:40 +00:00Commented Oct 20, 2017 at 11:30
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1 Answer
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2 $(p-1)!=1\cdot2\cdots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdots(p-2)(p-1)$ and $(p-1)!\equiv 1\cdot(-1)\cdot2\cdot(-2)\cdots\frac{p-1}{2}\cdot\left(-\frac{p-1}{2}\right)\pmod p.$
From Wilson's theorem, $(p-1)!\equiv -1\pmod{p}$
Implies $\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$
- 2$\begingroup$ How to Determine the number of positive integers less than $\dfrac{p}{2}$ that are quadratic nonresidues of p ,when $p=2017$ $\endgroup$math110– math1102017-10-20 11:31:13 +00:00Commented Oct 20, 2017 at 11:31
- $\begingroup$ Since $p \equiv 1 \pmod 4$, $a$ is quadratic non-residue iff so is $p-a$. Combine it with fact that there's equal amount of residues and non-residues from $1$ to $p-1$ to get the answer $504$. $\endgroup$G. Strukov– G. Strukov2017-10-31 23:10:30 +00:00Commented Oct 31, 2017 at 23:10