Proving a congruence using Wilson Theorem

Question

If $p \equiv 1 \pmod{4}$ and $p-1 =m+n,$ prove that $m!n! \equiv (-1)^{n+1} \pmod{p}.$

How to prove that using Wilson Theorem $(p-1)! \equiv -1 \pmod{p}$?

Answer 1

Hints:

• Show that if $p$ is prime then $\binom{p-1}{k}\equiv (-1)^{k}\pmod{p}$ for $1\leq k \leq p-1$.

• Use the fact that $m!\cdot n! \cdot \binom{m+n}{m} =(m+n)!=(p-1)!$

Note you have $m+n=p-1$ And using this in the second hint you get $$m!\cdot n! \cdot \binom{p-1}{m} \equiv m!\cdot n! \cdot (-1)^m \equiv (p-1)!\equiv -1\pmod{p}$$ Multiply both sides by $(-1)^{m}$ and you get $$m!\cdot n! =(-1)^{m+1}\pmod{p}$$

Answer 2

Hint: $-1 \equiv (p-1)! = m! (m+1)(m+2)\cdots(m+n)$ and $m+k \equiv -(n-k+1)$

Answer 3

p need not be $\equiv 1\pmod 4$. The statement you're attempting to prove is true for all primes. A famous result about primes equivalent to 1 modulo 4 is Fermat's Theorem on Sums of Two Squares

Note the following:
- $p| {p\choose k}$ for all primes $p$ and $k \in [1,p-1]$
- ${p\choose k} = {p-1\choose k-1} + {p-1\choose k}$
- ${p-1\choose 0} = 1$

It follows that ${p-1\choose i} \equiv \pm1\pmod p$

The hints above will guide you well.