number theory

• Oct 20th 2013, 11:37 AM
virebbala90
number theory
Given p≡1(mod 4), prove [((p-1)/2)!]2≡-1 (mod p)
• Oct 20th 2013, 12:41 PM
SlipEternal
Re: number theory
Use induction? $p = 4k+1$ for some integer $k$. So, $\dfrac{p-1}{2} = 2k$. Obviously, $[(0)!]^2 = 1 \equiv -1 \pmod{1}$ is true, so assume it is true for all nonnegative integers less than or equal to k. Then $p = 4(k+1)+1 = 4k+5$ implies $\dfrac{p-1}{2} = 2k+2$. So, $[(2k+2)!]^2 = (2k+2)^2(2k+1)^2[(2k)!]^2$. By the induction assumption, you have $[(2k)!]^2 \equiv -1 \pmod{p-4}$, so there exists an integer $q$ with $[(2k)!]^2 = -1+q(p-4)$. Plug that in and see if you can get $-1\pmod{p}$.
• Oct 26th 2013, 12:59 PM
virebbala90
Re: number theory
i'm sorry , i could'nt get the final answer after substitution
• Oct 26th 2013, 04:24 PM
SlipEternal
Re: number theory
There is a good reason you are not finding it easy to prove. $9 \equiv 1 \pmod{4}$. $\left[\left(\dfrac{9-1}{2}\right)!\right]^2 = 24^2 = 576 = 64\cdot 9 \equiv 0 \pmod{9}$. Since this is a counterexample, the proposition is disproven.

Edit: If the problem added the constraint that $p$ is prime, it will be true.

Assuming $p$ is prime and $p \equiv 1 \pmod{4}$, then
\begin{align*}\left[\left(\dfrac{p-1}{2}\right)!\right]^2 & \equiv [(-1)1(p-1)][(-1)2(p-2)][(-1)3(p-3)]\cdots \left[(-1)\left(\dfrac{p-1}{2}\right) \left(p-\dfrac{p-1}{2}\right)\right] \pmod{p} \\ & \equiv (-1)^{(p-1)/2}(p-1)! \pmod{p} \end{align*}

Since $p \equiv 1 \pmod{4}$, $(-1)^{(p-1)/2} = 1$. So, you are just trying to show that $(p-1)! \equiv -1 \pmod{p}$ where $p$ is a prime. Since $\mathbb{Z} / p\mathbb{Z}\setminus \{[0]\}$ is a group under multiplication, each nonzero element has a unique multiplicative inverse. Since given any prime $p$, if $k^2 \equiv 1 \pmod{p}$, then $k \equiv \pm 1 \pmod{p}$, the product $(p-2)! \equiv 1 \pmod{p}$, so $(p-1)! \equiv -1 \pmod{p}$.