Given p≡1(mod 4), prove [((p-1)/2)!]^{2}≡-1 (mod p)

Printable View

- October 20th 2013, 11:37 AMvirebbala90number theory
Given p≡1(mod 4), prove [((p-1)/2)!]

^{2}≡-1 (mod p) - October 20th 2013, 12:41 PMSlipEternalRe: number theory
Use induction? for some integer . So, . Obviously, is true, so assume it is true for all nonnegative integers less than or equal to k. Then implies . So, . By the induction assumption, you have , so there exists an integer with . Plug that in and see if you can get .

- October 26th 2013, 12:59 PMvirebbala90Re: number theory
i'm sorry , i could'nt get the final answer after substitution

- October 26th 2013, 04:24 PMSlipEternalRe: number theory
There is a good reason you are not finding it easy to prove. . . Since this is a counterexample, the proposition is disproven.

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

Assuming is prime and , then

Since , . So, you are just trying to show that where is a prime. Since is a group under multiplication, each nonzero element has a unique multiplicative inverse. Since given any prime , if , then , the product , so .