- #1
johng1
- 235
- 0
A recent challenge problem by Kaliprasad was: x=2*4*6...*(p-1) and y=1*3*5...*(p-2); find x-y mod p. I first thought of Wilson's theorem: for any prime p, (p-1)! is -1 mod p. So I then thought about the exact values of x and y mod p. By writing the factors of x in reverse order, one gets x is congruent to $(-1)^{(p-1)/2}y$. So for p a prime congruent to 1 mod 4, x is y mod p, and so $x^2=xy\equiv-1$; i.e. $x=\sqrt{-1}$. Good luck with finding x. However for p congruent to 3 mod 4, $x^2\equiv1$ and so x is 1 or -1. Now my question is: for what primes p with p=4k+3 is x congruent to 1? Experimentally, I found for the first 2,117 primes that are 3 mod 4, there are 1033 of these that have x=1. But I could find no pattern.