Let k be a positive integer and let m = 4k + 3.
a Show that the even numbers 2k +2, 2k +4, . . . , 4k, 4k +2 are congruent
modulo m to the negatives of the odd numbers 2k + 1, 2k − 1, . . . , 3, 1,
and deduce that
2^k+1(k+1)(k+2) · · (2k+1) = (congruent)
(−1)^k+1(2k+1)(2k−1) · · 3·1 (mod m).
ii) Use Part (i) and the fact that 2^k(k!) is the product of the even numbers
from 2 to 2k to deduce that 2^2k+1(2k +1)!=(−1)^k+1(2k+1)!(mod m)
[ Done ]
Assume now that
m is prime. Use Part (ii) to show that if k is odd then
2^0.5(m-1)=2^2k+1=1 mod m
I have no idea...
Do i use my original m=4k-3 and make m-1 = 4k +2 so 0.5(m-1) = 2k+1
now I from part ii) that 2^2k+1 = 2^0.5(m-1) = (-1)^k+1
k = 1,3,5 ...
hence we get (-1)^k+1 = 1 at all times... btw if your wondering what happend to (2k+1)! i just divided it by both side.. because I can.