
Primitive root
p prime, If p=1 ( mod 3) then Zp contains primitive cube
roots of unity. Now I am considering which p does Zp contains primitive fourth
roots of unity.
I can prove that if Zp contains primitive fourth roots of
unity, then 4(p1) . but how about the opposite way? I mean if p=1(mod4) then
Zp contains primitive fourth roots of unity?? I know this statements true if q
prime instead of 4. And what values of p does Zp contains primitive fourth roots
of unity???

Re: Primitive root
Hi,
By a primitive 4th root of unity, I assume you mean a y such that y^{4} is 1 mod p, but no smaller power is 1. Now if you know about cyclic groups, your problem is easy. However, even without this here's a proof.
There is a primitive 4th root of unity mod p if and only if 4 divides p1.
Proof of only if:
Let x be a primitive root mod p; i.e. every nonzero element of Z_{p} is a power of x. (You need to know such exist.) So suppose y=x^{m} is a primitive 4th root. I need the following fact: the order of x^{m} mod p is (p1)/gcd(m,p1). (Order is the least positive power k with (x^{m})^{k} = 1 mod p; this is a standard result from cyclic groups. But it's not hard to prove separately.) So p1=4gcd(m,p1) or 4 divides p1. QED.