Artin's Conjecture on Primitive Roots: Perfect Squares

[setkeroppi]

If a is a perfect square then a is not a primitive root modulo p (p is an odd prime). (from Artin's conjecture on primitive roots) http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots

This is what I know: suppose a = b^2
a is a primitive root mod p when , a^(p-1) congruent to 1 (mod p)
that means b^2^(p-1) congruent to 1 (mod p)..,... I got stuck from here. Someone kindly gives me a hint ?

thank you

 

[ramsey2879]

If a is a perfect square then a is not a primitive root modulo p (p is an odd prime). (from Artin's conjecture on primitive roots) http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots

This is what I know: suppose a = b^2
a is a primitive root mod p when , a^(p-1) congruent to 1 (mod p)
that means b^2^(p-1) congruent to 1 (mod p)..,... I got stuck from here. Someone kindly gives me a hint ?

thank you

Actually the test you give is not correct. The test is that every number $$n$$ such that $$n^{p-1} = 1 \mod p$$ must equal some power of $$a$$ modulus $$p$$. In other words if p is prime and a is a primitive root then the set of $$a^i \mod p$$ equals the congruence set {1,2,3 ... p-1}. But a = b^2 = a primitive root implies b is also a primitive root. Could b and b^2 both be primitive roots?

 

[robert Ihnot]

The matter is easy enough. Every element is such that a^(p-1)==1 Mod p, for p prime.

But to be a primitive root, a must be such that a, a^2, a^3...a^(p-1) all generate
different elements.

However should a ==b^2 Mod p, then a^(p-1)/2 ==b^(p-1) ==1 Mod p. So that a is then capable of generating no more than half the multiplicative group.