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(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Primitive root !

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