For any odd prime $p$ there exists at least one prime $q < p$ such that $q$ is a primitive root $\text{mod } p$ ; is this true?

  • $\begingroup$ I recommend you read this page: mathworld.wolfram.com/PrimitiveRoot.html then come back and let us know if you're still unsure. $\endgroup$ – Robert Soupe Dec 20 '14 at 2:26
  • $\begingroup$ Without additional assumptions (GRH) the question is currently open. Please see this. $\endgroup$ – André Nicolas Dec 20 '14 at 3:20
  • $\begingroup$ @RobertSoupe: André Nicolas' comment suggests that OP will still be unsure. $\endgroup$ – daniel Dec 20 '14 at 6:23
  • $\begingroup$ This question has a simple solution, I think. Say all the primes < = p are Q(1) to Q(n) where n='phi'(p-1). Any residue R = [(Q(1)^A(1))(Q(2)^A(2))..(Q(n)^A(n))] mod(p) where A(i) are elements of integers. If each Q(i) prime < p is not a primitive root then for any i , (Q(i)^((p-1)/2)) = 1 mod(p). Therefore R^((p-1)/2) = 1 mod(p) for any residue R | 1 < R < (p-1). This contradiction proves the initial assertion. $\endgroup$ – 201044 Dec 24 '14 at 5:32
  • $\begingroup$ Let all primes < p be q(1) to q(n) where n = $\phi{p-1}$. Any residue R= ($\q(1)^a(1)$) $\endgroup$ – 201044 Jan 9 '15 at 0:07

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