3 is a primitive root modulo primes of the form 16 q^4 + 1, where q>3 is prime

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Subject: Number Theory
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Recomm. for undergrads: no
Posted by: princeps
on: February 25th, 2012

\begin{conjecture} $3~$ is a \Def[primitive root]{Primitive_root_modulo_n} modulo $~p$ for all primes $~p=16\cdot q^4+1$, where $~q>3$ is prime. \end{conjecture}

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% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)


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group theory answer

Using group theory, the multiplicative group of order p=16q^4+1 has order p-1=16q^4. Using lagrange's theorem, the order of any element divides the order of the group. Therefore, any element is either a primitive root, a quadratic residue, or a qth power residue mod 16q^4+1. Using the laws of quadratic reciprocity, 3 is a quadratic residue modulo a prime if and only if the prime is congruent to plus or minus 1 mod 12. Since q>3 is a prime and therefore not divisible by 3, 16q^4=1(mod 3), so 16q^4+1=2(mod 3). That means that 16q^4+1=5(mod 12), and therefore 3 is not a quadratic residue mod p. Therefore the only thing left to prove is that 3 is not a qth power residue.

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