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

Definitions and mathematical statement

For a prime $p$ and an integer $a$ not divisible by $p$, the multiplicative order $\operatorname{ord}_p(a)$ is the least positive integer $r$ such that $a^r\equiv1\pmod p$. A primitive root modulo $p$ is an element of order $p-1$, so its powers generate the entire group $(\mathbb Z/p\mathbb Z)^\times$.

The source's conjecture concerns the thin family \[ q>3\text{ prime},\qquad p=16q^4+1\text{ prime}: \quad \operatorname{ord}_p(3)=16q^4. \] Both primality assumptions are hypotheses. No assertion that this family contains infinitely many primes $p$ is required. Since the only prime divisors of $p-1$ are $2$ and $q$, the order condition is equivalent to the simultaneous noncongruences \[ 3^{8q^4}\not\equiv1\pmod p, \qquad 3^{16q^3}\not\equiv1\pmod p. \] Indeed a proper divisor of $p-1$ divides $(p-1)/r$ for some prime divisor $r$ of $p-1$. This equivalent finite criterion fixes exactly what must be verified for a proposed example. Prove the assertion for every admissible $q$, or exhibit primes $q,p$ of the specified form for which the order is smaller.

Short English statement

Whenever q and 16q to the fourth power plus one are both prime, with q greater than three, must three generate all nonzero residues modulo the latter prime?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3356 (OPG-37396), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, 3 is a primitive root modulo primes of the form 16 q^4 + 1, where q>3 is prime. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/primes_p_such_that_3_is_a_primitive_root_modulo_p.