Let $p$ and $q$ be odd primes. If $p - 1 = 2q$, and $5$ is a primitive root $\bmod q$, then $5$ is a primitive root $\bmod p$.
Thanks to Álvaro Lozano-Robledo, here is what i have; valid?
If:
(A) p and q are odd primes; and
(B) p - 1 = 2q <-> p = 2q + 1 ; and
(C) 5 is a primitive root mod q.
then:
(1) There exist phi(p-1) = phi(2q) = (1*(q-1)) = (q-1) primitive roots mod p
(2) There exist (p-1)/2 = q quadratic non-residues mod p
(3) Since ((odd*2) + 1) is always congruent to 3 mod 4; -1 mod p is a quadratic non-residue (Law Of Quadratic Reciprocity)
(4) (-1 mod p) i.e. (p-1 mod p) cannot be a primitive root because ((p-1)^2 mod p) = ((p^2) - 2p + 1 mod p) = 1 mod p. p is also >= 7
(5) From (1),(2),(3),(4) Every Quadratic-NonResidue other than p-1 is a primitive root.
(6) Legendre(5/q) = -1, Legendre(q/5) = -1, i.e. q^(4/2) is congruent to (4 mod 5); q is congruent to (+ or -) (2 mod 5)
(7) 2q is congruent to (4 or 1) mod 5
(8) (substitute for 2q, p-1) p is congruent to (0 or 2) mod 5
(9) p cannot be 5 or any of its multiples, because its a prime and must be greater than 6.
(10) p must be congruent to 2 mod 5
(11) p^(5-1 /2) mod 5 is -1,
(12) Legendre(p/5) = -1, Legendre(5/p) = -1
(13) 5 is a quadratic non-residue mod p
(14) 5 is a primitive root mod p fom (13),(1),(2),(3)
cheers arun