1. ## Primitive Root proof

Let p = 2^(k) + 1 be a prime. Prove that if a is a natural number and a is a quadratic non-residue modulo p, then a is a primitive root modulo p.

thanks

2. Originally Posted by Tand
Let p = 2^(k) + 1 be a prime. Prove that if a is a natural number and a is a quadratic non-residue modulo p, then a is a primitive root modulo p.

thanks
Euler's Criterion says $\displaystyle -1 = \left(\frac{a}{p}\right) \equiv a^{(p-1)/2} \bmod{p}$

But $\displaystyle \frac{p-1}{2} = 2^{k-1}$, so we know $\displaystyle a^{2^{k-1}}\equiv -1 \bmod{p}$.

Now all the divisors of $\displaystyle \phi(p)$ are of the form $\displaystyle 2^b$ where $\displaystyle b\leq k$. So suppose there exists $\displaystyle b < k-1$ such that $\displaystyle a^{2^b}\equiv 1\bmod{p}$. This would however imply $\displaystyle a^{2^{k-1}}\equiv 1 \bmod{p}$ since $\displaystyle 2^b\mid2^{k-1}$, which is a contradiction.

Thus we see for every proper divisor $\displaystyle b$ of $\displaystyle \phi(p)$, $\displaystyle a^b\not\equiv 1 \bmod{p}\implies a$ is a primitive root modulo $\displaystyle p$.