Show that $2$ is not a primitive root of $8k + 7$

Question

I'm attempting to show that $2$ is not a primitive root of primes of the form $p = 8k + 7$. I know that, to do so, I must show that $2$ has order less than $\phi(p)$ modulo $p$ (where $\phi$ denotes Euler's Totient function), however I've found myself a bit stuck.

I've begun by way of contradiction. If we assume $2$ to be a primitive root of $p$, then:

$$2^{\phi(8k+7)} \equiv 1 \pmod{8k + 7}$$

Since $p$ is prime, $\phi(p) = (8k+7)-1 = 8k+6$, hence:

$$2^{8k+6} \equiv 1 \pmod{8k+7}$$

...it isn't clear to me where to go from here.

Answer 1

Since $p\equiv -1 \bmod 8$, the Legendre symbol is equal to $1$, i.e., $$ \left( \frac{2}{p}\right)=1. $$ On the other hand, by Euler's criterion we have $$ 2^{(p-1)/2}\equiv\left(\dfrac2p\right)=1\bmod p, $$ hence $2$ is not a primive root modulo $p$.

Reference:

Solution to $x^2 = 2$ in field of $p$ element with $p \equiv \pm 1 \bmod 8$ ($p$ prime)