# $2$ is a primitive root of $p = 2q+1$ when $p = 8t+3$

I'm working through exercises in Ireland & Rosen textbook. I'm having trouble understanding the idea behind Exercise 4.7:

Suppose that $p$ is a prime of the form $8t + 3$ and that $q = (p - 1)/2$ is also a prime. Show that $2$ is a primitive root modulo $p$.

One of the previous exercises is to show that for $p = 4k+3$, $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1)/2$. In our case, since $p = 2q + 1$, possible values for order of $-2$ are $2, q$, and $2q$. We can easily rule out the case of order $2$, and so we pretty much need to show that $-2$ is a quadratic residue modulo $p$. Since $-1$ is a quadratic nonresidue (as $p = 8t+3$), this is equivalent to showing that $2$ is a quadratic nonresidue modulo $p$.

It is known when $2$ is a quadratic nonresidue modulo $p$ -- it is the case when $p = 8t+3$ or $p = 8t+5$. In fact, Ireland and Rosen do prove it, but only in the next chapter. Since they give this exercise before they show when $2$ is a quadratic residue, it suggests they expect the reader to solve this exercise without knowing the quadratic character of $2$.

So, my question is the following: if $p = 2q+1$, and $q = 4k+1$ is prime, is there a more straightforward way to show that $2$ is a quadratic nonresidue modulo $p$ than to appeal to a general fact about quadratic character of $2$?