About the parity of $ord_p(7)$ Recently I had the final exam of a number theory course I was taking, there were 6 questions and I know I got at least 3 right, the fourth one I didn't quite solve it, because I used that $ord_p(7)$ was odd for infinitely many primes, and I don't know if this is true, it appears to be true at least, could someone tell me if this is true or not?
For the people curious about the problem, it stated:
Prove that for infinitely many primes $p$, the polynomial $x^2-x+2$ is irreducible over $\mathbb{F}_p$
I figured out that I had to prove that $\left(\frac{-7}{p}\right)=-1$ for infinitely many primes, you can prove this by quadratic reciprocity, but I want to see how to prove or disprove what I stated earlier.
 A: A reference to this answer is here: This is a paper by Lagarias. 
Your problem is first solved by Hasse as mentioned by Lagarias' paper. Lagarias tried Hasse's method in some examples of Lucas sequences. Their methods rely on Chebotarev Density Theorem. 
I will summarize Lagarias' exposition of Hasse's method:
Theorem

Let $\mathcal{P}$ be the set of all prime numbers. Let $a\geq 3$ be a square-free number. Then $\mathrm{ord}_p(a)$ is odd for infinitely many primes $p$. Moreover, the natural density of primes with this property is $1/3$. That is, 
  $$
\frac{\#\{p\leq x | p\in \mathcal{P}, \ \mathrm{ord}_p(a) \textrm{ is odd} \}}{\#\{p\leq x | p\in \mathcal{P}\}}\rightarrow \frac13 \ \mathrm{as} \ x\rightarrow\infty.
$$

We begin the proof of the theorem. 
For each $j\geq 1$, let $S_j$, $\overline{S}_j$ be the primes with  following properties. 

$$\overline{S}_j=\{p\leq x| p\in\mathcal{P}, \ 2^j ||p-1, \ \mathrm{ord}_p(a) \ \mathrm{is} \ \mathrm{odd}\},$$
$$S_j=\{p\leq x| p\in\mathcal{P}, \ 2^j ||p-1, \ \mathrm{ord}_p(a) \ \mathrm{is} \ \mathrm{even}\}.$$

Then we have for primes $p\leq x$, 

$$\begin{align}p\in \overline{S}_j &\Longleftrightarrow a^{(p-1)/2^j}\equiv 1 \ \mathrm{mod}\ p\\ & \Longleftrightarrow y^{2^j}\equiv a \ \mathrm{mod}\ p \textrm{ is solvable in }\mathbb{Z}/p\mathbb{Z}\\
& \ \ \ \  \ \ \mathrm{and} \ p\equiv 1 + 2^j \ \mathrm{mod} \ 2^{j+1}\end{align}$$

Define the number fields:

$$K_j=\mathbb{Q}(1^{1/2^j}, a^{1/2^j}), \ \ L_j=K_j(1^{1/2^{j+1}}).$$

We have $[K_j:\mathbb{Q}]=2^{2j-1}$ and $[L_j:\mathbb{Q}]=2^{2j}$. 
Then for primes $p\leq x$, 

$$p\in\overline{S}_j \Longleftrightarrow p \ \textrm{splits completely in} \  K_j \ \textrm{ but not in } \ L_j.$$

Then the number $N_x$ of primes $p\leq x$ with $\mathrm{ord}_p(a)$ is odd satisfies for any $m\geq 1$, 
$$
|\cup_{j=1}^m \overline{S}_j|\leq N_x \leq |\mathcal{P}_x-\cup_{j=1}^m S_j|
$$
Here, $\mathcal{P}_x=\mathcal{P}\cap [1,x]$.
Now, divide by $\pi(x)=|\mathcal{P}_x|$. Then as we take the limit $x\rightarrow\infty$, 
$$
\sum_{j=1}^m \left(\frac 1{2^{2j-1}}-\frac1{2^{2j}}\right)\leq \liminf \frac{N_x}{\pi(x)} $$
$$\leq \limsup \frac{N_x}{\pi(x)} \leq 1-\sum_{j=1}^m \left(\frac 1{2^j}-\left(\frac 1{2^{2j-1}}-\frac1{2^{2j}}\right)\right)
$$
Letting $m\rightarrow\infty$, we obtain 
$$
\lim \frac{N_x}{\pi(x)} = \sum_{j=1}^{\infty} \left(\frac 1{2^{2j-1}}-\frac1{2^{2j}}\right) = \frac13.
$$
Remark : The case with $a\geq 3$ and square free is easier than the case $a=2$ considered in the linked paper. This is because the cyclotomic fields $\mathbb{Q}(1^{1/2^j})$ contains $\sqrt 2$ for $j\geq 3$. Thus, when $a=2$, the density turns out to be $7/24$. 
