$a+a^{-1}=1$ over $\mathbb F_{p}$ For what values of $p$ prime does the equation $a+a^{-1}=1$ hold true for some $a\in \mathbb F_{p}$ such that $a \ne0$? I tried counting field elements that are inverses of each other and playing with modular arithmetic, but I can't seem to work it out.
 A: We note that $a + a^{-1} = 1$ if and only if $a^2 - a + 1 = 0$.
Now note that $a^3 + 1 = (a+1)(a^2 - a + 1) = 0$ thus $a^3 = -1$, but $a\neq -1$  if $p\neq 3$, as otherwise $-2 = (-1)  + (-1)^{-1} = 1$ and so $p=3$.
Conversly, since $a^3 + 1 = (a+1)(a^2 - a + 1)$, we get that if $a^3 = -1$ with $a\neq -1$, then $a^2 - a + 1 = 0$.
Observation 1: If $p\neq 3$: $a + a^{-1} = 1$ occurs if and only if $a^3 = -1$ and $a\neq -1$.
I will give you indications for the rest, when $p\neq 3$:
Question: What is the order of $a$ ? What does that imply by recalling Fermat's Little Theorem?
For the converse : Remember that ${\mathbb F}_p^\times$ is cyclic (the multiplicative group), use this to show for the $p$ as you found, there is at least one $a$ satisfying $a+a^{-1}=1$.
Observation 2 This proof actually works finite fields $F$ with a minor change. Indeed, change $p\neq 3$ above for $\text{char}(F) \neq 3$, since for characteristic $3$ we can take $a=-1$.
Edit A more routinary way would probably be the following: $a^2 - a + 1 = 0$ if and only if $(2a-1)^2 = -3$, so $-3$ must be a nonzero quadratic residue if $p\neq 3$. Next what is $\left(\tfrac{-1}{p}\right)$? and use quadratic reciprocity to find $\left(\tfrac{3}{p}\right)$, don't forget $\left(\tfrac{p}{3}\right)\equiv p (\bmod. 3)$ too. You will get a congruence for $p(\bmod. 3)$, and turn it into one $\bmod.6$ by observing that $p$ can't be even ;)
