When does this system of congruences hold? Let $\alpha,\beta,\gamma$ be quadratic irrationalities of the form $(n\pm\sqrt{n^2-4})/2$ for some integer $n$ (the $n$ is different for each of the three numbers). What are the solutions to the system of congruences
$$\begin{cases}\alpha+\alpha^{-1}+\beta+\beta^{-1}\equiv\gamma+\gamma^{-1}\\\alpha^p+\alpha^{-p}+\beta^p+\beta^{-p}\equiv\gamma^p+\gamma^{-p}\end{cases}\pmod{p}?$$

First, note that the question is well-defined, and $\alpha+\alpha^{-1}$ and $\alpha^p+\alpha^{-p}$ are both integers. This is because
$$\alpha=\frac{n+\sqrt{n^2-4}}{2}\implies\alpha^{-1}=2\cdot\frac{n-\sqrt{n^2-4}}{n^2-(n^2-4)}=\frac{n-\sqrt{n^2-4}}2,$$
so $\alpha,\alpha^{-1}$ are conjugate. Further, $\alpha^n+\alpha^{-n}=(\alpha+\alpha^{-1})(\alpha^{n-1}+\alpha^{-(n-1)})-(\alpha^{n-2}+\alpha^{-(n-2)})$, and it is easily verified that $\alpha^2+\alpha^{-2}\in\mathbb Z$. These two facts together imply the sequence $(\alpha^n+\alpha^{-n})_{n\in\mathbb N}\in\mathbb Z$, and in particular, $\alpha^p+\alpha^{-p}\in\mathbb Z$. The identical argument of course shows the same thing for $\beta$ and $\gamma$, so even though we seem to have weird irrationalities in the congruences, all of the quantities involved are actually integers. So the congruences are just the normal relations defined on $\mathbb Z$.
Naturally, I tried to set $\alpha+\alpha^{-1}=n_1$, $\beta+\beta^{-1}=n_2$, $\gamma+\gamma^{-1}=n_3$. Then the first congruence becomes a nice $n_1+n_2\equiv n_3$ mod $p$, but unfortunately the second congruence has no nice representation in $n_1,n_2,n_3$. So unless I'm missing something, this approach cannot work. Any thoughts or partial solutions would be greatly appreciated!
 A: Galois theory of finite fields sheds some light on this question. In fact, it follows that the latter congruence will always be a consequence of the former! Basically because they are each others Galois conjugates!
As you observed $\alpha^{\pm1}=(n\pm\sqrt{n^2-4})/2$ are the solutions of the quadratic equation
$$
m(x)=x^2-nx+1=0.
$$
Everything takes place in the ring $R$ of algebraic integers of 
$\Bbb{Q}(\sqrt{n^2-4})$. From basic algebraic number theory we infer that
$R$ has a prime ideal $\mathfrak{p}$ such that $\mathfrak{p}\cap\Bbb{Z}=p\Bbb{Z}$.
The quotient ring $R/\mathfrak{p}$ is then a finite field $K$. If $n^2-4$ is
a quadratic residue modulo $p$ then $|K|=p$. But if $m(x)$ is irreducible modulo $p$
then $|K|=p^2$.
The idea is that for integers a congruence modulo $p$ is equivalent to congruence modulo $\mathfrak{p}$, and the latter can be decided by projecting everything to the quotient field $K$.
If $m(x)$ factors modulo $p$, then the images of $\alpha^{\pm1}$ in $K$ are residue classes of integers modulo $p$. So Little Fermat says that $\alpha^p\equiv\alpha\pmod{\mathfrak{p}}$ and also $\alpha^{-p}\equiv\alpha^{-1}\pmod{\mathfrak{p}}.$
Consequently 
$$\alpha^p+\alpha^{-p}\equiv\alpha+\alpha^{-1}\pmod{\mathfrak{p}}$$
and this implies the same congruence of integers modulo $p$.
If $m(x)$ is irreducible modulo $p$ then its zeros in $K$ are Frobenius conjugates of each other. As those zeros are the projections of $\alpha^{\pm1}$, it follows that
$$
\alpha^p\equiv\alpha^{-1}\pmod{\mathfrak{p}}.
$$
Applying the Frobenius again then gives the congruence
$$
\alpha^p+\alpha^{-p}\equiv\alpha^{-1}+\alpha\pmod{\mathfrak{p}}
$$
and we are done by repeating the earlier argument.

The conclusion is that the condition $$n_1+n_2\equiv n_3\pmod p$$ is all you need for both of the congruences to hold.

A: Jyrki Lahtonen already provided a great solution to the problem, reaching the conclusion that $\alpha+\alpha^{-1}=\alpha^p+\alpha^{-p}$ mod $p$ for each $\alpha$. I believe I found an even easier way to get this conclusion, starting from Fermat's Little Theorem.
Fermat's Little Theorem asserts that $x^p=x$ mod $p$ for all $x\in\mathbb Z$. Then set $\alpha+\alpha^{-1}=n$, and apply the theorem to $n$. Since the binomial coefficients $\binom{p}{k}$ are divisible by $p$ for $k=1,2,\dots,p-1$, and the terms of the form $\alpha^n+\alpha^{-n}$ are integral, we have
$$\alpha+\alpha^{-1}=(\alpha+\alpha^{-1})^p=\alpha^p+\alpha^{-p}+p(...)=\alpha^p+\alpha^{-p}\pmod{p}.$$
