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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!

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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.

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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

$$\alpha+\alpha^{-1}=(\alpha+\alpha^{-1})^p=\alpha^p+\alpha^{-p}+p(...)=\alpha^p+\alpha^{-p}\pmod{p}.$$

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