Question about a kind of generalized Fermat numbers The present question is directly inspired by this one. 
Let $\alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words: 
$$\alpha=\frac{a\pm\sqrt{a^2\pm4}}{2},  $$ for some natural number $a$. 
Let $\bar\alpha$ be the conjugate of $\alpha$, that is:  $\overline{\frac{a\pm\sqrt{a^2\pm4}}{2}}=\frac{a\mp\sqrt{a^2\pm4}}{2} $.
Now, for $n$ a natural number, let $F_n(\alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(\alpha):=  \alpha^{2^n}+\bar\alpha^{2^n}.  $$
$F_n(\alpha)$ is a natural number, since in the expansion of $F_n(\alpha)$ all the square roots cancel each other. 

Let $p$ be an odd prime divisor of $F_n(\alpha)$. Is it true that  $$ p^2 \equiv 1 \pmod {2^{n+1} } $$
Edit 21/12/18: More pecisely, is it true that
$$ p\equiv \left(\frac{a^2\pm 4}{p}\right) \pmod {2^{n+1} } $$
  where $ \left(\frac{\cdot}{p}\right)$ is the Legendre symbol.

As an example, with $\alpha = 2+\sqrt5$, we have: 
$$ F_4(\alpha)= 10749957122=2\cdot769\cdot2207\cdot3167 $$
and 
\begin{align*} 769 &\equiv  \left(\frac{5}{769}\right) = +1\pmod {2^5}\\
2207 &\equiv \left(\frac{5}{2207}\right)=-1 \pmod {2^5}\\
3167 &\equiv \left(\frac{5}{3167}\right)=-1 \pmod {2^5}
\end{align*}
I have checked many other $\alpha$ and $n$, but I could not find a counterexample to this.
Edit 21/12/18: a proof should be like in the answer to the above related question.

The conjecture is true.
Specifically this is a general case of $\alpha=\frac 12 \sqrt{a+2} \pm \frac 12 \sqrt{a-2}$.  We then have $\alpha^2+\alpha^{-2}=a$ and $\alpha\alpha'=1$.
From this we generate a series of $\alpha^n+\alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$.  When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.
We first see that A(-n)=A(n).  
We can demonstrate that A(x+n)=A(n)A(x)-A(x-n).  This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.  
Once this is done we consider the modular case, relative some prime $\pmod{p}$
The trick here is to show that $A(p)=a \pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.
This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies.  That is, if p | C(n), then p | C(mn).  This is why the fibonacci series looks like a base.
We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie,  $p \mod n = \pm 1$
The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors.  But the condition is perfectly general and in keeping with ordinary base theory.
PS.  I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).  
