Dtyjlui

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:
$$alpha=frac{apmsqrt{a^2pm4}}{2}, $$ for some natural number $a$.

Let $baralpha$ be the conjugate of $alpha$, that is: $overline{frac{apmsqrt{a^2pm4}}{2}}=frac{ampsqrt{a^2pm4}}{2} $.

Now, for $n$ a natural number, let $F_n(alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(alpha):= alpha^{2^n}+baralpha^{2^n}. $$

$F_n(alpha)$ 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
$$ pequiv left(frac{a^2pm 4}{p}right) pmod {2^{n+1} } $$
where $ left(frac{cdot}{p}right)$ is the Legendre symbol.

As an example, with $alpha = 2+sqrt5$, we have:

$$ F_4(alpha)= 10749957122=2cdot769cdot2207cdot3167 $$
and
begin{align*} 769 &equiv left(frac{5}{769}right) = +1pmod {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 $alphaalpha'=1$.

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).