# 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.

• I see: when you choose $\frac{a+ \sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$4,18, 322, 103682, 1074995712,...$$ Dec 11, 2018 at 0:04
• But in your example with $\alpha = 2+\sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree... Dec 11, 2018 at 0:09
• Rene, see my answer at the question you link. Dec 11, 2018 at 2:28
• @Servaes You are right, the squarefree condition is not necessary. I removed it. Dec 13, 2018 at 17:44

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).