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.

  • $\begingroup$ 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 $$ 4,18, 322, 103682, 1074995712,... $$ $\endgroup$
    – Will Jagy
    Dec 11, 2018 at 0:04
  • $\begingroup$ But in your example with $\alpha = 2+\sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree... $\endgroup$
    – Servaes
    Dec 11, 2018 at 0:09
  • $\begingroup$ Rene, see my answer at the question you link. $\endgroup$
    – Will Jagy
    Dec 11, 2018 at 2:28
  • $\begingroup$ @Servaes You are right, the squarefree condition is not necessary. I removed it. $\endgroup$
    – René Gy
    Dec 13, 2018 at 17:44

1 Answer 1


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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .