For integer $n$, let $P_n$ be a Pell number, and $Q_n$ its companion. Is there a characterization of the prime numbers $p$ which don't divide any $Q_n$?
By brute-force search, I found that this appears to be true for all primes $p \equiv 5\!\pmod{8}$. I was going to conjecture [and eventually prove] that this was the complete characterization… until I discovered that it's also true for $p=137$. Are there any other such primes (i.e. $\not\equiv 5\!\pmod{8}$?
Note: I have skimmed through the relevant section (IV) of Ribenboim's The Little Book of Bigger Primes, but the answer isn't explicit there, and I wanted to make sure I wasn't reinventing the wheel before I go reinvent this wheel.
EDIT: In Ribenboim [p. 57], it says
if $p \nmid 2DQ$, $(Q\mid p)=1$, and $(D \mid p) = -(-1 \mid p)$, then $p$ [is such a prime]. This does not determine, without a further analysis, whether a prime $p$, such that $p \nmid 2DQ$, $(Q\mid p)=1$, and $(D \mid p) = (-1 \mid p)$ belongs, or does not belong, to [the set of such primes].
This is a description in the context of general Lucas sequences. In the particular case of the Pell numbers (and companions), the relevant constants are $Q=-1$ and $D=8$. So Ribenboim is saying that primes $p \equiv 5\!\pmod{8}$ are certainly that in set (hence a proof of my initial ``conjecture''), but that nothing immediately eliminates or characterizes the primes $p=137,\dots(?)$ that are also in that set. I was/am just hoping someone else has already done the heavy lifting on this one.
Well, $x^2 - 2 y^2 = \pm 1.$ We want to know about primes dividing $x,$ which will then divide $2 y^2 \pm 1.$ If prime $p \equiv 5 \pmod 8,$ then $2y^2 - 1 \equiv 0 \pmod p$ is impossible, as $2y^2 \equiv 1 \pmod p$ implies $(2|p) = 1,$ which is false. Also $2y^2 + 1 \equiv 0 \pmod p$ is impossible, as $2y^2 \equiv -1 \pmod p$ implies $(-2|p) = 1,$ which is false.
Other than that, I would emphasize projective Pisano periods; mod 5, the sequence $1,3,7,17,41,99$ becomes $1,3,2,2,4,...$ where the quadruple $3,2,2,4$ is of type $-A,A,A,3A \pmod 5$ for $A=2.$ So we have a nonzero multiple of $1,3$ before getting a $0,$ which is a hands on way to show it never equals $0.$ Mod 137, within two dozen steps we get $37,100,100,26 \pmod {137}$ before seeing any $0,$ so that does it. We need only note the pair $100,100$ to have our conclusion.
Maybe there is a way to show all $p \equiv 3 \pmod 4$ do eventually divide some $x,$ not sure.
Evidently the misses are 5 mod 8 and a selection of 1 mod 8. So, worth looking at proof for 3 mod 4.
It appears that the peculiar primes are all of the form $p = x^2 + 128 y^2,$ about two thirds of those. I am, so far, unable to find a quadratic form that represents the peculiar primes $137, 521, 569,593,$ and nothing else. The figure $2/3$ suggests Chebotarev density, I suspect there is a polynomial with integer coefficients, call it degree $n,$ such that these numbers are precisely those for which there are $n$ distinct roots $\pmod p.$ Not so easy to find the polynomial, though.
Meanwhile, it appears that if a prime is either $(4,4,9)$ or $(4,4,33)$ it is ordinary, does divide some Pell companion. Put another way, if either $p = x^2 + 8 y^2$ with both $x,y$ odd, or $p = x^2 + 32 y^2$ with both $x,y$ odd, then $p$ divides some Pell companion.