I have a question concering the family of Pellian equations
$$x^2 – (k^2+1)y^2 = k^2. \qquad (*)$$
For an integer $k\geq 2$, the equation (*) has at least three classes of solutions
in integers, corresponding to the fundamental solutions
$(x_0,y_0) = (k, 0)$, $(k^2-k+1, k-1)$, $(k^2-k+1, -(k-1))$.
Each fundamental solution induces a sequence (class) of solutions by
$$ x+\sqrt{k^2+1}y = (x_0+\sqrt{k^2+1}y_0)(k+\sqrt{k^2+1})^{2m}. $$
For certain values of $k$, there are two additional classes
of solutions. E.g. for $k=2t^2$, we have also fundamental
solutions $(x_0,y_0) = (2t^3+t, \pm t)$, and additional
solutions occur also for $k=4s^3-4s^2+3s-1$,
for $k=F_{2n}$, and other polynomial or exponential subfamilies.
However, I am not able to find any example such that (*)
has more than five classes of solutions.
So, I am wondering does it make sense to state the conjecture
that for $k\geq 2$, equation (*) always have
exactly three or five classes of solutions
(e.g. 3 or 5 fundamental solutions).
Is there an obvious reason why should (or should not)
the number of fundamental solutions of (*) be bounded
by an absolute constant (independent on $k$)?
It is easy to see that each fundamental solution $(x_0,y_0)$
has to satisfy $|y_0| < k$, so the conjecture actually says
that there is at most one solution of equation (*) with
$0 < y < k-1$.
This question is related to the conjecture which
says that there does not exist a set of four
positive integer with the property that the
product of any two of them is 1 greater than a square
(see e.g. Section 3.1 of Diophantine m-tuples page
and references given there).
Best Answer
Oh, I think the answer is definitely yes!
Let $\{k \to x,y\}$ be any solution of $x^2 - (k^2+1)y^2 = k^2$, and let $K$ be the set of $k$ for which a solution has $0 < k < y-1$. In a paper recently submitted to Glasnik Matematicki we call these solutions exceptional solutions. Andrej's conjecture is that for any $k$ there is at most 1 exceptional solution.
One interesting result we obtain is that, if $k \in K$, then $y < (2 - \sqrt{3})k$.
A particular feature of this Pell equation is its symmetry wrt $k$ and $y$. These are interchangeable, so for any solution $\{k \to x,y\}$ there is a corresponding solution $\{y \to x,k\}$.
It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.
Now, for any $k \geq 2$, we have 3 particular solution classes $(x_n, y_n)$ with $y_0 = \{0, k-1, -(k-1)\}$. For any $n > \{0, 0, 1\}$ we have $y_n > k-1$ and so $\{y_n \to x_n,k\}$ is exceptional, ie. $y_n \in K$.
We also need to consider $k=1$, for which there is just the one class with $(x_0, y_0) = (1, 0)$ , $(x_1, y_1) = (3, 2)$ and so for any $n > 1$ we have $y_n \in K$. For example $(x_2, y_2) = (17, 12)$ from which we obtain exceptional solution $\{12 \to 17, 1\}$, and so $12 \in K$.
In our paper we call the corresponding set of exceptional solutions "Type 1". But here let us simply define the set $K_1$ to be all of these $y_n > k$ that we find from these 3 classes for any $k > 1$, and from the one class for $k = 1$.
One property shared by all type 1 solutions, ie $\{k \to x, y\}$ with $k \in K_1$ is that either $(y^2 +1) | (x+y)$ or $(y^2 + 1) |(x-y)$.
Now, for any $k \in K_1$ we have a corresponding $\{k \to x,y\}$ for which our Pell eqn has 2 additional solution classes, with fundamental solutions $(x_0, y_0) = (x, \pm y)$. For any $n > \{0,1\}$ we then have $y_n > k-1$ and so $\{y_n \to x_n, k\}$ is exceptional, ie $y_n \in K$.
And of course we can apply the same process to any of these new $y_n$ ad infinitum, each $y_n$ seeding a forest of others. For example, just considering $n = 1$ alone in each case, from $\{8 \to 18,\pm{2}\}, 8 \in K_1$ we obtain $\{546 \to 4402,8\}$ and $\{30 \to 242,8\}$, so $546, 30 \in K$, and from $\{30 \to 242, \pm{8}\}$ we get $\{28928 \to 868322,242\}$ and $\{112 \to 3362,30\}$, so $28928, 112 \in K$.
We call these "Type 2" solutions, so let's define $K_2$ to be all of the $y_n$ found this way. These do not have the divisibility property that was noted above for the $y_n \in K_1$.
In the paper we show that all exceptional solutions can be enumerated recursively in this fashion, ie. that $K = K_1 \cup K_2$. This is done by showing any $k \in K$ can be traced back to a root in $K_1$.
The enumeration algorithm is given below. Solution classes are referred to as $0, -1, +1$, the interpretation of which I hope is reasonably clear!
Proc Enum_K:
Proc Enum_K1(k, class):
Proc Enum_K2(k, class):
To generate the solution sequences in any class, we note that each class has the same recurrence relation:
but of course have different initial conditions:
Now if Andrej's conjecture is true, and we believe it is, then each operation "add $y_n$" always adds a new $y_n$ to its list, and the two lists $K_1, K_2$ have no common elements.
An implementation of Enum_K with a bailout parameter finds that with $k <10^6$ we have $|K_1| = 882, |K_2| = 163, |K| = 1045$, and that every $k$ enumerated was unique. This agrees with Andrej's figure.