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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 $y_n \in K_1$ is that either $(y_n^2 +1) | (x_n+y_n)$ or $(y_n^2 + 1) |(x_n-y_n)$.

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!

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!

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 $y_n \in K_1$ is that either $y_n | (x_n+y_n)$ or $y_n |(x_n-y_n)$.

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!

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 $y_n \in K_1$ is that either $(y_n^2 +1) | (x_n+y_n)$ or $(y_n^2 + 1) |(x_n-y_n)$.

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!

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 $y_n \in K_1$ is that either $y_n | (x_n+y_n)$ or $y_n |(x_n-y_n)$.

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 which I hope is reasonably clear!

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 $y_n \in K_1$ is that either $y_n | (x_n+y_n)$ or $y_n |(x_n-y_n)$.

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!