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I will illustrate the enumeration process with some examples in order to make clear the structure described above.

We start with $k = 1$, the only case with a single solution class $(k, 0)$. We have $k^2+1 = 2$ and $k^2 = 1$. Here is a partial enumeration of all solutions to $x^2 - 2y^2 = 1$:

       n            x            y
0            1            0
1            3            2
2           17           12
3           99           70
4          577          408
5         3363         2378


Because of the symmetry of the equation wrt $k$ and $y$ we know that each pair $(x_n, y_n)$ for $n > 1$ means that $\{y_n \to x_n, 1\}$ is an exceptional solution. For example, we can see that $17^2 - (12^2 + 1).1^2 = 12^2$. Clearly $1 < 12-1$, and so $k = 12$ has an exceptional solution, and because it came from an enumeration of a root class $(1, 0)$ it is a type-1 solution and so we add $12$ to the set $K_1$.

For all $k > 1$ we have 3 root classes, $(k, 0)$, $(k^2-k+1, k-1)$ and $(k^2-k+1, -k+1)$. Partial enumerations for $k=2$ are shown below:

       n            x            y
0            2            0
1           18            8
2          322          144
3         5778         2584
4       103682        46368
5      1860498       832040

n            x            y
0            3            1
1           47           21
2          843          377
3        15127         6765
4       271443       121393

n            x            y
0            3           -1
1            7            3
2          123           55
3         2207          987
4        39603        17711
5       710647       317811


Each $y_n$ where $n>0$ (or $n>1$ for the 3rd class) provides an exceptional solution $\{y_n \to x_n, 2\}$, and so each $y_n$ is added to $K_1$.

Now every value we add to $K_1$ is an exceptional solution of the form $\{k \to x,y\}$, so for each $k$ in $K_1$ we have an additional pair of conjugate solution classes $(x, \pm{y})$. Assuming the Dujella conjecture is true, these will always be the 4th and 5th classes.

We simply enumerate these classes in similar fashion, except we add the new $y_n$ values to the list $K_2$, since they come from these additional classes for $k \in K_1$, not from the 3 root classes. For example taking exceptional solution $\{18 \to 8,2\}$, we enumerate the classes $(8, 2)$ and $(8, -2)$ for $k=18$:

       n            x            y
0            18           2
1          4402         546
2       1135698      140866
3     293005682    36342882

n            x            y
0            18          -2
1           242          30
2         62418        7742
3      16103602     1997406


Again, every $(x_n, y_n)$ for $n>0$ gives a new exceptional solution $\{y_n \to x_n,18\}$, and so we add each $y_n$ to $K_2$. And every item $k$ we add to $K_2$ represents 2 new classes for that $k$, so we can apply the same procedure recursively to each and every one.

The reason that I have kept $K_1$ and $K_2$ as two distinct lists is that the members of $K_1$ have properties not shared by $K_2$. The divisibility property noted above is one such property, another is the fact that all of the root classes for any $k$, from which we poulate $K_1$, have explicit polynomial descriptions, which lend themselves to the sort of analysis that we can't readily apply to $K_2$.

For example, we can (I believe) deduce from the properties of these polynomials that every operation "add $y_n$ to $K_1$" provides a unique value. I also believe that I will be able It remains to demonstrate that no value in $K_2$ be seen whether we can also occur in $K_1$. If this is in fact true, then prove the only remaining hurdle is a proof that every operation "add $y_n$ to same holds for $K_2$" adds a unique value.K_2$. 3 deleted 64 characters in body; deleted 23 characters in body; deleted 86 characters in body I will illustrate the enumeration process with some examples in order to make clear the structure described above. We start with$k = 1$, the only case with a single solution class$(k, 0)$. We have$k^2+1 = 2$and$k^2 = 1$. Here is a partial enumeration of all solutions to $x^2 - 2y^2 = 1$:  n x y 0 1 0 1 3 2 2 17 12 3 99 70 4 577 408 5 3363 2378  Because of the symmetry of the equation wrt$k$and$y$we know that each pair$(x_n, y_n)$for $n > 1$ means that $\{y_n \to x_n, 1\}$ is an exceptional solution. For example, we can see that $17^2 - (12^2 + 1).1^2 = 12^2$. Clearly $1 < 12-1$, and so$k = 12$has an exceptional solution, and because it came from an enumeration of a root class$(1, 0)$it is a type-1 solution and so we add$12$to the set$K_1$. For all $k > 1$ we have 3 root classes,$(k, 0)$,$(k^2-k+1, k-1)$and$(k^2-k+1, -k+1)$. Partial enumerations for$k=2$are shown below:  n x y 0 2 0 1 18 8 2 322 144 3 5778 2584 4 103682 46368 5 1860498 832040 n x y 0 3 1 1 47 21 2 843 377 3 15127 6765 4 271443 121393 n x y 0 3 -1 1 7 3 2 123 55 3 2207 987 4 39603 17711 5 710647 317811  Each$y_n$where $n>0$ (or $n>1$ for the 3rd class) provides an exceptional solution $\{y_n \to x_n, 2\}$, and so each$y_n$is added to$K_1$. Now every value we add to$K_1$is an exceptional solution of the form $\{k \to x,y\}$, so for each$k$in$K_1$we have an additional pair of conjugate solution classes $(x, \pm{y})$. Assuming the Dujella conjecture is true, these will always be the 4th and 5th classes. We simply enumerate these classes in similar fashion, except we add the new$y_n$values to the list$K_2$, since they come from these additional classes for$k \in K_1$, not from the 3 root classes. For example taking exceptional solution $\{18 \to 8,2\}$, we enumerate the classes$(8, 2)$and$(8, -2)$for$k=18$:  n x y 0 18 2 1 4402 546 2 1135698 140866 3 293005682 36342882 n x y 0 18 -2 1 242 30 2 62418 7742 3 16103602 1997406  Again, every$(x_n, y_n)$for $n>0$ gives a new exceptional solution $\{y_n \to x_n,18\}$, and so we add each$y_n$to$K_2$. And every item$k$we add to$K_2$represents 2 new classes for that$k$, so we can apply the same procedure recursively to each and every one. The reason that I have kept$K_1$and$K_2$as two distinct lists is that the members of$K_1$have properties not shared by$K_2$. The divisibility property noted above is one such property, another is the fact that all of the root classes for any$k$, from which we poulate$K_1$, have explicit polynomial descriptions, which lend themselves to the sort of analysis that we can't readily apply to$K_2$. For example, we can (I believe) deduce from the properties of these polynomials that every operation "$add y_n add $y_n$ to K_1$"$K_1$" provides a unique value. I am also confident believe that I will be able to demonstrate that no value in$K_2$can also occur in$K_1$. If this is in fact true, then the only remaining hurdle is a proof that every operation "$add y_n add $y_n$ to K_2$"$K_2$" adds a unique value. 2 added 41 characters in body I will illustrate the enumeration process with some examples in order to make clear the structure described above. We start with$k = 1$, the only case with a single solution class$(k, 0)$. We have$k^2+1 = 2$and$k^2 = 1$. Here is a partial enumeration of all solutions to $x^2 - 2y^2 = 1$:  n x y 0 1 0 1 3 2 2 17 12 3 99 70 4 577 408 5 3363 2378  Because of the symmetry of the equation wrt$k, k$and$y$we know that each pair$(x_n, y_n)$for $n > > 1$ means that $\{y_n \to x_n, 1\}$ is an exceptional solution. For example, we can see that $17^2 - (12^2 + 1).1^2 = 12^2$. Clearly $1 < 12-1$, and so$k = 12$has an exceptional solution, and because it came from an enumeration of a root class$(1, 0)$it is a type-1 solution and so we add$12$to the set$K_1$. For all $k > > 1$ we have 3 root classes,$(k, 0)$,$(k^2-k+1, k-1)$and$(k^2-k+1, -k+1)$. Partial enumerations for$k=2$are shown below:  n x y 0 2 0 1 18 8 2 322 144 3 5778 2584 4 103682 46368 5 1860498 832040   n x y 0 3 1 1 47 21 2 843 377 3 15127 6765 4 271443 121393   n x y 0 3 -1 1 7 3 2 123 55 3 2207 987 4 39603 17711 5 710647 317811  Each$y_n$where $n>0$n>0$ ($n>1$ or $n>1$ for the 3rd class) provides an exceptional solution $\{y_n \to x_n, 2\}$, and so each $y_n$ is thus added to $K_1$.<br><br>

Now every value we add to $K_1$ is an exceptional solution of the form ${k \{k \to x,y}$x,y\}$, so for each$k$in$K_1$we have an additional pair of conjugate solution classes $(x, \pm{y})$. pm{y})$. Assuming the Dujella conjecture is true, these will always be the 4th and 5th classes.<br><br>

We simply enumerate these classes in similar fashion, except we add the new $y_n$ values to the list $K_2$, since they come from these additional classes for $k \in K_1$, not from the 3 root classes. For example taking exceptional solution ${18 \{18 \to 8,2}$8,2\}$, we enumerate the classes$(8, 2)$and$(8, -2)$for$k=18$:  n x y 0 18 2 1 4402 546 2 1135698 140866 3 293005682 36342882  n x y 0 18 -2 1 242 30 2 62418 7742 3 16103602 1997406 Again, every$(x_n, y_n)$for $n>0$n>0$ gives a new exceptional solution $\{y_n \to x_n,18\}$`, and so we add each $y_n$ to $K_2$. And every item $k$ we add to $K_2$ represents 2 new classes for that $k$, so we can apply the same procedure recursively to each and every one.
The reason that I keep have kept $K_1$ and $K_2$ as two distinct lists is that the members of $K_1$ have properties not shared by $K_2$. The divisibility property noted above is one such property, another is the fact that all of the root classes for any $k$, from which we poulate $K_1$, have explicit polynomial descriptions, which lend themselves to the sort of analysis that we can't yet readily apply to $K_2$.

For example, we can (I believe) deduce from the properties of these polynomials that every operation "$add y_n to K_1$" provides a unique value. I am also confident that I will be able to demonstrate that no value in $K_2$ can also occur in $K_1$. If this is in fact true, then the only remaining hurdle is a proof that every operation "$add y_n to K_2$" adds a unique value.

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