2 Some intermediate results

Throughout this section, R denotes a nonzero subring of ${\mathbb C}$ with unity. Let $D \in R[x]$ . The following theorem of Dubickas and Steuding [Reference Dubickas and Steuding4] gives a structure of the solutions in $R[x]$ of the polynomial Pell equations of the form $P^2-DQ^2=1$ .

Theorem DS Let $D \in R[x]$ . If the polynomial Pell equation

(9) $$ \begin{align} P^2-DQ^2=1 \end{align} $$

has a nontrivial solution in $R[x]$ , then it has precisely four minimal solutions. If $(p,q)$ is one of these, then all the nontrivial solutions of (9) in $R[x]$ are obtained as $(\pm P, \pm Q)$ , where

$$ \begin{align*}P+Q \sqrt D= (p + q\sqrt D)^k \text{ for } k \in {\mathbb N}. \end{align*} $$

We prove the following key proposition, where we derive a few properties of the solutions of (9) in terms of those of the minimal solutions. This proposition is of independent interest.

Proposition 1 Let $D \in R[x]$ . For a solution $(p,q)$ of (9) in $R[x]$ , consider $P,Q\in R[x]$ , defined by $P+Q \sqrt D:= (p + q\sqrt D)^k$ for some $k \in {\mathbb N}$ . Suppose the leading coefficients of p and q are denoted by a and b, respectively. Then, the leading coefficients of P and Q are $2^{k-1} a^k$ and $2^{k-1} a^{k-1} b$ , respectively. Moreover, we have

$$ \begin{align*}\deg P= k \deg p, \ \ \deg Q= k \deg p- \frac{\deg D}{2}. \end{align*} $$

Proof We prove this by induction on k. Clearly, the statement is true for $k=1$ . We set, $P'+Q' \sqrt D:= (p + q\sqrt D)^{k-1}$ . By the induction hypothesis,

$$ \begin{align*}\deg P'= (k-1)\deg p, \ \ \deg Q'= (k-1) \deg p- \frac{\deg D}{2},\end{align*} $$

and the leading coefficients of $P'$ and $Q'$ are $2^{k-2} a^{k-1}$ and $2^{k-2} a^{k-2} b$ , respectively. Now, we have

$$ \begin{align*}P+Q \sqrt D= (P'+Q' \sqrt D) (p + q\sqrt D). \end{align*} $$

So $P=pP'+qQ'D$ and $Q=pQ'+qP'$ . If c denotes the leading coefficient of D, then from the relation $p^2-Dq^2=1$ , we have $a^2=b^2 c$ .

By the induction hypothesis, $pP'$ is of degree $k\deg p$ and its leading coefficient is $2^{k-2} a^{k}$ . Similarly, $qQ'D$ is of degree $k\deg p$ and its leading coefficient is $2^{k-2} a^{k-2} b^2c=2^{k-2} a^{k}$ . So $\deg P= k\deg p$ and the leading coefficient of P is $2^{k-1} a^k$ . Similarly, the degree of Q and its leading coefficient can be computed from the formula $Q=pQ'+qP'$ , using the induction hypothesis.

We now assume n is a nonzero integer and R is a subring of ${\mathbb C}$ containing $\sqrt n$ and its inverse. For the generalized polynomial Pell equation $P^2-DQ^2=n$ , there is a one–one correspondence between the solutions in $R[x]$ of (9) and those of (5). Multiplying a solution $(P,Q)$ of (9) by $\sqrt n$ we get a solution of (5). Similarly, multiplying a solution of (5) by $1/\sqrt n$ gives a solution of (9). So we have the following as an immediate corollary of Theorem DS.

Proposition 2 Let n be a nonzero integer, R a subring of ${\mathbb C}$ containing $\sqrt n$ and its inverse. If the generalized polynomial Pell equation (5) has a nontrivial solution in $R[x]$ , then it has precisely four minimal solutions. If $(p,q)$ is one of these, then all the nontrivial solutions of (5) in $R[x]$ are obtained as $(\pm P, \pm Q)$ , where

$$ \begin{align*}P+Q \sqrt D= \sqrt n \left( \frac{p}{\sqrt n} + \frac{q}{\sqrt n}\sqrt D\right)^k \text{ for some } k \in {\mathbb N}. \end{align*} $$

3 Proof of Theorem 1

We begin by noting that a minimal solution in ${\mathbb C}[x]$ of the generalized polynomial Pell equation (6) is given by

(10) $$ \begin{align} P=\frac{2nx+nc}{\sqrt{\Delta}}, \ Q=\frac{2n}{\sqrt{\Delta}}. \end{align} $$

If D is not irreducible in ${\mathbb Z}[x]$ , then $\Delta $ is a square in ${\mathbb Z}$ and therefore the solution (10) is in ${\mathbb Q}[x]$ . Now, if $\alpha ,\beta $ denote the roots of D in ${\mathbb Z}$ and $\alpha> \beta $ , then we can write

$$ \begin{align*}P=\frac{2nx+nc}{\alpha-\beta}, \ Q=\frac{2n}{\alpha-\beta}. \end{align*} $$

We first prove the case (a). Suppose $\alpha -\beta $ divides $2n$ . Then clearly, $Q \in {\mathbb Z}[x]$ . Note that if $\alpha -\beta $ is even, we get $c= -(\alpha +\beta )=\alpha -\beta -2\alpha $ is also even and therefore $P \in {\mathbb Z}[x]$ . Now, if $\alpha -\beta $ is odd, we have $\alpha -\beta $ divides n and hence $P \in {\mathbb Z}[x]$ .

Next, suppose that (6) has a solution in ${\mathbb Z}[x]$ . Since D is not irreducible in ${\mathbb Z}[x]$ , we already know that (10) is a minimal solution in ${\mathbb Q}[x]$ and hence by Proposition 2, a solution of (6) in ${\mathbb Z}[x]$ must be of the form $(\pm P_k, \pm Q_k),$ where

$$ \begin{align*}P_k + Q_k \sqrt D= n \left( \frac{P}{n} + \frac{Q}{n} \sqrt D\right)^k, \text{ for some } k \in {\mathbb N}. \end{align*} $$

To have $P_k,Q_k \in {\mathbb Z}[x]$ , we must have their leading coefficients in ${\mathbb Z}$ . Using Proposition 1, we therefore have

$$ \begin{align*}n 2^{k-1} \left( \frac{2}{\alpha-\beta} \right)^k = \left(\frac{2}{n}\right)^{k-1} \left( \frac{2n}{\alpha-\beta} \right)^{k} \in {\mathbb Z}. \end{align*} $$

If $(\alpha -\beta ) \nmid 2n$ , then there exists a prime number p such that $p \mid (\alpha -\beta )$ and the p-adic valuation $v_p({2n}/{(\alpha -\beta )})\le -1$ . So $v_p(({2n}/{(\alpha -\beta )})^k) \le -k$ , but $p^k \nmid 2^{k-1}$ . Hence, we must have $(\alpha -\beta ) \mid 2n$ . This completes the proof of the case (a).

Next, we prove the case (b). If D is irreducible in ${\mathbb Z}[x]$ , we have $\sqrt \Delta \notin {\mathbb Q}$ . Hence, the solution (10) is not in ${\mathbb Q}[x]$ . Note that $\tilde P, \tilde Q$ , given by

$$ \begin{align*}\tilde P + \tilde Q \sqrt D= n \left( \frac{P}{n} + \frac{Q}{n} \sqrt D\right)^2, \end{align*} $$

is a solution of (6), by Proposition 2. It is easy to check that

(11) $$ \begin{align} \tilde P=\frac{n(8x^2+8cx+c^2+4d)}{\Delta}, \ \tilde Q=\frac{n(8x+4c)}{\Delta}. \end{align} $$

Hence, $(\tilde P, \tilde Q)$ is a minimal solution in ${\mathbb Q}[x]$ of (6). Now suppose $\Delta \mid 8n$ . If c is even, we see that $\tilde Q \in {\mathbb Z}[x]$ , moreover, writing $n(c^2+4d)=n(c^2-4d+8d)=n \Delta + 8nd$ , we see that $\tilde P \in {\mathbb Z}[x]$ . Now, if c is odd then $\Delta $ is odd and therefore $\Delta \mid n$ . Hence, we have both $\tilde P,\tilde Q \in {\mathbb Z}[x]$ .

For the converse, note that by Proposition 2, a solution of (6) in ${\mathbb Z}[x]$ must be of the form $(\pm P_k, \pm Q_k),$ where

$$ \begin{align*}P_k + Q_k \sqrt D= n \left( \frac{\tilde P}{n} + \frac{\tilde Q}{n} \sqrt D\right)^k, \text{ for some } k \in {\mathbb N}. \end{align*} $$

Note that by Proposition 1, the leading coefficient of $P_k$ is

$$ \begin{align*}n 2^{k-1} \left(\frac{8}{\Delta} \right)^k=\left(\frac{2}{n}\right)^{k-1}\left(\frac{8n}{\Delta}\right)^{k}. \end{align*} $$

Now if this coefficient is in ${\mathbb Z}$ for some $k \in {\mathbb N}$ and $\Delta \nmid 8n$ , then there exists a prime number p such that $p \mid \Delta $ and the p-adic valuation $v_p({8n}/{\Delta })\le -1$ . So ${v_p(({8n}/{\Delta })^k) \le -k}$ , but $p^k \nmid 2^{k-1}$ . Hence, we must have $\Delta \mid 8n$ . This completes the proof of Theorem 1.

4 Proof of Theorem 2

We begin by noting that a minimal solution in ${\mathbb C}[x]$ of (7) is given by

(12) $$ \begin{align} P=\frac{\sqrt n (2x+c)}{\sqrt{\Delta}}, \ Q=\frac{2\sqrt n}{\sqrt{\Delta}}. \end{align} $$

Now, if $\sqrt {n/\Delta }$ is a rational number then the solution (12) is in ${\mathbb Q}[x]$ . Now suppose $\sqrt {n/\Delta }$ is a half-integer. Then, clearly $Q \in {\mathbb Z}[x]$ . If $\sqrt {n/\Delta }$ is an integer itself, then P is also in ${\mathbb Z}[x]$ . So we suppose $\sqrt {n/\Delta }=a/2$ for some odd integer a. By taking squares on both sides, we observe that $\Delta $ must be even, and therefore c must be even. Hence, in this case, also $P \in {\mathbb Z}[x]$ .

Now suppose (7) has a solution in ${\mathbb Q}[x]$ . By Proposition 2, a solution of (7) in ${\mathbb Q}[x]$ must be of the form $(\pm P_k, \pm Q_k),$ where

(13) $$ \begin{align} P_k + Q_k \sqrt D= \sqrt n \left( \frac{P}{\sqrt n} + \frac{Q}{\sqrt n} \sqrt D\right)^k, \end{align} $$

for some $k \in {\mathbb N}$ . By Proposition 1, the leading coefficient of $P_k$ is

(14) $$ \begin{align} \sqrt n 2^{k-1} \left( \frac{2}{\sqrt \Delta} \right)^k. \end{align} $$

If k is even, (14) is not in ${\mathbb Q}$ . So k must be an odd integer. If $k=2l+1$ , we have

$$ \begin{align*}\sqrt n 2^{k-1} \left( \frac{2}{\sqrt \Delta} \right)^k=2 \sqrt{\frac{n}{\Delta}} \left( \frac{16}{\Delta} \right)^l. \end{align*} $$

So to have this coefficient in ${\mathbb Q}$ , we must have $\sqrt {{n}/{\Delta }}$ is in ${\mathbb Q}$ .

Now, if (7) has a solution in ${\mathbb Z}[x]$ , we already have $\sqrt {{n}/{\Delta }}\in {\mathbb Q}$ . We write $\sqrt {{n}/{\Delta }}=a/b$ for integers $a,b$ with $\gcd (a,b)=1$ . Then, the coefficient (14) becomes

(15) $$ \begin{align} {\frac{2a}{b}} \left( \frac{16}{\Delta} \right)^l. \end{align} $$

As this coefficient is in ${\mathbb Z}$ , we must have $b=2^t$ for some $t \ge 0$ . Now, if $t \ge 2$ , we must have the $2$ -adic valuation $v_2(\Delta ) \ge 4$ . But in that case, the coefficient (15) cannot be in ${\mathbb Z}$ , as a is odd here. Hence, $b=1$ or $2$ , i.e., $\sqrt {{n}/{\Delta }}$ is a half-integer. This completes the proof of Theorem 2.

5 Recurrence relations satisfied by the solutions in ${\mathbb Q}[x]$

The proof of Theorem 1 tells us that (6) always has a nontrivial solution in ${\mathbb Q}[x]$ . Further Theorem 2 identifies the non-square monic quadratic polynomials D such that (7) has a nontrivial solution in ${\mathbb Q}[x]$ . In this section, we first derive the recurrence relations satisfied by these solutions.

We first consider equation (6), where $D=x^2+cx+d$ is non-square and not irreducible. Then, a minimal solution of (6) in ${\mathbb Q}[x]$ is given by (10). As in §3, let $\alpha ,\beta $ denote the roots of D in ${\mathbb Z}$ with $\alpha> \beta $ . Then, using Proposition 2, we get that all the solutions of (6) in ${\mathbb Q}[x]$ are given by $(\pm P_k, \pm Q_k),$ where $k \ge 0$ is an integer and

(16) $$ \begin{align} &P_0:=n, \ \ Q_0:=0, \nonumber\\ &P_k:=\frac{1}{\alpha-\beta} \left((2x+c) P_{k-1} + 2D Q_{k-1}\right), Q_k:=\frac{1}{\alpha-\beta} \left(2P_{k-1} + (2x+c) Q_{k-1}\right). \end{align} $$

Now, we consider equation (6), where $D=x^2+cx+d$ is irreducible. Here, a minimal solution of (6) in ${\mathbb Q}[x]$ is given by (11). Using Proposition 2, in this case, we get that all the solutions of (6) in ${\mathbb Q}[x]$ are given by $(\pm P_k, \pm Q_k),$ where $k \ge 0$ is an integer and

(17) $$ \begin{align} &P_0:=n, \ \ Q_0:=0, \nonumber\\ &P_k:=\frac{1}{\Delta} \left((8x^2+8cx+c^2+4d) P_{k-1} + (8x+4c)D Q_{k-1}\right),\nonumber\\ &Q_k:=\frac{1}{\Delta} \left((8x+4c)P_{k-1} + (8x^2+8cx+c^2+4d) Q_{k-1}\right). \end{align} $$

For equation (7), a minimal solution in ${\mathbb C}[x]$ is given by (12). In the proof of Theorem 2, we have seen that for $P_k, Q_k$ as in (13), we have $P_k, Q_k \notin {\mathbb Q}[x]$ for even values of k. In fact, when $\sqrt {n/\Delta } \in {\mathbb Q}$ , for the odd values of k, we have $P_k, Q_k \in {\mathbb Q}[x]$ . This is because, for

$$ \begin{align*}P^\circ=\frac{(2x+c)}{\sqrt{\Delta}}, \ Q^\circ=\frac{2}{\sqrt{\Delta}}, \end{align*} $$

the polynomials $\tilde P, \tilde Q$ are in ${\mathbb Q}[x]$ , where

$$ \begin{align*}\tilde P + \tilde Q \sqrt D:= (P^\circ+Q^\circ\sqrt D)^2. \end{align*} $$

Hence, in this case, we get that all the solutions of (7) in ${\mathbb Q}[x]$ are given by $(\pm P_k, \pm Q_k),$ where $k \ge 0$ is an integer and

(18) $$ \begin{align} &P_0:=\frac{\sqrt n (2x+c)}{\sqrt{\Delta}}, \ \ Q_0:=\frac{2\sqrt n}{\sqrt{\Delta}},\nonumber \\ &P_k:=\frac{1}{\Delta} \left((8x^2+8cx+c^2+4d) P_{k-1} + (8x+4c)D Q_{k-1}\right),\\ &Q_k:=\frac{1}{\Delta} \left((8x+4c)P_{k-1} + (8x^2+8cx+c^2+4d) Q_{k-1}\right).\nonumber \end{align} $$

The above recurrence relations will be useful in proving Theorem 3.

6 Proof of Theorem 3

Let $n \in {\mathbb Z}$ be a nonzero integer and D be Pellian over ${\mathbb Z}$ with norm n. We first prove that if $\Delta \nmid 16$ , then there are only finitely many solutions of (5) in ${\mathbb Z}[x]$ . We first suppose that D is not irreducible and (6) has a nontrivial solution in ${\mathbb Z}[x]$ . We have seen in the proof of Theorem 1 that for $P_k$ , as in (16), its leading coefficient is

$$ \begin{align*}n 2^{k-1} \left( \frac{2}{\alpha-\beta} \right)^k=\frac{n}{2}\left( \frac{4}{\alpha-\beta} \right)^k. \end{align*} $$

So if $\alpha -\beta \nmid 4$ , i.e., $\Delta \nmid 16$ , the above coefficient can be in ${\mathbb Z}$ for only finitely many values of k. Next, let D be irreducible and (6) has a nontrivial solution in ${\mathbb Z}[x]$ . We have seen that for $P_k$ as in (17), its leading coefficient is

$$ \begin{align*}n 2^{k-1} \left(\frac{8}{\Delta} \right)^k=\frac{n}{2}\left(\frac{16}{\Delta}\right)^{k}. \end{align*} $$

Hence, again if $\Delta \nmid 16$ , the above coefficient can be in ${\mathbb Z}$ for only finitely many values of k. Now, if D be such that (7) has a nontrivial solution in ${\mathbb Z}[x]$ , then for $P_k$ as in (18), its leading coefficient is

$$ \begin{align*}2 \sqrt{\frac{n}{\Delta}} \left( \frac{16}{\Delta} \right)^k. \end{align*} $$

So the above coefficient can be in ${\mathbb Z}$ for only finitely many values of k, if $\Delta \nmid 16$ .

For the converse, it is enough to prove that if $\Delta \mid 16$ , then the nontrivial solutions of (5) in ${\mathbb Z}[x]$ (if exists) are the same as those in ${\mathbb Q}[x]$ . Note that if $\Delta =1,4,16$ , then D is not irreducible and for the remaining values of $\Delta $ , D is irreducible. Note that the values $\Delta =-1,\pm 2$ are not admissible.

If $\Delta =1$ , it is immediate that all the solutions in (16) are in ${\mathbb Z}[x]$ . Similarly, if $\Delta =4$ , we further have c is even and since $\alpha -\beta =2$ , all the solutions in (16) are in ${\mathbb Z}[x]$ .

Now, if $\Delta =-4, \pm 8$ , we have c even. So writing $c^2+4d=\Delta +8d$ , we see that the solutions given by (17) and (18) are in ${\mathbb Z}[x]$ , provided the initial solutions are there in ${\mathbb Z}[x]$ , which is clearly true under the given hypothesis.

Next, we consider $\Delta =\pm 16$ . The proof of Theorem 3 then follows from the lemma below.

Lemma 1 Let $\Delta \in \{\pm 16\}$ and (5) has a nontrivial solution in ${\mathbb Z}[x]$ . For $k \ge 0$ and $P_k,Q_k$ as in (16), or (17), or (18), we have

$$ \begin{align*}P_k,Q_k,\frac{1}{2}\left(P_k+\left(x+\frac{c}{2}\right)Q_k\right) \in {\mathbb Z}[x]. \end{align*} $$

Proof We prove this by induction on k. This is true for $k=0$ , as n in (5) is even here.

For $\Delta =16$ , we have $\alpha -\beta =4$ and c is even, say $2e$ . Then $D=(x+e)^2-e^2+d=(x+e)^2-4$ . So the recurrence relation (16) gives

$$ \begin{align*} &P_k=\frac{1}{\alpha-\beta} \left((2x+c) P_{k-1} + 2D Q_{k-1}\right)=\frac{x+e}{2}\left(P_{k-1}+(x+e)Q_{k-1}\right) - 2Q_{k-1}, \\ &Q_k=\frac{1}{\alpha-\beta} \left(2P_{k-1} + (2x+c) Q_{k-1}\right)=\frac{1}{2}\left(P_{k-1}+(x+e)Q_{k-1}\right). \end{align*} $$

Hence, by the induction hypothesis, we have $P_k, Q_k \in {\mathbb Z}[x]$ . Now, using (16), we also get

$$ \begin{align*} \frac{1}{2}({P_k} + (x+e) {Q_k}) &= \frac{1}{2} \left( \frac{x+e}{2} P_{k-1} + \frac{(x+e)^2-4}{2} Q_{k-1} + \frac{x+e}{2} P_{k-1} + \frac{(x+e)^2}{2} Q_{k-1}\right) \\ &= \frac{x+e}{2} \left(P_{k-1} + (x+e) Q_{k-1}\right) - Q_{k-1}, \end{align*} $$

which therefore is in ${\mathbb Z}[x]$ , by the induction hypothesis.

Next, let $\Delta =-16$ . Here, also c is even, say $2e$ . So $D=(x+e)^2-e^2+d=(x+e)^2+4$ . Now, using (17) or (18) (whichever is applicable), we have

$$ \begin{align*} P_k&= - \frac{1}{16} \left((8x^2+16ex+4e^2+4d) P_{k-1} + (8x+8e)\left((x+e)^2+4\right) Q_{k-1}\right) \\ &= -\left(\frac{(x+e)^2}{2}+ 1\right) P_{k-1} - \frac{(x+e)((x+e)^2+4)}{2} Q_{k-1}\\ &=-\frac{(x+e)^2}{2}\left({P_{k-1}} + (x+e) {Q_{k-1}}\right) - P_{k-1} - 2(x+e)Q_{k-1}, \end{align*} $$

and

$$ \begin{align*} Q_k&= - \frac{1}{16} \left((8x+8e)P_{k-1} + (8x^2+16ex+4e^2+4d) Q_{k-1}\right)\\ &=-\frac{x+e}{2}P_{k-1}-\left(\frac{(x+e)^2}{2}+ 1\right) Q_{k-1} =-\frac{x+e}{2}\left({P_{k-1}} + (x+e) {Q_{k-1}}\right) - Q_{k-1}. \end{align*} $$

Hence, by the induction hypothesis, we have $P_k, Q_k \in {\mathbb Z}[x]$ . And finally,

$$ \begin{align*}&\frac{1}{2}\left({P_k} + (x+e) {Q_k}\right)\\ & \quad = -\frac{(x+e)^2}{2} \left(P_{k-1} + (x+e) Q_{k-1}\right) - \frac{1}{2} \left(P_{k-1} + (x+e) Q_{k-1}\right) - (x+e)Q_{k-1} \end{align*} $$

is also in ${\mathbb Z}[x]$ , by the induction hypothesis.

7 Concluding remarks

Our theorems here give concrete descriptions of integer and rational solutions of the generalized polynomial Pell equation (5). Depending on whether n in (5) is a square or not, there are different facets of the rational solutions in comparison with the complex solutions. For instance, when D is not irreducible in ${\mathbb Z}[x]$ , the solutions of (6) in ${\mathbb Q}[x]$ are the same as those in ${\mathbb C}[x]$ , whereas in this case, there is no solution of (7) in ${\mathbb Q}[x]$ . Further, if D is irreducible in ${\mathbb Z}[x]$ , the nontrivial solutions of (6) in ${\mathbb Q}[x]$ are obtained by considering the positive even integer powers of a minimal solution of (9) in ${\mathbb C}[x]$ and multiplying $\pm n$ to them. On the other hand, for irreducible D, the nontrivial solutions of (7) in ${\mathbb Q}[x]$ are obtained by considering the positive odd integer powers of a minimal solution of (9) in ${\mathbb C}[x]$ and multiplying $\pm \sqrt n$ to them.

The contrasting features encountered in dealing with the generalized polynomial Pell equations, as opposed to the polynomial Pell equations, lead to interesting questions, for example, the one answered in Theorem 3. If we consider the negative polynomial Pell equation (8), then for any D such that (8) has a nontrivial integer solution, we see that not only (8) has infinitely many integer solutions, but all the rational solutions of (8) are in ${\mathbb Z}[x]$ , as $\Delta =-4$ here. On the other hand, if we consider the generalized polynomial Pell equation (5), with $n=3$ , we see that (5) has nontrivial integer solutions if and only if $\Delta =12$ . Since $12 \nmid 16$ , we will have only finitely many such integer solutions in that case.

There are several other questions that we can consider. For example, we can consider D that are non-monic. This study does not follow from analogous arguments, primarily because for a minimal solution $(P,Q)$ , the leading coefficients of P and Q are going to be different and thereby for all the solutions, using Proposition 1. This makes the study much more challenging and it is a subject of our forthcoming work. We have also been studying analogous questions where D is a quartic polynomial, following the work of Scherr and Thompson [Reference Scherr and Thompson15]. Further, Proposition 2 allows us to study the generalized polynomial Pell equations over more general rings. But since in this article our focus is to study the integral solutions in depth, we relegate this to future work.