Proof of Theorem 1.2.

A minimal solution of (1.4) in ${\mathbb {C}}[x]$ is

$$ \begin{align*}(P,Q)=\bigg(\dfrac{n(2f+a)}{\sqrt{\Delta}},\dfrac{2n}{\sqrt{\Delta}}\bigg).\end{align*} $$

By Theorem 1.9, all nontrivial solutions of (1.4) in ${\mathbb {C}}[x]$ are $(\pm P_k,\pm Q_k)$ , where

(2.1) $$ \begin{align}P_k+Q_k\sqrt{f^2+af+b}=n\bigg(\dfrac{2f+a+2\sqrt{f^2+af+b}}{\sqrt{\Delta}}\bigg)^k,\quad k\geq 1. \end{align} $$

Since $\alpha $ is the leading coefficient of f, by Theorem 1.9, the leading coefficients of $P_k$ and $Q_k$ are $n2^{k-1}{(2\alpha )^k}/{\sqrt {\Delta }^k}$ and $n2^{k-1}{(2\alpha )^{k-1}}/{\sqrt {\Delta }^{k}}$ , respectively. Suppose (1.4) has a nontrivial solution in ${\mathbb {Z}}[x]$ . Then, there exists a positive integer k such that $P_k, Q_k\in {\mathbb {Z}}[x]$ . Therefore,

(2.2) $$ \begin{align} n2^{k-1}\dfrac{(2\alpha)^k}{\sqrt{\Delta}^k}\in {\mathbb{Z}} \quad\text{and}\quad n2^{k-1}\dfrac{(2\alpha)^{k-1}}{\sqrt{\Delta}^{k}}\in {\mathbb{Z}}. \end{align} $$

Case 1: $\sqrt {\Delta }\in {\mathbb {Z}}$ . Then, $x^2+ax+b$ can be factorised as $x^2+ax+b=(x-u)(x-v)$ , with $u,v\in {\mathbb {Z}}$ and $\sqrt {\Delta }=u-v>0$ . By (2.2),

(2.3) $$ \begin{align} \dfrac{n2^{2k-2}\alpha^{k-1}}{(u-v)^k}=n2^{k-1}\dfrac{(2\alpha)^{k-1}}{\sqrt{\Delta}^{k}}\in {\mathbb{Z}}. \end{align} $$

We claim that $(u-v)\mid 2n$ . The case $u-v=1$ is immediate. Consider the case ${u-v\geq 2}$ . Let p be a prime factor of $u-v$ . We consider three subcases.

Subcase (i): $p=2$ . Because $2\nmid \alpha $ , it follows from (2.3) that

$$ \begin{align*}kv_2(u-v)\leq 2k-2+v_2(n)<2k+v_2(n).\end{align*} $$

Hence,

(2.4) $$ \begin{align}v_2(u-v)\leq 1+v_2(n).\end{align} $$

Subcase (ii): $p>2$ and $p\mid n$ . Then, $p\nmid \alpha $ and it follows from (2.3) that

(2.5) $$ \begin{align}v_p(u-v)\leq kv_p(u-v)\leq v_p(n).\end{align} $$

Subcase (iii): $p>2$ and $p\mid \alpha $ . Since $\alpha $ is square-free and $\gcd (n,\alpha )=1$ , we have $v_p(\alpha )=1$ and $v_p(n)=0$ . It follows from (2.3) that

$$ \begin{align*}k\leq kv_p(u-v)\leq v_p(\alpha^{k-1})=k-1, \end{align*} $$

which is impossible.

Inequalities (2.4) and (2.5) show that $(u-v)\mid 2n$ . Conversely, assume ${\sqrt {\Delta }=(u-v)\mid 2n}$ . Then, (1.4) has a nontrivial solution in ${\mathbb {Q}}[x]$ with $Q\in {\mathbb {Z}}[x]$ :

$$ \begin{align*} (P,Q)=\bigg(\dfrac{2nf+na}{\sqrt{\Delta}},\dfrac{2n}{\sqrt{\Delta}}\bigg). \end{align*} $$

If $2\mid a$ , then $P\in {\mathbb {Z}}[x]$ . If $2\nmid a$ , then since $a^2-4b=\Delta =(u-v)^2\mid 4n^2$ , we have ${u-v\mid {n}}$ . Thus, $P\in {\mathbb {Z}}[x]$ .

Case 2: $\sqrt {\Delta }\not \in {\mathbb {Z}}$ . It follows from (2.2) that k is even. Let $k=2m$ , with $m\in {\mathbb {Z}}_{>0}$ . Then,

(2.6) $$ \begin{align} \dfrac{n2^{4m-1}\alpha^{2m-1}}{\Delta^m}=n2^{k}\dfrac{(2\alpha)^{k-1}}{\sqrt{\Delta}^{k}}\in {\mathbb{Z}}. \end{align} $$

Let p be a prime divisor of $\Delta $ . We claim that

(2.7) $$ \begin{align}v_p(\Delta)\leq v_p(8n\alpha).\end{align} $$

Subcase (i): $p=2$ . Then, $p\nmid \alpha $ . By (2.6), we have $mv_2(\Delta )\leq v_2(n)+(4m-1)$ . Hence, $v_2(\Delta )<4+v_2(n)$ and it follows that

(2.8) $$ \begin{align}v_2(\Delta)\leq 3+v_2(n).\end{align} $$

Subcase (ii): $p>2$ and $p\nmid \alpha $ . Then, $p\mid n$ . By (2.6),

(2.9) $$ \begin{align}v_p(\Delta)\leq mv_p(\Delta)\leq v_p(n).\end{align} $$

Subcase (iii): $p>2$ and $p\mid \alpha $ . Then, $p\nmid 2n$ . Since $\alpha $ is square-free, $v_p(\alpha )=1$ . By (2.6), $mv_p(\Delta )\leq (2m-1)v_p(\alpha )=2m-1$ . Hence,

(2.10) $$ \begin{align}v_p(\Delta)=1=v_p(\alpha).\end{align} $$

Inequalities (2.8), (2.9) and (2.10) show that (2.7) holds. Hence,

(2.11) $$ \begin{align}\Delta\mid8n\alpha.\end{align} $$

In ${\mathbb {F}}_p[x]$ , we have $f^2+af+b= (f+{a}/{2})^2$ . Note that $P_{2m}\in {\mathbb {Z}}[x]$ and $p\nmid 2n$ . We use (2.1) and work in ${\mathbb {F}}_p[x]$ :

$$ \begin{align*} 0=\dfrac{2\Delta^{m}P_{2m}}{n} &=\bigg(2f+a+2\sqrt{f^2+af+b}\bigg)^{2m}+\bigg(2f+a-2\sqrt{f^2+af+b}\bigg)^{2m}\\ &=(2f+a+2f+a)^{2m}+(2f+a-2f-a)^{2m}\\ &=4^k(2f+a)^{2m}. \end{align*} $$

Hence,

(2.12) $$ \begin{align}f+a/2=0 \text{ in } {\mathbb{F}}_p[x] \quad \text{for all primes} \ p\mid\Delta.\end{align} $$

Conversely, assume that both (2.11) and (2.12) hold. In (2.1), let $k=2$ . Then, (1.4) has a nontrivial solution in ${\mathbb {Q}}[x]$ :

$$ \begin{align*}(P_2,Q_2)= \bigg(\frac{n(8f^{2} + 8af + a^{2} + 4b)}{\Delta}, \frac{n(8f + 4a)}{\Delta}\bigg). \end{align*} $$

To show $P_2,Q_2\in {\mathbb {Z}}[x]$ , it suffices to show that

(2.13) $$ \begin{align}P_2,Q_2\in {\mathbb{Z}}_p[x]\end{align} $$

for all prime divisors p of $\Delta $ , where ${\mathbb {Z}}_p$ is the ring of p-adic integers.

Case (i): $p=2$ . Then, $p\nmid \alpha $ . By (2.8), $v_2(\Delta )\leq 3+v_2(n)$ . If a is even, then

$$ \begin{align*}P_2=\frac{n(8f^{2} + 8af + \Delta +8b)}{\Delta}\in {\mathbb{Z}}_2[x] \quad\mbox{and}\quad Q_2=\dfrac{8n(f+{a}/{2})}{\Delta}\in {\mathbb{Z}}_2[x].\end{align*} $$

Case (ii): $p>2$ and $p\nmid \alpha $ . By (2.9), $v_p(\Delta )\leq v_p(n)$ . Thus, $P_2,Q_2\in {\mathbb {Z}}_p[x]$ .

Case (iii): $p>2$ and $p\mid \alpha $ . We use (2.12) and work in ${\mathbb {F}}_p[x]$ :

$$ \begin{align*}8f^2+8af+a^2+4b=2(2f+a)^2+4b-a^2=0.\end{align*} $$

By (2.10), $v_p(\Delta )=1$ . It follows that $P_2,Q_2\in {\mathbb {Z}}_p[x]$ . Thus, (2.13) holds.

Theorem 1.2 is proved.

Proof of Theorem 1.3.

Equation (1.5) has a minimal solution in ${\mathbb {C}}[x]$ :

$$ \begin{align*}(P,Q)=\bigg(\dfrac{2\sqrt{n}f+\sqrt{n}a}{\sqrt{\Delta}},\dfrac{2\sqrt{n}}{\sqrt{\Delta}}\bigg).\end{align*} $$

Hence, all nontrivial solutions of (1.5) in ${\mathbb {C}}[x]$ are $(\pm P_k,\pm Q_k)$ , where

(2.14) $$ \begin{align}P_k+Q_k\sqrt{f^2+af+b}=\sqrt{n}\bigg(\dfrac{2f+a+2\sqrt{f^2+af+b}}{\sqrt{\Delta}}\bigg)^k,\quad k\in {\mathbb{Z}}_{>0}. \end{align} $$

Since $\alpha $ is the leading coefficient of f, by Theorem 1.9, the leading coefficients of $P_k$ and $Q_k$ in (2.14) are $\sqrt {n}2^{k-1}{(2\alpha )^k}/{\sqrt {\Delta }^k}$ and $\sqrt {n}2^{k}{(2\alpha )^{k-1}}/{\sqrt {\Delta }^{k}}$ , respectively. Assume that (1.5) has a nontrivial solution in ${\mathbb {Z}}[x]$ . Then, there exists a positive integer k such that $P_k,Q_k\in {\mathbb {Z}}[x]$ . Therefore,

$$ \begin{align*}\sqrt{n}\dfrac{2^{2k-1}\alpha^{k-1}}{\sqrt{\Delta}^k}\in {\mathbb{Z}}.\end{align*} $$

Since $\sqrt {n}\not \in {\mathbb {Z}}$ , we must have k odd. Let $k=2m+1$ , with $m\in {\mathbb {Z}}_{\geq 0}$ . Then,

(2.15) $$ \begin{align} \dfrac{\sqrt{n}}{\sqrt{\Delta}}\cdot\dfrac{2^{4m+1}\alpha^{2m-1}}{\Delta^m}\in {\mathbb{Z}}.\end{align} $$

Let ${\sqrt {n}}/{\sqrt {\Delta }}={r}/{s}$ , where $r,s\in {\mathbb {Z}}_{>0}$ and $\gcd (r,s)=1$ . By (2.15),

(2.16) $$ \begin{align} \dfrac{2r}{s}\cdot \dfrac{16^m\alpha^{2m-1}}{\Delta^m}\in {\mathbb{Z}}.\end{align} $$

We claim that

(2.17) $$ \begin{align} \dfrac{2r}{s}\in {\mathbb{Z}}. \end{align} $$

If $s=1$ , then (2.17) is obvious. Assume that $s>1$ . Let p be a prime factor of s.

Case (i): $p=2$ . Let $v_2(s)=t\in {\mathbb {Z}}_{>0}$ . If $t\geq 2$ , then $v_2(\Delta )\geq 2t\geq 4$ . Thus,

$$ \begin{align*}v_2(s\Delta^m)\geq 2+4m>v_2(2r\cdot 16^m\alpha^{2m-1}).\end{align*} $$

Hence, (2.16) is impossible. Therefore,

(2.18) $$ \begin{align} v_2(s)=1\leq v_2(2r).\end{align} $$

Case (ii): $p>2$ . Since $p\nmid r$ , by (2.16), we must have $p\mid \alpha $ . Since $\alpha $ is square-free, we have $v_p(\alpha )=1$ . Since $p\mid s$ , $p^2\mid \Delta $ . Thus,

$$ \begin{align*}v_p(\Delta^m)\geq 2m>v_p(\alpha^{2m-1})=v_p(2r\cdot 16^m\alpha^{2m-1}).\end{align*} $$

Hence, (2.16) is impossible. Therefore,

(2.19) $$ \begin{align} v_p(s)=0\quad\mbox{for all prime} \ p>2. \end{align} $$

Now (2.17) follows from (2.18) and (2.19), which shows that $4n/\Delta $ is a nonzero integer square. Conversely, assume that $4n/\Delta $ is a nonzero integer square. Then, (1.5) has a nontrivial solution in ${\mathbb {Q}}[x]$ with $Q\in {\mathbb {Z}}[x]$ :

$$ \begin{align*}(P,Q)=\bigg(\dfrac{2\sqrt{n}f+\sqrt{n}a}{\sqrt{\Delta}},\dfrac{2\sqrt{n}}{\sqrt{\Delta}}\bigg).\end{align*} $$

If a is even, then $P\in {\mathbb {Z}}[x]$ . If a is odd, since $4n/\Delta $ is an integer square, $n/\Delta $ is also an integer square. Hence, $P\in {\mathbb {Z}}[x]$ .

Theorem 1.3 is proved.

Proof. By (3.1),

(3.2) $$ \begin{align} \dfrac{Q_k}{\sqrt{n}}&=\dfrac{1}{2} \bigg(\bigg(\dfrac{2f+a}{\sqrt{\Delta}}+\dfrac{2}{\sqrt{\Delta}}\sqrt{f^2+af+b}\bigg)^k-\bigg(\dfrac{2f+a}{\sqrt{\Delta}}-\dfrac{2}{\sqrt{\Delta}}\sqrt{f^2+af+b}\bigg)^k\bigg) \notag\\ &=\sum\limits_{j=0}^{{[{(k-1)}/{2}]}}\binom{k}{2j+1}\bigg(\dfrac{2f+a}{\sqrt{\Delta}}\bigg)^{k-2j-1}\bigg(\dfrac{4(f^2+af+b)}{\Delta}\bigg)^j.\end{align} $$

If $b=0$ , then $a\neq 0$ because $a^2-4b\neq 0$ . Reducing (3.2) mod f gives

$$ \begin{align*}\dfrac{Q_k}{\sqrt{n}}\equiv k\bigg(\dfrac{a}{\sqrt{\Delta}}\bigg)^{k-1}\not\equiv 0\pmod{f}, \quad k\geq 1.\end{align*} $$

If $b\neq 0$ , reducing (3.2) mod f gives

(3.3) $$ \begin{align} \dfrac{Q_k}{\sqrt{n}}&\equiv \sum\limits_{j=0}^{{[{(k-1)}/{2}]}}\binom{k}{2j+1}\bigg(\dfrac{a}{\sqrt{\Delta}}\bigg)^{k-2j-1}\bigg(\dfrac{4b}{\Delta}\bigg)^j\pmod{f} \notag\\ &\equiv \dfrac{1}{2} \bigg(\bigg(\dfrac{a+2\sqrt{b}}{\sqrt{\Delta}}\bigg)^k-\bigg(\dfrac{a-2\sqrt{b}}{\sqrt{\Delta}}\bigg)^k \bigg) \pmod{f}. \end{align} $$

Suppose there exists a positive integer k such that $f\mid Q_k$ . Then, (3.3) yields

(3.4) $$ \begin{align}\bigg(\dfrac{a+2\sqrt{b}}{a-2\sqrt{b}}\bigg)^k=1.\end{align} $$

If $a=0$ , then (3.4) forces $2\mid k$ . Consider now $a\neq 0$ .

Case 1: $b>0$ . In this case, both $a+2\sqrt {b}$ and $a-2\sqrt {b}$ are real numbers. Since $ab\neq 0$ , $|a+2\sqrt {b}| \neq |a-2\sqrt {b}|$ . Hence, (3.4) is impossible.

Case 2: $b<0$ . Let $\omega = {(a+2\sqrt {b})}/{(a-2\sqrt {b})}$ . Then,

(3.5) $$ \begin{align}\omega^k=1 \quad\text{and}\quad \omega=\dfrac{a^2+4b}{a^2-4b}+\dfrac{4a\sqrt{b}}{a^2-4b^2}.\end{align} $$

Hence, $\omega $ is a root of unity of degree $2$ over ${\mathbb {Q}}$ . Therefore, $\omega \in \{\pm i, \tfrac 12(\pm 1\pm \sqrt {-3})\}$ .

Subcase (i): $\omega =\pm i$ . Then, (3.5) forces $4\mid k$ and $a^2+4b=0$ . Thus, $a=2c$ and ${b=-c^2}$ , where $c\in {\mathbb {Z}}-\{0\}$ .

Subcase (ii): $\omega =\tfrac 12 (1\pm \sqrt {-3})$ . Then, (3.5) forces $6\mid k$ and

$$ \begin{align*}\dfrac{a^2+4b}{a^2-4b}=\dfrac{1}{2}.\end{align*} $$

Hence, $4b=-a^2/3$ . Thus, $a=6c$ and $b=-3c^2$ , where $c\in {\mathbb {Z}}-\{0\}$ .

Subcase (iii): $\omega =\tfrac 12 (-1\pm \sqrt {-3})$ . Then, (3.5) forces $3\mid k$ and

$$ \begin{align*}\dfrac{a^2+4b}{a^2-4b}=-\dfrac{1}{2}.\end{align*} $$

Hence, $4b=-3a^2$ . Thus, $a=2c$ and $b=-3c^2$ , where $c\in {\mathbb {Z}}-\{0\}$ .

Proof of Theorem 1.5.

By Lemma 3.2 and a ${\mathbb {Z}}$ -linear change of variable, it suffices to consider nontrivial solutions in ${\mathbb {Z}}[x]$ of

(3.6) $$ \begin{align}P^2-(f^2+af+b)Q^2=n^2\end{align} $$

with $f\mid Q$ . Equation (3.6) has a minimal solution in ${\mathbb {C}}[x]$ :

$$ \begin{align*}(P,Q)=\bigg(\dfrac{n(2f+a)}{\sqrt{\Delta}},\dfrac{2n}{\sqrt{\Delta}}\bigg).\end{align*} $$

Hence, all nontrivial solutions in ${\mathbb {C}}[x]$ of (3.6) are $(\pm P_k,\pm Q_k)$ , where

$$ \begin{align*}P_k+\sqrt{f^2+af+b} Q_k=n\bigg(\dfrac{2f+a}{\sqrt{\Delta}}+\dfrac{2}{\sqrt{\Delta}}\sqrt{f^2+af+b}\bigg)^k,\quad k\geq 1.\end{align*} $$

Assume that there exists a positive integer k such that $P_k,Q_k\in {\mathbb {Z}}[x]$ and $f\mid Q_k$ .

Case 1: $\sqrt {\Delta }\in {\mathbb {Z}}$ . By Theorem 1.2, $\sqrt {\Delta }\mid 2n$ .

Case 1.1: $a=0$ . Then, $b=-c^2$ , where c is an integer divisor of n. Lemma 3.1 shows that $2\mid k$ . Let $k=2m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{2m}$ is $n2^{2m-1}(\alpha /c)^{2m}$ . Since $c\mid n$ and $\gcd (n,\alpha )=1$ , it follows that

(3.7) $$ \begin{align}\dfrac{n2^{2m-1}}{c^{2m}}\in {\mathbb{Z}}.\end{align} $$

We claim $c^2\mid 2n$ . This is obvious if $m=1$ . Assume $m\geq 2$ . Let p be a prime divisor of c.

Subcase (i): $p=2$ . From (3.7),

(3.8) $$ \begin{align} v_2(n)&\geq 2m(v_2(c)-1)+1 \geq 4(v_2(c)-1)+1 \notag\\ &\geq 2v_2(c)-1+2v_2(c)-2 \geq 2v_2(c)-1. \end{align} $$

Subcase (ii): $p>2$ . From (3.7),

(3.9) $$ \begin{align} v_p(n)\geq v_p(c^2). \end{align} $$

Inequalities (3.8) and (3.9) show that $c^2\mid 2n$ . Conversely, if $(a,b)=(0,-c^2)$ , where $c\in {\mathbb {Z}}-\{0\}$ and $c^2\mid 2n$ , then (3.6) has a nontrivial solution in ${\mathbb {Z}}[x]$ :

$$ \begin{align*}(P_2,Q_2)=\bigg(\dfrac{n(2f^2-c^2)}{c^2},\dfrac{2nf}{c^2}\bigg).\end{align*} $$

Case 1.2: $a\neq 0$ . Since $\sqrt {\Delta }\in {\mathbb {Z}}$ , by Lemma 3.1, $(a,b)=(2c,-3c^2)$ and $3\mid k$ , where $c\in {\mathbb {Z}}-\{0\}$ and $2c\mid n$ . Let $k=3m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{3m}$ is $n2^{3m-1}(\alpha /2c)^{3m}$ . Since $2c\mid n$ and $(n,\alpha )=1$ , it follows that $2c^{3m}\mid n$ . Hence, $2c^{3}\mid n$ . Conversely, if $(a,b)=(2c,-3c^2)$ , where $c\in {\mathbb {Z}}-\{0\}$ and $2c^3\mid n$ , we verify in MAGMA [Reference Bosma, Cannon and Playoust2] that (3.6) has a nontrivial solution in ${\mathbb {Z}}[x]$ :

$$ \begin{align*}(P_3,Q_3)=\bigg( \dfrac{n(f+c)(f^2+2cf-2c^2)}{2c^3},\dfrac{nf(f+2c)}{2c^3}\bigg).\end{align*} $$

Case 2: $\sqrt {\Delta }\not \in {\mathbb {Z}}$ . By Theorem 1.2, $\Delta \mid 8n\alpha $ and for any prime divisor p dividing both $\Delta $ and $\alpha $ , $f+a/2=0$ in ${\mathbb {F}}_p[x]$ .

Case 2.1: $a=0$ . Then, $b\mid 2n\alpha $ and $-b$ is not a square in ${\mathbb {Z}}$ , and $f=0$ in ${\mathbb {F}}_p[x]$ for any prime p dividing both b and $\alpha $ . Conversely, in this case, (3.6) has a nontrivial solution in ${\mathbb {Z}}[x]$ :

$$ \begin{align*}(P_2,Q_2)= \bigg(\frac{n(2f^{2} + b)}{b}, \frac{2nf}{b}\bigg).\end{align*} $$

Case 2.2: $a\neq 0$ . By Lemma 3.1, $(a,b)$ is $(6c,-3c^2)$ and $6\mid k$ or $(2c,-c^2)$ and $4\mid k$ , where c is a nonzero integer.

Subcase (i): $(a,b)=(6c,-3c^2)$ and $6\mid k$ . Since $48c^2=\Delta \mid 8n\alpha $ and $\alpha $ is square-free, we have $6c^2\mid n$ . Let $k=6m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{6m}$ is

$$ \begin{align*}2^{6m-1}n\bigg(\dfrac{\alpha}{2\sqrt{3}c}\bigg)^{6m}=\dfrac{n\alpha^{6m}}{2\cdot 3^{3m}\cdot c^{6m}}.\end{align*} $$

Since $6c^2\mid n$ and $\gcd (n,\alpha )=1$ , we have $\gcd (2\cdot 3^{3m}\cdot c^{6m},\alpha )=1$ . It then follows that $2\cdot 3^{3m}\cdot c^{6m}\mid n$ . Hence, $54c^6\mid n$ . Conversely, if $(a,b)=(6c,-3c^2)$ , where $c\in {\mathbb {Z}}-\{0\}$ and $54c^6\mid n$ , we verify in MAGMA [Reference Bosma, Cannon and Playoust2] that (3.6) has a nontrivial solution in ${\mathbb {Z}}[x]$ :

$$ \begin{align*} (P_6,Q_6) & =\bigg(\dfrac{n(-54 c^6 + 324 c^4 f^2 + 324 c^3 f^3 + 117 c^2 f^4 + 18 c f^5 + f^6)}{54 c^6}, \\ & \qquad \dfrac{nf (3 c + f) (6 c + f) (6 c^2 + 6 c f + f^2)}{54 c^6}\bigg). \end{align*} $$

Subcase (ii): $(a,b)=(2c,-c^2)$ and $3\mid k$ . Since $8c^2=\Delta \mid 8n\alpha $ and $\alpha $ is square-free, we have $c^2\mid n$ . Let $k=4m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{4m}$ is

$$ \begin{align*}n2^{4m-1}\bigg(\frac{\alpha}{\sqrt{2}c}\bigg)^{4m}=\dfrac{n2^{2m-1}\alpha^{4m}}{c^{4m}}.\end{align*} $$

Since $c^2\mid n$ and $\gcd (n,\alpha )=1$ , we have $\gcd (c^{4m},\alpha )=1$ . It follows that

(3.10) $$ \begin{align}\dfrac{n2^{2m-1}}{c^{4m}}\in {\mathbb{Z}}.\end{align} $$

We claim $c^4\mid 2n$ . This is obvious if $m=1$ , so we assume $m\geq 2$ . Let p be a prime divisor of c.

Subcase (ii)(a): $p=2$ . From (3.10),

(3.11) $$ \begin{align} v_2(n)\geq 4mv_2(c)-(2m-1) \geq 8v_2(c)-(2m-1)>4v_2(c). \end{align} $$

Subcase (ii)(b): $p>2$ . From (3.10),

(3.12) $$ \begin{align}v_p(n)\geq 4mv_p(c)>4v_p(c).\end{align} $$

Inequalities (3.11) and (3.12) show that $c^4\mid 2n$ . Conversely, if $(a,b)=(2c,-c^2)$ , where $c\in {\mathbb {Z}}-\{0\}$ and $c^4\mid 2n$ , we verify in MAGMA [Reference Bosma, Cannon and Playoust2] that (3.6) has a nontrivial solution in ${\mathbb {Z}}[x]$ :

$$ \begin{align*} (P_4,Q_4)= \bigg(\frac{n(2f^4+8cf^3+8c^2f^2-c^4)}{c^4}, \frac{2nf(f+c)(f+2c)}{c^4}\bigg).\\[-40pt] \end{align*} $$

Proof of Theorem 1.6.

By Lemma 3.2 and a ${\mathbb {Z}}$ -linear change of variable, it suffices to consider nontrivial solutions in ${\mathbb {Z}}[x]$ of

(3.13) $$ \begin{align}P^2-(f^2+af+b)Q^2=n\end{align} $$

with $f\mid Q$ . Equation (3.13) has a minimal solution in ${\mathbb {C}}[x]$ :

$$ \begin{align*}(P,Q)=\sqrt{n}\bigg(\dfrac{2f+a}{\sqrt{\Delta}},\dfrac{2}{\sqrt{\Delta}}\bigg).\end{align*} $$

Hence, all nontrivial solutions in ${\mathbb {C}}[x]$ of (3.13) are $(\pm P_k,\pm Q_k)$ , where

$$ \begin{align*}P_k+\sqrt{f^2+af+b} Q_k=\sqrt{n}\bigg(\dfrac{2f+a}{\sqrt{\Delta}}+\dfrac{2}{\sqrt{\Delta}}\sqrt{f^2+af+b}\bigg)^k,\quad k\geq 1.\end{align*} $$

Assume that there exists a positive integer k such that $P_k,Q_k\in {\mathbb {Z}}[x]$ and $f\mid Q_k$ .

Case 1: $a=0$ . Lemma 3.1 shows that $2\mid k$ . Let $k=2m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{2m}$ is $\sqrt {n}2^{2m-1}(\alpha /\sqrt {-b})^{2m}$ , which is an irrational number. Hence, ${P_{k}\not \in {\mathbb {Z}}[x]}$ .

Case 2: $a\neq 0$ . By Lemma 3.1, $(a,b)$ is one of:

• • $(2c,-c^2)$ , where $c\in {\mathbb {Z}}$ , $c\neq 0$ and $4\mid k$ ;

• • $(6c,-3c^2)$ , where $c\in {\mathbb {Z}}$ , $c\neq 0$ and $6\mid k$ ;

• • $(2c,-3c^2)$ , where $c\in {\mathbb {Z}}$ , $c\neq 0$ and $3\mid k$ .

Case 2.1: $(2c,-c^2)$ and $4\mid k$ . Let $k=4m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{4m}$ is $\sqrt {n}2^{4m-1}(\alpha /\sqrt {2}c)^{4m}$ , an irrational number. Hence, $P_{k}\not \in {\mathbb {Z}}[x]$ .

Case 2.2: $(6c,-3c^2)$ and $6\mid k$ . Let $k=6m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{6m}$ is $\sqrt {n}2^{6m-1}(\alpha /2\sqrt {3}c)^{6m}$ , an irrational number. Hence, $P_{k}\not \in {\mathbb {Z}}[x]$ .

Case 2.3: $(2c,-3c^2)$ and $3\mid k$ . Let $k=3m$ , where $m\in {\mathbb {Z}}_{>0}$ . The leading coefficient of $P_{3m}$ is $\sqrt {n}2^{3m-1}(\alpha /2c)^{3m}$ , an irrational number. Hence, $P_{k}\not \in {\mathbb {Z}}[x]$ .

In all cases, $P_k\not \in {\mathbb {Z}}[x]$ . Hence, (3.13) has no nontrivial solutions in ${\mathbb {Z}}[x]$ . Theorem 1.9 is proved.