# Generic diagonal conic bundles revisited

Alexei N. Skorobogatov and Efthymios Sofos

###### Abstract

We prove a stronger form of our previous result that Schinzel’s Hypothesis
holds for 100% of $n$-tuples of integer polynomials satisfying the usual
necessary conditions, where the primes represented by the polynomials are
subject to additional constraints in terms of Legendre symbols, as well as
upper and lower bounds. We establish the triviality of the Brauer group of
generic diagonal conic bundles over the projective line. Finally, we give an
explicit lower bound for the probability that diagonal conic bundles in
certain natural families have rational points.

## Introduction

In our previous paper we proved that Schinzel’s Hypothesis (H) holds for 100%
of $n$-tuples of integer polynomials satisfying the usual necessary conditions
[SS22, Thm. 1.2]. Here we give some improvements, complements and further
applications of the results of [SS22] relevant to diagonal conic and quadric
bundles over the projective line.

In the first section we prove a stronger form of [SS22, Thm. 1.2] where the
primes represented by the polynomials are required to satisfy additional
conditions in terms of Legendre symbols, as well as upper and lower bounds,
see Theorem 1.1. Using this result, in Corollary 1.3 we give a simplified
proof of a weaker form of the Hasse principle for random diagonal conic
bundles over the projective line [SS22, Thm. 6.1], with a bound for the least
solution. We prove an analogous statement for diagonal quadric bundles of
relative dimension 2, see Corollary 1.4. (It is well-known that quadric
bundles of relative dimension at least 3 over the projective line satisfy the
Hasse principle [CTSS87, Prop. 3.9].)

The absence of Brauer–Manin conditions in Corollary 1.3 is due to the
triviality of the Brauer group of generic diagonal conic bundles mentioned in
[SS22, Remark 6.2] and proved in the second section of this note, see Theorem
2.1 and Corollary 2.2.

In the last section we give an explicit lower bound for the density of pairs
of integer polynomials $P_{1}(t)$, $P_{2}(t)$ of arbitrary fixed degrees such
that the equation

$P_{1}(t)x^{2}+P_{2}(t)y^{2}=z^{2}$ (1)

is soluble in ${\mathbb{Z}}$. When the degrees of $P_{1}(t)$ and $P_{2}(t)$
are large, this density is close to one third. We also estimate the height of
the smallest integer solution of (1).

The authors have been partly supported by the EPSRC New Horizons grant “Local-
to-global principles for random Diophantine equations” (EP/V019066/1).

##  1 Schinzel hypothesis on average with quadratic residue conditions

Non-constant polynomials $P_{1}(t),\ldots,P_{n}(t)\in{\mathbb{Z}}[t]$ are
called a Schinzel $n$-tuple if the leading coefficient of each $P_{i}(t)$ is
positive, and for every prime $\ell$ the product $\prod_{i=1}^{n}P_{i}(t)$ is
not divisible by $t^{\ell}-t$ modulo $\ell$.

The height of a polynomial $P(t)\in{\mathbb{Z}}[t]$ is defined as the maximum
of the absolute values of the coefficients, and is denoted by $|P|$. The
height of an $n$-tuple of polynomials ${\bf
P}=(P_{1}(t),\ldots,P_{n}(t))\in({\mathbb{Z}}[t])^{n}$ is defined as $|{\bf
P}|=\max_{i=1,\ldots,n}(|P_{i}|)$

The following result is [SS22, Thm. 1.2] with additional properties (2), (3)
and (6). The proof of (6) uses [SS22, Prop. 6.5] based on Heath-Brown’s bound
for character sums [HB95, Cor. 4].

###### Theorem 1.1

Fix any $(d_{1},\ldots,d_{n})\in\mathbb{N}^{n}$, $\varepsilon>0$,
$M\in\mathbb{N}$, $m_{0}\in{\mathbb{Z}}/M$ and ${\bf
Q}\in(({\mathbb{Z}}/M)[t])^{n}$ such that ${\rm{deg}}(Q_{i})\leq d_{i}$ and
$\gcd(Q_{i}(m_{0}),M)=1$ for all $i=1,\ldots,n$. For every
$(i,j)\in(\mathbb{N}\cap[1,n])^{2}$ with $i<j$ let
$\epsilon_{ij}\in\\{1,-1\\}$. Then for $100\%$ of Schinzel $n$-tuples ${\bf
P}\in({\mathbb{Z}}[t])^{n}$ of respective degrees $d_{1},\ldots,d_{n}$ such
that ${\bf P}\equiv{\bf Q}\bmod M$, there exists a natural number $m$ with the
following properties:

$\displaystyle\min\\{P_{i}(m):1\leq i\leq n\\}>|{\bf P}|(\log|{\bf
P}|)^{\varepsilon/2},$ (2) $\displaystyle
m\leq(\log|\mathbf{P}|)^{n+\varepsilon},$ (3) $\displaystyle m\equiv
m_{0}\bmod M,$ (4) $\displaystyle P_{1}(m),\ldots,P_{n}(m)\text{ are distinct
primes},$ (5) $\displaystyle\text{the Legendre symbol
}\left(\frac{P_{i}(m)}{P_{j}(m)}\right)\text{ equals }\epsilon_{ij}\ \text{
for all }i<j.$ (6)

Proof. Define

$C_{\bf P}(x):=\sum_{\begin{subarray}{c}m\in\mathbb{N}\cap[1,x]\\\
\eqref{eq:1}-\eqref{eq:3}\end{subarray}}\prod_{t=1}^{n}\log
P_{t}(m),\quad\quad\widetilde{\theta}_{\bf
P}(x):=\sum_{\begin{subarray}{c}m\in\mathbb{N}\cap[1,x]\\\
\eqref{eq:1}-\eqref{eq:2}\end{subarray}}\prod_{t=1}^{n}\log P_{t}(m).$

Let us write $\Omega=\\{(i,j):1\leq i<j\leq n\\}$. For any $S\subset\Omega$
define

$T_{S,{\bf P}}(x):=\sum_{\begin{subarray}{c}m\in\mathbb{N}\cap[1,x]\\\
\eqref{eq:1}-\eqref{eq:2}\end{subarray}}\prod_{t=1}^{n}(\log
P_{t}(m))\prod_{(i,j)\in S}\left(\frac{P_{i}(m)}{P_{j}(m)}\right).$

We follow the standard convention that a product indexed by the elements of an
empty set is equal to $1$. In particular, we have $T_{\emptyset,{\bf
P}}(x)=\widetilde{\theta}_{\bf P}(x)$. Assuming (5), the following function
takes the value $1$ if (6) holds and the value $0$ otherwise:

$2^{-\\#\Omega}\prod_{(i,j)\in\Omega}\left(1+\epsilon_{ij}\left(\frac{P_{i}(m)}{P_{j}(m)}\right)\right)=2^{-\\#\Omega}\sum_{S\subset\Omega}\prod_{(i,j)\in
S}\epsilon_{ij}\left(\frac{P_{i}(m)}{P_{j}(m)}\right),$

which leads to

$C_{\bf P}(x)=2^{-\\#\Omega}\sum_{S\subset\Omega}T_{S,{\bf
P}}(x)\prod_{(i,j)\in S}\epsilon_{ij}.$ (7)

For the rest of the proof we restrict attention only to the range

$(\log H)^{A_{1}}<x\leq(\log H)^{A_{2}},$ (8)

where $A_{1},A_{2}$ are arbitrary fixed constants satisfying $n<A_{1}<A_{2}$.
By [SS22, Eq. (6.7)] the term corresponding to $S=\emptyset$ equals

$\frac{\widetilde{\theta}_{\bf P}(x)}{2^{\\#\Omega}}=\frac{\theta_{\bf
P}(x)}{2^{\\#\Omega}}+O((\log H)^{n})=\frac{\theta_{\bf
P}(x)}{2^{\\#\Omega}}+O(x^{n/A_{1}}),$ (9)

where $\theta_{\bf P}(x)$ is defined similarly to $\widetilde{\theta}_{\bf
P}(x)$ by dropping the condition that the primes $P_{1}(m),\ldots,P_{n}(m)$
are necessarily distinct. Let us define

$\texttt{Poly}(H):=\left\\{{\bf
P}\in({\mathbb{Z}}[t])^{n}:\begin{array}[]{l}{\rm{deg}}(P_{i})=d_{i},P_{i}\equiv
Q_{i}\bmod M\text{ for }i=1,\ldots,n,\ |{\bf P}|\leq H\end{array}\right\\}.$

We next show that uniformly in the range (8) and for all $S\neq\emptyset$ one
has

$\frac{1}{H^{d+n}}\sum_{{\bf P}\in\texttt{Poly}(H)}|T_{S,{\bf P}}(x)|\ll
x^{\frac{1}{2}+\frac{n}{2A_{1}}}.$ (10)

Fix $h<k$ such that $(h,k)\in S$. Letting

$\mathcal{F}_{1}:=\prod_{\begin{subarray}{c}(i,j)\in S\\\
\\{i,j\\}\cap\\{h,k\\}=\\{h\\}\end{subarray}}\left(\frac{P_{i}(m)}{P_{j}(m)}\right),\
\mathcal{F}_{2}:=\prod_{\begin{subarray}{c}(i,j)\in S\\\
\\{i,j\\}\cap\\{h,k\\}=\\{k\\}\end{subarray}}\left(\frac{P_{i}(m)}{P_{j}(m)}\right),\
\mathcal{G}:=\prod_{\begin{subarray}{c}(i,j)\in S\\\
\\{i,j\\}\cap\\{h,k\\}=\emptyset\end{subarray}}\left(\frac{P_{i}(m)}{P_{j}(m)}\right)$

makes it plain that $T_{S,{\bf P}}$ takes the form of the function $\eta_{\bf
P}$ from [SS22, Def. 6.4]. This allows us to apply [SS22, Prop. 6.5] to verify
(10). Feeding (9)-(10) into (7) yields

$\sum_{{\bf P}\in\texttt{Poly}(H)}\frac{|C_{\bf
P}(x)-2^{-\\#\Omega}\theta_{\bf P}(x)|}{\\#\texttt{Poly}(H)}\ll
x^{n/A_{1}}+x^{\frac{1}{2}+\frac{n}{2A_{1}}}\ll
x^{\frac{1}{2}+\frac{n}{2A_{1}}}.$

Hence the number of $\bf P$ in $\texttt{Poly}(H)$ with $|C_{\bf
P}(x)-2^{-\\#\Omega}\theta_{\bf P}(x)|>x^{\frac{1}{2}+\frac{n}{2A_{1}}}\log x$
is

$\leq\sum_{{\bf P}\in\texttt{Poly}(H)}\frac{\left|C_{\bf
P}(x)-2^{-\\#\Omega}\theta_{\bf
P}(x)\right|}{x^{\frac{1}{2}+\frac{n}{2A_{1}}}\log
x}\ll\frac{\\#\texttt{Poly}(H)}{\log x}=o(\\#\texttt{Poly}(H)).$

Therefore, for $100\%$ of $\bf P$ in $\texttt{Poly}(H)$ one has

$|C_{\bf P}(x)-2^{-\\#\Omega}\theta_{\bf P}(x)|\leq
x^{\frac{1}{2}+\frac{n}{2A_{1}}}\log x,$

which implies

$C_{\bf P}(x)\geq\frac{2^{-\\#\Omega}\beta_{0}x}{2(\log\log
x)^{d-n}}-x^{\frac{1}{2}+\frac{n}{2A_{1}}}\log x>\frac{x}{\log x},$

where we used [SS22, Eq. (4.10)] for the last deduction. Hence, there exists
$m\leq x$ satisfying (4)-(6). To verify (3), we take
$A_{2}=n+\frac{7}{10}\varepsilon$ and $A_{1}=A_{2}-\frac{\varepsilon}{10}$.
Note that $x\leq(\log H)^{A_{2}}$ and that for $100\%$ of ${\bf
P}\in\texttt{Poly}(H)$ one has $H\leq|{\bf P}|^{2}$. Therefore, when $|{\bf
P}|$ is sufficiently large,

$m\leq x\leq(\log H)^{A_{2}}\ll(\log|{\bf P}|)^{A_{2}}\leq(\log|{\bf
P}|)^{n+\varepsilon},$

which proves (3). To prove (2) we note that if the largest integer $m\leq x$
that satisfies (4)-(6) is $m_{1}$ then

$C_{\bf P}(x)=\sum_{\begin{subarray}{c}m\in\mathbb{N}\cap[1,x]\\\
\eqref{eq:1}-\eqref{eq:3}\end{subarray}}\prod_{t=1}^{n}\log
P_{t}(m)=\sum_{m\in\mathbb{N}\cap[1,m_{1}]}O((\log H)^{n})=O(m_{1}(\log
H)^{n}).$

For almost all $\mathbf{P}$ with $|{\bf P}|\leq H$ we have shown that
$C_{\mathbf{P}}(x)>\frac{x}{\log x}$. Combined with the upper bound $C_{\bf
P}(x)=O(m_{1}(\log H)^{n})$ this gives

$m_{1}\gg\frac{x}{(\log H)^{n}\log x}\gg\frac{(\log H)^{A_{1}-n}}{\log\log
H}=\frac{(\log H)^{\frac{3}{5}\varepsilon}}{\log\log H}.$

Since $P_{i}(m_{1})=\sum_{j=1}^{d_{i}}c_{ij}m_{1}^{j}$ for some integer vector
${\bf c}$ with $|c_{ij}|\leq H$, we see that
$P_{i}(m_{1})=c_{i{d_{i}}}m_{1}^{d_{i}}+O(Hm_{1}^{d_{i}-1})$. Note that for
almost all $\bf P$ with $|{\bf P}|\leq H$ one has
$\min_{i,j}|c_{ij}|\geq\frac{H}{\log\log H}$, hence, for all $i$ we have

$P_{i}(m_{1})=Hm_{1}^{d_{i}-1}\left(\frac{c_{i{d_{i}}}}{H}m_{1}+O(1)\right)\gg\frac{Hm_{1}^{d_{i}}}{\log\log
H}\geq\frac{Hm_{1}}{\log\log H}$

because $m_{1}/\log\log H\geq(\log H)^{A_{1}-n}(\log\log H)^{-2}\to+\infty$.
Hence, for all $i$ one has

$P_{i}(m_{1})\gg\frac{Hm_{1}}{\log\log H}\gg H\frac{(\log
H)^{\frac{3}{5}\varepsilon}}{\log\log H}>|{\bf P}|\frac{(\log|{\bf
P}|)^{\frac{3}{5}\varepsilon}}{\log\log|{\bf P}|}.$

In particular, $\min_{i}P_{i}(m_{1})>|{\bf P}|(\log|{\bf
P}|)^{\varepsilon/2}$, which concludes the proof of (2). $\Box$

### Generic diagonal conic bundles

As an application we give a simplified proof of [SS22, Thm. 6.1] with an added
value of a bound for the least solution. Finite search bounds for Diophantine
equations are not well-studied but are nevertheless relevant to the theory as
they are intimately related to Hilbert’s 10th problem for $\mathbb{Q}$. We
give a search bound that is of polynomial growth in the size of the
coefficients.

We need the following special case of a theorem of Cassels [Cas55].

######  Proposition 1.2 (Cassels)

If $f_{1},f_{2},f_{3}$ are non-zero integers such that the quadratic form
$\sum_{i=1}^{3}f_{i}x_{i}^{2}$ represents zero in ${\mathbb{Q}}$, then there
exists a solution $(x_{1},x_{2},x_{3})\in\mathbb{N}^{3}$ such that

$\max\\{x_{i}:i=1,2,3\\}\leq 40\max\\{|f_{i}|:i=1,2,3\\}.$

For $\mathbf{m}=(m_{0},m_{1},\ldots,m_{r})\in{\mathbb{Z}}^{r}$ we write
$P(t,\mathbf{m})$ for the polynomial $\sum_{k=0}^{r}m_{k}t^{k}$.

###### Corollary 1.3

Let $n_{1},n_{2},n_{3}$ be integers such that $n_{1}>0$, $n_{2}>0$, and
$n_{3}\geq 0$, and let $n=n_{1}+n_{2}+n_{3}$. Let $a_{1},a_{2},a_{3}$ be non-
zero integers not all of the same sign. Let $d_{ij}$ be natural numbers, for
$i=1,2,3$ and $j=1,\ldots,n_{i}$. Define $d_{i}=\sum_{j=1}^{n_{i}}d_{ij}$ and
$d=\sum_{i=1}^{3}d_{i}$. Let $\mathcal{P}$ be the set of ${\bf m}=({\bf
m}_{ij})\in{\mathbb{Z}}^{d+n}$ such that the $n$-tuple $(P_{ij}(t,{\bf
m}_{ij}))$ is Schinzel. Let $\mathcal{M}$ be the set of ${\bf
m}\in\mathcal{P}$ such that for each $p|2a_{1}a_{2}a_{3}$ the equation

$a_{1}\prod_{j=1}^{n_{1}}P_{1,j}(t)\,x_{1}^{2}+a_{2}\prod_{k=1}^{n_{2}}P_{2,k}(t)\,x_{2}^{2}+a_{3}\prod_{l=1}^{n_{3}}P_{3,l}(t)\,x_{3}^{2}=0,$
(11)

has a solution in ${\mathbb{Z}}_{p}$ for which the value of each polynomial
$P_{ij}(t)$ is a $p$-adic unit. Then there is a subset
$\mathcal{M}^{\prime}\subset\mathcal{M}$ of density $1$ such that for every
${\bf m}\in\mathcal{M}^{\prime}$ the equation $(\ref{cinque})$ has a solution
$(x_{1},x_{2},x_{3},t)\in\mathbb{N}^{4}$ with

$\max\\{x_{1},x_{2},x_{3},t\\}\leq\max\\{|a_{i}|\\}(\log|{\bf{P}}|)^{(n+1)\max\\{d_{i}\\}}|{\bf
P}|^{\max\\{n_{i}\\}}.$

The set $\mathcal{M}^{\prime}$ has positive density in ${\mathbb{Z}}^{d+n}$
ordered by height.

Proof. By absorbing primes into variables $x,y,z$ we can assume that
$a_{1}a_{2}a_{3}$ is square-free. Write $M=8a_{1}a_{2}a_{3}$. Local solubility
of (11) with $t=m$ at an odd prime $p|M$ with an additional condition that the
value of each $P_{ij}(m)$ is a $p$-adic unit depends only on the value of
$P_{ij}(m)$ modulo $p$. For the prime $2$ the same holds modulo $8$. Thus
$\mathcal{M}$ is a finite disjoint union of subsets given by the condition
${\bf P}\equiv{\bf Q}\bmod M$, where ${\bf Q}$ is an $n$-tuple of polynomials
in $({\mathbb{Z}}/M)[t]$ for which there exists an $m_{0}\in{\mathbb{Z}}$ such
that (11) with $t=m_{0}$ has a solution in ${\mathbb{Z}}_{p}$ for each $p|M$,
and $\gcd(Q_{ij}(m_{0}),M)=1$. Let us fix such a ${\bf Q}$ and such an
$m_{0}$. Then for any ${\bf P}\equiv{\bf Q}\bmod M$ and any $m\equiv
m_{0}\bmod M$ the equation (11) with $t=m$ is soluble in ${\mathbb{Z}}_{p}$
for each $p|M$.

Suppose that $p_{ij}:=P_{ij}(m)$, where $i=1,2,3$ and $j=1,\ldots,n_{i}$, are
distinct primes, where $P_{ij}(t)\equiv Q_{ij}(t)\bmod M$ and $m\equiv
m_{0}\bmod M$. Thus $p_{ij}\equiv Q_{ij}(m_{0})\bmod M$, hence $p_{ij}$ does
not divide $M$. The local solubility of (11) with $t=m$ at the primes not
dividing $M$ and not equal to one of the $p_{ij}$ is clear, since the conic
has good reduction modulo such a prime. It remains to show that we can choose
$m\equiv m_{0}\bmod M$ so that (11) is solvable at each of the primes
$p_{ij}=P_{ij}(m)$.

For $i=1,2,3$ define $\pi_{i}=\prod_{j=1}^{n_{i}}p_{ij}$. Let
$\lambda_{ij}\in{\mathbb{F}}_{2}$ be such that

$(-1)^{\lambda_{ij}}=\left(\frac{-a_{i^{\prime}}a_{i^{\prime\prime}}}{p_{ij}}\right)=(p_{ij},-a_{i^{\prime}}a_{i^{\prime\prime}})_{p_{ij}},$

where $\\{i,i^{\prime},i^{\prime\prime}\\}=\\{1,2,3\\}$. Here the middle term
is the Legendre symbol and the right hand term is the Hilbert symbol. By
global reciprocity we obtain

$(-1)^{\lambda_{ij}}=\prod_{p|M}(p_{ij},-a_{i^{\prime}}a_{i^{\prime\prime}})_{p}.$
(12)

Define $\widetilde{\lambda}_{ij}\in{\mathbb{F}}_{2}$ as follows:

  * •

$\widetilde{\lambda}_{1,j}=\lambda_{1j}$, for $j=1,\ldots,n_{1}$;

  * •

$(-1)^{\widetilde{\lambda}_{2,k}}=(-1)^{\lambda_{2,k}}(\pi_{1},p_{2,k})_{2}$,
for $k=1,\ldots,n_{2}$;

  * •

$(-1)^{\widetilde{\lambda}_{3,l}}=(-1)^{\lambda_{3,l}}(\pi_{1}\pi_{2},p_{3,l})_{2}$,
for $l=1,\ldots,n_{3}$.

Since $p_{ij}\equiv Q_{ij}(m_{0})\bmod M$, we see from (12) that the
$\lambda_{ij}$ depend only on ${\bf Q}$ and $m_{0}$. Thus the same is true for
the $\widetilde{\lambda}_{ij}$.

The solubility of (11) at $p_{ij}$ is equivalent to the condition

$\left(\frac{-a_{i^{\prime}}a_{i^{\prime\prime}}\prod_{k=1}^{n_{i^{\prime}}}p_{i^{\prime}k}\prod_{l=1}^{n_{i^{\prime\prime}}}p_{i^{\prime\prime}l}}{p_{ij}}\right)=1,$
(13)

where $\\{i,i^{\prime},i^{\prime\prime}\\}=\\{1,2,3\\}$. Consider the
lexicographic order on the set of pairs $(ij)$, thus $(1,j)<(2,k)<(3,l)$ for
any $j,k,l$. For pairs $(ij)<(i^{\prime}j^{\prime})$ define
$x_{ij,i^{\prime}j^{\prime}}\in{\mathbb{F}}_{2}$ so that

$(-1)^{x_{ij,i^{\prime}j^{\prime}}}=\left(\frac{p_{i^{\prime}j^{\prime}}}{p_{ij}}\right).$

If $(ij)>(i^{\prime}j^{\prime})$, we define $x_{ij,i^{\prime}j^{\prime}}$ to
be equal to $x_{i^{\prime}j^{\prime},ij}$. Thus for any distinct pairs
$(a,b)\neq(c,d)$ we have $x_{ab,cd}=x_{cd,ab}$. Using the quadratic
reciprocity law we see that (13) is equivalent to the equation

$\sum_{k=1}^{n_{i^{\prime}}}x_{ij,i^{\prime}k}+\sum_{l=1}^{n_{i^{\prime\prime}}}x_{ij,i^{\prime\prime}l}=\widetilde{\lambda}_{ij}.$
(14)

By assumption we have $n_{1}>0$ and $n_{2}>0$, so we have at least one
variable $x_{ij,i^{\prime}j^{\prime}}$ and $n\geq 2$ equations (14). The sum
of the left hand sides of all equations (14) is zero, and it is easy to see
that no other non-trivial linear combination of these is zero.

We claim that the sum of all $\widetilde{\lambda}_{ij}$ is zero. Indeed, since
$a_{1},a_{2},a_{3}$ are not of the same sign we obtain from global reciprocity

$\prod_{p|M}(-a_{1}a_{2},-a_{1}a_{3})_{p}=1.$

Using that (11) is solvable for each $p|M$, we have

$(-a_{1}a_{2}\pi_{1}\pi_{2},-a_{1}a_{3}\pi_{1}\pi_{3})_{p}=1\quad\text{for
each}\ p|M.$

Taking the product over all $p|M$ and using that each $\pi_{i}$ is coprime to
$M$ we prove our claim.

We conclude that the system of linear equations (14) is solvable for arbitrary
$\widetilde{\lambda}_{ij}$. Let us choose a solution
$\varepsilon_{ij,i^{\prime}j^{\prime}}\in{\mathbb{F}}_{2}$. Applying Theorem
1.1, for ${\bf P}\equiv{\bf Q}\bmod M$ in a subset of density 1 we find a
natural number $m\equiv m_{0}\bmod M$, $m\leq(\log|{\bf P}|)^{n+1/2}$, such
that the numbers $P_{ij}(m)$ are distinct primes satisfying

$\left(\frac{P_{i^{\prime}j^{\prime}}(m)}{P_{ij}(m)}\right)=(-1)^{\varepsilon_{ij,i^{\prime}j^{\prime}}},$

whenever $(ij)<(i^{\prime}j^{\prime})$. Then the conic (11) with $t=m$ is
everywhere locally solvable, and so is solvable in ${\mathbb{Z}}$.
Furthermore, Proposition 1.2 ensures the existence of a solution
$(x_{1},x_{2},x_{3})\in\mathbb{N}^{3}$ such that

$\max\\{x_{1},x_{2},x_{3}\\}\leq\max\left\\{|a_{i}\mathcal{H}_{i}|:i=1,2,3\right\\},$

where $\mathcal{H}_{i}=40\prod_{j=1}^{n_{i}}P_{ij}(m)$. Using $m\leq(\log|{\bf
P}|)^{n+1/2}$ we obtain

$|\mathcal{H}_{i}|\leq 40\prod_{j=1}^{n_{i}}\left((1+d_{ij})|{\bf
P}|m^{d_{ij}}\right)\ll_{d_{ij}}|{\bf
P}|^{n_{i}}(\log|{\bf{P}}|)^{(n+1/2)d_{i}},$

which is at most $|{\bf P}|^{n_{i}}(\log|{\bf{P}}|)^{(n+1)d_{i}}$ for all
sufficiently large $|{\bf P}|$. $\Box$

### Generic diagonal quadric bundles of relative dimension 2

Let $a_{0},a_{1},a_{2},a_{3}$ be non-zero integers not all of the same sign
and let $a=a_{0}a_{1}a_{2}a_{3}$. Let $d_{1},\ldots,d_{n}$ be positive
integers and let $d=\sum_{i=1}^{n}d_{i}$. For $i=1,\ldots,n$ let $P_{i}(t,{\bf
m}_{i})=\sum_{j=0}^{d_{i}}m_{ij}t^{j}$ be the generic polynomial of degree
$d_{i}$. Let $S_{0},S_{1},S_{2},S_{3}$ be subsets of $\\{1,\ldots,n\\}$. The
equation

$a_{0}\prod_{i\in S_{0}}P_{i}(t)\,x_{0}^{2}+a_{1}\prod_{i\in
S_{1}}P_{i}(t)\,x_{1}^{2}+a_{2}\prod_{i\in
S_{2}}P_{i}(t)\,x_{2}^{2}+a_{3}\prod_{i\in S_{3}}P_{i}(t)\,x_{3}^{2}=0$ (15)

defines a family of quadrics $Q_{t}$ parametrised by the affine line with
coordinate $t$ over the field ${\mathbb{Q}}({\bf m})$, where ${\bf m}=({\bf
m}_{1},\ldots,{\bf m}_{n})$. The generic fibre $Q_{\eta}$ is a quadric of
dimension 2 over ${\mathbb{Q}}(t,{\bf m})$. We can multiply (15) by a non-zero
element of ${\mathbb{Q}}({\bf m})$ and absorb squares into coordinates $x_{i}$
without affecting the isomorphism class of $Q_{\eta}$. This allows us to
assume without loss of generality that each $a_{i}$ is square-free with no
prime dividing more than two of the $a_{i}$, and that no element of
$\\{1,\ldots,n\\}$ belongs to more than two of the sets $S_{i}$. Moreover, we
can assume that if a prime divides two of the $a_{i}$ then one of them is
$a_{0}$, and if an element of $\\{1,\ldots,n\\}$ belongs to two of the sets
$S_{i}$ then one of them is $S_{0}$. Define

$\delta=a\prod_{i\in S_{0}}P_{i}(t)\prod_{i\in S_{1}}P_{i}(t)\prod_{i\in
S_{2}}P_{i}(t)\prod_{i\in S_{3}}P_{i}(t).$

###### Corollary 1.4

In the above notation assume that $\delta$ is not a square in
$\overline{\mathbb{Q}}(t,{\bf m})$. Let $\mathcal{P}$ be the set of ${\bf
m}=({\bf m}_{i})\in{\mathbb{Z}}^{d+n}$ such that the $n$-tuple $(P_{i}(t,{\bf
m}_{i}))$ is Schinzel. Let $\mathcal{M}$ be the set of ${\bf m}\in\mathcal{P}$
such that for each $p|2a$ the equation $(\ref{q1})$ has a solution in
${\mathbb{Z}}_{p}$ for which the value of each polynomial $P_{i}(t)$ is a
$p$-adic unit. Then there is a subset $\mathcal{M}^{\prime}\subset\mathcal{M}$
of density $1$ such that for every ${\bf m}\in\mathcal{M}^{\prime}$ the
equation $(\ref{q1})$ has a solution in ${\mathbb{Z}}$. The set
$\mathcal{M}^{\prime}$ has positive density in ${\mathbb{Z}}^{d+n}$ ordered by
height.

Proof. We follow the beginning of proof of Corollary 1.3. Let $M=8a$. Local
solubility of (15) at $t=m$ at an odd prime $p|a$ with an additional condition
that the value of each $P_{ij}(m)$ is a $p$-adic unit depends only on the
value of $P_{ij}(m)$ modulo $p$. For the prime $2$ the same holds modulo $8$.
Thus $\mathcal{M}$ is a finite disjoint union of subsets given by the
condition ${\bf P}\equiv{\bf Q}\bmod M$, where ${\bf Q}$ is an $n$-tuple of
polynomials in $({\mathbb{Z}}/M)[t]$ for which there exists an
$m_{0}\in{\mathbb{Z}}$ such that (15) with $t=m_{0}$ has a solution in
${\mathbb{Z}}_{p}$ for each $p|M$, and $\gcd(Q_{ij}(m_{0}),M)=1$. Let us fix
such a ${\bf Q}$ and such an $m_{0}$. Then for any ${\bf P}\equiv{\bf Q}\bmod
M$ and any $m\equiv m_{0}\bmod M$ the equation (15) with $t=m$ is solvable in
${\mathbb{Z}}_{p}$ for each $p|M$.

Suppose that $p_{i}:=P_{i}(m)$, where $i=1,\ldots,n$, are distinct primes,
where $P_{i}(t)\equiv Q_{i}(t)\bmod M$ and $m\equiv m_{0}\bmod M$. Thus
$p_{i}\equiv Q_{i}(m_{0})\bmod M$, hence $p_{i}$ does not divide $M$. The
local solubility of (15) with $t=m$ at the primes not dividing $M$ and not
equal to one of the $p_{i}$ is clear, since the quadric has good reduction
modulo such a prime.

Suppose that every element of $\\{1,\ldots,r\\}$ belongs to exactly one of the
sets $S_{i}$, and every element of $\\{r+1,\ldots,n\\}$ belongs to two of
these sets. The condition on $\delta$ implies $r\geq 1$. The local solubility
of (15) with $t=m$ at the primes $p_{i}$, where $i=1,\ldots,r$, is automatic:
this follows from the fact that the conic obtained by setting $x_{j}=0$ in
(15), where $i\in S_{j}$, has good reduction modulo $p_{i}$. It remains to
show that for ${\bf P}\equiv{\bf Q}\bmod M$ in a subset of density 1 we can
choose $m\equiv m_{0}\bmod M$ so that (15) is solvable at each of the primes
$p_{r+1},\ldots,p_{n}$.

We set $\epsilon_{k,l}=1$ for all $k\geq 2$ and $l>k$. We also set
$\epsilon_{1,i}=1$ for $i=2,\ldots,r$. Now for each $i\in\\{r+1,\ldots,n\\}$
we can find a value of $\epsilon_{1,i}=\left(\frac{p_{1}}{p_{i}}\right)$ such
that (15) is solvable in ${\mathbb{Z}}_{p_{i}}$. An application of Theorem 1.1
finishes the proof since the resulting quadric $Q_{m}$ is everywhere locally
solvable, hence solvable in ${\mathbb{Z}}$. $\Box$

The case when $\delta$ is a square in ${\mathbb{Q}}({\bf m})$ can be reduced
to the case of conic bundles. Indeed, let $C\to{\mathbb{A}}^{1}$ be the conic
bundle defined by setting $x_{0}=0$ in (15). Then every smooth fibre $Q_{t}$
is isomorphic to $C_{t}\times C_{t}$, see [CTS21, Prop. 7.2.4 (c′′)], so
$Q_{t}$ has a rational point if and only if $C_{t}$ has a rational point. Thus
Corollary 1.3 gives a similar statement for generic diagonal quadric bundles
for which $\delta$ is a square in ${\mathbb{Q}}({\bf m})$.

When $\delta$ is a square in $\overline{\mathbb{Q}}(t,{\bf m})$ but not a
square in ${\mathbb{Q}}({\bf m})$, prime values of polynomials are not enough
to prove the analogue of Corollary 1.4. Indeed, consider the following
particular case of (15):

$P(t)(x_{0}^{2}+x_{1}^{2})+x_{2}^{2}-2x_{3}^{2}=0,$ (16)

where $P(t)$ is the generic polynomial of degree $d$. Since $\delta$ is not a
square in ${\mathbb{Q}}_{2}$, this equation is solvable in ${\mathbb{Z}}_{2}$
for any non-zero value of $P(t)$, see [Cas78, Ch. 4, Lemma 2.6]. When $P(m)=p$
is prime, equation (16) with $t=m$ is solvable in ${\mathbb{Z}}_{p}$ if and
only if either $-1$ or $2$ is a square modulo $p$. Thus solvability in
${\mathbb{Z}}_{p}$ does not hold for 100% of Schinzel polynomials congruent to
$Q(t)$ modulo 8, where $Q(t)$ is an arbitrary polynomial such that $Q(m_{0})$
is odd for some $m_{0}$. (Take $Q(t)$ to be the constant polynomial $3$. Then
$p\equiv 3\bmod 8$, thus neither $-1$ nor $2$ is a square modulo $p$.)

##  2 Brauer group of generic diagonal conic bundles

In this section we prove the triviality of the Brauer group of generic
diagonal conic bundles as was mentioned but not proved in [SS22, Remark 6.2].

Let $n_{1},n_{2},n_{3}$ be non-negative integers such that $n_{1}$ and $n_{2}$
are positive. Suppose that we have positive integers $d_{ij}$, where $i=1,2,3$
and $j=1,\ldots n_{i}$. Let $m_{ijk}$ be independent variables, where
$i=1,2,3$, $j=1,\ldots n_{i}$, and $k=0,\ldots,d_{ij}$. Write ${\bf
m}_{ij}=(m_{ijk})$. Consider the generic polynomials of degree $d_{ij}$:

$P_{ij}(t,{\bf m}_{ij})=\sum_{k=0}^{d_{ij}}m_{ijk}t^{k}\ \in\
{\mathbb{Z}}[t,{\bf m}_{ij}].$

Let $K$ be a field of characteristic zero, and let $F$ be the purely
transcendental extension of $K$ obtained by adjoining all variables $m_{ijk}$,
for $i=1,2,3$, $j=1,\ldots n_{i}$, and $k=0,\ldots,d_{ij}$. Define
$X\to{\mathbb{P}}^{1}_{F}$ as the conic bundle surface which is a natural
compactification of the conic bundle in
${\mathbb{P}}^{2}_{F}\times{\mathbb{A}}^{1}_{F}$ given by (11), where
$a_{1},a_{2},a_{3}\in K^{\times}$.

The main result of this section is

###### Theorem 2.1

The natural map ${\rm{Br}}(F)\to{\rm{Br}}(X)$ is an isomorphism.

We need to introduce some notation. Let $\overline{F}$ be an algebraic closure
of $F$.

Each polynomial $P_{ij}(t)$ is irreducible over $F$, thus

$F_{ij}:=F[t]/(P_{ij}(t))$

is a field extension of $F$ of degree $d_{ij}$. We write ${\rm N}_{ij}\colon
F_{ij}\to F$ for the norm map. The norm map ${\rm N}_{ij}$ gives rise to a
homomorphism $F_{ij}^{\times}/(F_{ij}^{\times})^{2}\to
F^{\times}/(F^{\times})^{2}$, which we shall denote also by ${\rm N}_{ij}$.

If the pairs $(ij)$ and $(rs)$ are different, then $P_{ij}(t)$ and $P_{rs}(t)$
have no common root in $\overline{F}$. Thus the resultant of $P_{ij}(t)$ and
$P_{rs}(t)$, which we denote by $R_{ij,rs}$, is an element of $F^{\times}$. We
note that each $R_{ij,rs}$ is (absolutely) irreducible as a polynomial in the
variables $m_{ijk}$ over $K$, see [GKZ94, p. 398]. The polynomials $R_{ij,rs}$
and $R_{i^{\prime}j^{\prime},r^{\prime}s^{\prime}}$ differ by an element of
$K^{\times}$ if and only if either $i=i^{\prime}$, $j=j^{\prime}$,
$r=r^{\prime}$, $s=s^{\prime}$ or $i=r^{\prime}$, $j=s^{\prime}$,
$r=i^{\prime}$, $s=j^{\prime}$. In the second case we have
$R_{ij,rs}=(-1)^{d_{ij}d_{rs}}R_{rs,ij}$.

To simplify notation, we write $p_{ij}=m_{i,j,d_{ij}}$ for the leading
coefficient of $P_{ij}(t)$. Write ${\rm N}_{ij}(P_{rs})$ for the norm of the
image of $P_{rs}(t)$ in $F_{ij}$. Then we have

$R_{ij,rs}=p_{ij}^{d_{rs}}{\rm
N}_{ij}(P_{rs})=(-1)^{d_{ij}d_{rs}}p_{rs}^{d_{ij}}{\rm N}_{rs}(P_{ij})\ \in\
F^{\times}.$ (17)

For $i=1,2,3$ we define $P_{i}(t)=\prod_{j=1}^{n_{i}}P_{ij}(t)$ and write
$p_{i}=\prod_{j=1}^{n_{i}}p_{ij}$ for the leading coefficient of $P_{i}(t)$.
(When $n_{3}=0$, we write $p_{3}=1$ for the leading coefficient of the
constant polynomial $1$.) Let $d_{i}=\sum_{j=1}^{n_{i}}d_{ij}$ and let
$n=n_{1}+n_{2}+n_{3}$.

Proof of Theorem 2.1. Let $A\in{\rm{Br}}(F(t))$ be the class of the conic (11)
over $F(t)$. Let $\alpha_{ij}$ be the residue of $A$ at the closed point of
${\mathbb{P}}^{1}_{F}$ given by $P_{ij}(t)=0$. It is clear that $\alpha_{ij}$
is the image of
$-a_{i^{\prime}}a_{i^{\prime\prime}}P_{i^{\prime}}(t)P_{i^{\prime\prime}}(t)$
in $F_{ij}^{\times}/(F_{ij}^{\times})^{2}$, where
$\\{i,i^{\prime},i^{\prime\prime}\\}=\\{1,2,3\\}$. From (17) we obtain

${\rm
N}_{ij}(-a_{i^{\prime}}a_{i^{\prime\prime}}P_{i^{\prime}}(t)P_{i^{\prime\prime}}(t))=(-1)^{d_{ij}}a_{i^{\prime}}^{d_{ij}}a_{i^{\prime\prime}}^{d_{ij}}\prod_{j^{\prime}=1}^{n_{i^{\prime}}}p_{ij}^{-d_{i^{\prime}j^{\prime}}}R_{ij,i^{\prime}j^{\prime}}\prod_{j^{\prime\prime}=1}^{n_{i^{\prime\prime}}}p_{ij}^{-d_{i^{\prime\prime}j^{\prime\prime}}}R_{ij,i^{\prime\prime}j^{\prime\prime}}.$

Whenever $1\leq j\leq n_{i}$ and $1\leq j^{\prime}\leq n_{i^{\prime}}$, this
polynomial over $K$ in the variables $m_{ijk}$ is simply divisible by the
irreducible polynomial $R_{ij,i^{\prime}j^{\prime}}$. Therefore in this case
$\alpha_{ij}\in F_{ij}^{\times}/(F_{ij}^{\times})^{2}$ has non-trivial norm
${\rm N}_{ij}(\alpha_{ij})$, hence $\alpha_{ij}$ is non-trivial. By assumption
$n_{1}\geq 1$ and $n_{2}\geq 1$, thus each $\alpha_{ij}$ is non-trivial. In
particular, $\alpha_{1,1}$ and $\alpha_{2,1}$ are two non-trivial residues of
$A$. This implies that $A$ does not belong to the image of the natural map
${\rm{Br}}(F)\to{\rm{Br}}(F(t))$.

The above discussion also shows that all fibres of $X\to{\mathbb{P}}^{1}_{F}$
over the points of ${\mathbb{A}}^{1}_{F}$ are reduced and irreducible. The
fibre at infinity is smooth if $d_{1},d_{2},d_{3}$ have the same parity. Let
us assume that this is the case. By [CTS21, Prop. 11.3.4], the map
${\rm{Br}}(F)\to{\rm{Br}}(X)$ is injective and its cokernel is the homology
group of the following complex:

${\mathbb{Z}}/2\longrightarrow({\mathbb{Z}}/2)^{n}\cong\oplus_{i=1,2,3}\oplus_{j=1}^{n_{i}}{\mathbb{Z}}/2\longrightarrow
F^{\times}/(F^{\times})^{2},$ (18)

where the first map sends the generator $1\in{\mathbb{Z}}/2$ to
$(1,\ldots,1)$, and the second map sends the generator of the $(i,j)$-summand
${\mathbb{Z}}/2$ to ${\rm N}_{ij}(\alpha_{ij})\in
F^{\times}/(F^{\times})^{2}$. (This sequence is a complex by Faddeev’s
reciprocity law [CTS21, Thm. 1.5.2].)

Let $\varepsilon_{ij}\in\\{0,1\\}$, for $i=1,2,3$ and $j=1,\ldots,n_{i}$, not
all of them zero, be such that

$\prod_{i=1,2,3}\prod_{j=1}^{n_{i}}{\rm
N}_{ij}(\alpha_{ij})^{\varepsilon_{ij}}=1\ \in\ F^{\times}/(F^{\times})^{2}.$

The factors given by pairs $(ij)$ and $(rs)$ contribute the irreducible
$R_{ij,rs}$ to the left hand side and no other factor does. Thus if
$\varepsilon_{ij}=1$ for some $i$ and $j$, then we must have
$\varepsilon_{rs}=1$ for all $r\neq i$ and all $s=1,\ldots,n_{r}$. Repeating
the argument, we obtain that $\varepsilon_{ij}=1$ for all possible values of
$i$ and $j$, proving Theorem 2.1 in the case when $d_{1},d_{2},d_{3}$ have the
same parity.

Now suppose that $d_{1},d_{2},d_{3}$ do not have the same parity. Write
$\\{1,2,3\\}=\\{i,i^{\prime},i^{\prime\prime}\\}$, where $d_{i}$ and
$d_{i^{\prime}}$ have the same parity. In this case the fibre at infinity is
not smooth. Let $\alpha_{\infty}$ be the residue of $A$ at infinity. We
calculate that $\alpha_{\infty}\in F^{\times}/(F^{\times})^{2}$ is the image
of $-a_{i}a_{i^{\prime}}p_{i}p_{i^{\prime}}\in F^{\times}$. At least one of
the factors $p_{i},p_{i^{\prime}}$ is an irreducible, and if both are
irreducibles, then they are not constant multiples of each other, thus
$\alpha_{\infty}$ is non-trivial. Now [CTS21, Prop. 11.3.4] says that the map
${\rm{Br}}(F)\to{\rm{Br}}(X)$ is injective and its cokernel is the homology
group of the following complex:

${\mathbb{Z}}/2\longrightarrow({\mathbb{Z}}/2)^{n+1}\cong{\mathbb{Z}}/2\oplus\big{(}\oplus_{i=1,2,3}\oplus_{j=1}^{n_{i}}{\mathbb{Z}}/2\big{)}\longrightarrow
F^{\times}/(F^{\times})^{2}.$ (19)

Here the first map sends the generator $1\in{\mathbb{Z}}/2$ to $(1,\ldots,1)$,
and the second map sends the generator of the $(i,j)$-summand ${\mathbb{Z}}/2$
to ${\rm N}_{ij}(\alpha_{ij})\in F^{\times}/(F^{\times})^{2}$ and sends the
generator of the extra copy of ${\mathbb{Z}}/2$ (given by the point at
infinity) to $\alpha_{\infty}$.

Let $\varepsilon_{ij}\in\\{0,1\\}$, for $i=1,2,3$ and $j=1,\ldots,n_{i}$, and
$\varepsilon_{\infty}\in\\{0,1\\}$, not all of them zero, be such that

$\alpha_{\infty}^{\varepsilon_{\infty}}\prod_{i=1,2,3}\prod_{j=1}^{n_{i}}{\rm
N}_{ij}(\alpha_{ij})^{\varepsilon_{ij}}=1\ \in\ F^{\times}/(F^{\times})^{2}.$

Since $\alpha_{\infty}$ is non-trivial, we must have $\varepsilon_{ij}=1$ for
some pair $(ij)$. Then the same argument as above implies that
$\varepsilon_{ij}=1$ for all pairs $(ij)$. From Faddeev’s reciprocity law (or
a direct calculation) we know that

$\alpha_{\infty}=\prod_{i=1,2,3}\prod_{j=1}^{n_{i}}{\rm
N}_{ij}(\alpha_{ij})\neq 1\ \in\ F^{\times}/(F^{\times})^{2},$

thus we must have $\varepsilon_{\infty}=1$. This finishes the proof of Theorem
2.1. $\Box$

Now let $K={\mathbb{Q}}$ and let $X_{\bf m}\to{\mathbb{P}}^{1}_{\mathbb{Q}}$
be the diagonal conic bundle surface over ${\mathbb{Q}}$ which is the natural
compactification of (11), where ${\bf m}=(m_{ijk})$ with
$m_{ijk}\in{\mathbb{Z}}$, for $i=1,2,3$, $j=1,\ldots n_{i}$, and
$k=0,\ldots,d_{ij}$.

###### Corollary 2.2

The natural map ${\rm{Br}}({\mathbb{Q}})\to{\rm{Br}}(X_{\bf m})$ is surjective
for 100% of ${\bf m}\in{\mathbb{Z}}^{d+n}$ ordered by height.

Proof. By Harari’s theorem [H94, Thm. 3.5.1] there is a hilbertian subset
${\mathcal{H}}\subset{\mathbb{Q}}^{d+n}$ such that
${\rm{Br}}({\mathbb{Q}})\to{\rm{Br}}(X_{\bf m})$ is surjective for ${\bf
m}\in{\mathcal{H}}$. The complement ${\mathbb{Q}}^{d+n}\setminus{\mathcal{H}}$
is a thin set [S97, Section 9.2, Prop. 1]. The classical theorem of S.D. Cohen
(see, e.g. [S97, Section 13.1, Thm. 1]) implies that ${\mathcal{H}}$ contains
100% of the points of ${\mathbb{Z}}^{d+n}$, when they are ordered by height.
$\Box$

##  3 Explicit densities

### Densities of Schinzel $n$-tuples

Let $\ell$ be a prime. For $P(t)\in{\mathbb{F}}_{\ell}[t]$ we denote by
$Z_{P}(\ell)$ the number of zeros of $P(t)$ in ${\mathbb{F}}_{\ell}$. For
$n\in\mathbb{N}$ and ${\bf d}=(d_{1},\ldots,d_{n})\in\mathbb{N}^{n}$, let

$\delta_{n}(\ell,{\bf d}):=\frac{\\#\\{{\bf
P}\in{\mathbb{F}}_{\ell}[t]^{n}:{\rm{deg}}(P_{i})\leq d_{i},Z_{P_{1}\cdots
P_{n}}(\ell)\neq\ell\\}}{\ell^{n+d_{1}+\ldots+d_{n}}}$

be the density of Schinzel $n$-tuples modulo $\ell$. Write
$d=d_{1}+\ldots+d_{n}$. An explicit expression for $\delta_{n}(\ell,{\bf d})$
is given in [SS22, Eq. (2.6)], but when $\ell$ is large or small compared to
the degrees $d_{i}$ it is easy to calculate $\delta_{n}(\ell,{\bf d})$
directly. For example, if $\ell>d$, then $Z_{P_{1}\cdots P_{n}}(\ell)\neq\ell$
if and only if each $P_{i}(t)$ is a non-zero polynomial, thus

$\delta_{n}(\ell,{\bf
d})=\prod_{i=1}^{n}\left(1-\frac{1}{\ell^{d_{i}+1}}\right).$

###### Lemma 3.1

If $\ell\leq 1+\min\\{d_{i}:1\leq i\leq n\\}$, then

$\delta_{n}(\ell,{\bf d})=1-\left(1-(1-1/\ell)^{n}\right)^{\ell}.$

Proof. The probability for a given polynomial of positive degree to have a
non-zero value at a given point of ${\mathbb{F}}_{\ell}$ is $1-1/\ell$. The
product of $n$ polynomials of positive degrees does not vanish at a given
point of ${\mathbb{F}}_{\ell}$ with probability $(1-1/\ell)^{n}$, so it
vanishes with probability $1-(1-1/\ell)^{n}$. If $\ell\leq d_{i}$ for all $i$,
these events are independent, so the product of polynomials vanishes
everywhere on ${\mathbb{F}}_{\ell}$ with probability $1-\delta_{n}(\ell,{\bf
d})=\big{(}1-(1-1/\ell)^{n}\big{)}^{\ell}$. $\Box$

###### Remark 3.2

For $\ell=2$ the assumptions of the lemma are always met, so we have
$\delta_{n}(2,{\bf d})=1/2^{n-1}-1/4^{n}$. In particular, the density of
Schinzel $n$-tuples is always at most $1/2^{n-1}$, hence goes to $0$ when
$n\to\infty$.

### The case $n=2$

Expression [SS22, Eq. (2.6)] is complicated to evaluate if
$1+\min\\{d_{i}\\}<\ell\leq d$. We now give a more practical lower bound in
the case $n=2$.

###### Lemma 3.3

We have

$\delta_{2}(\ell,{\bf
d})\geq\left(1-\frac{1}{\ell^{1+d_{1}}}\right)\left(1-\frac{1}{\ell^{1+d_{2}}}\right)-\frac{2^{\ell}}{\ell^{\ell}}.$

Proof. In the counting function in the definition of $\delta_{2}(\ell,{\bf
d})$ we can assume that neither polynomial is identically zero, thus

$\delta_{2}(\ell,{\bf d})=(\ell^{d_{1}+1}-1)(\ell^{d_{2}+1}-1)-N,$

where

$N=\\#\\{(P_{1},P_{2})\in({\mathbb{F}}_{\ell}[t]\setminus\\{0\\})^{2}:{\rm{deg}}(P_{i})\leq
d_{i},Z_{P_{1}P_{2}}(\ell)=\ell\\}.$

Since $P_{1}(t)P_{2}(t)$ vanishes everywhere on $F_{\ell}$, we see that

$N\leq\sum_{S\subset{\mathbb{F}}_{\ell}}\\#\\{(P_{1},P_{2})\in({\mathbb{F}}_{\ell}[t]\setminus\\{0\\})^{2}:{\rm{deg}}(P_{i})\leq
d_{i},\ P_{1}(S)=P_{2}(S^{c})=0\\},$

where $S^{c}={\mathbb{F}}_{\ell}\setminus S$. Since $\\#S\leq d_{1}$ and
$\\#S^{c}\leq d_{2}$, we obtain

$\displaystyle N\leq$
$\displaystyle\sum_{\begin{subarray}{c}S\subset{\mathbb{F}}_{\ell}\\\ \ell-
d_{2}\leq\\#S\leq
d_{1}\end{subarray}}\\#\\{(P_{1},P_{2})\in({\mathbb{F}}_{\ell}[t]\setminus\\{0\\})^{2}:{\rm{deg}}(P_{i})\leq
d_{i},\ P_{1}(S)=P_{2}(S^{c})=0\\}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}S\subset{\mathbb{F}}_{\ell}\\\ \ell-
d_{2}\leq\\#S\leq
d_{1}\end{subarray}}(-1+\ell^{1+d_{1}-\\#S})(-1+\ell^{1+d_{2}-\ell+\\#S})\leq\ell^{2+d_{1}+d_{2}-\ell}\,2^{\ell},$

which gives the desired bound. $\Box$

### Equation $P_{1}(t)x^{2}+P_{2}(t)y^{2}=z^{2}$

The main goal of this section is to give an explicit lower bound for the
density of pairs of integer polynomials $P_{1}(t)$, $P_{2}(t)$ of fixed
degrees $d_{1}$, $d_{2}$, respectively, such that the above equation is
solvable in ${\mathbb{Z}}$.

We use Corollary 1.3 with $n_{1}=n_{2}=1$, $n_{3}=0$, $a_{1}=a_{2}=1$, and
$a_{3}=-1$. We only need to deal with the prime $\ell=2$. If $a$ and $b$ are
odd integers, then the Hilbert symbol $(a,b)_{2}$ equals 1 if and only if $a$
or $b$ is $1\bmod 4$. Thus we need to calculate the probability $\sigma_{2}$
that a pair of polynomials $P_{1}(t),P_{2}(t)\in({\mathbb{Z}}/4)[t]$ of
degrees ${\rm{deg}}(P_{1})\leq d_{1}$, ${\rm{deg}}(P_{2})\leq d_{2}$ satisfies
the following conditions:

  * (a)

$P_{1}(0)$ and $P_{2}(0)$ are both odd, or $P_{1}(1)$ and $P_{2}(1)$ are both
odd, and

  * (b)

$P_{i}(n)\equiv 1\bmod 4$ for some $i\in\\{1,2\\}$ and some
$n\in{\mathbb{Z}}/4$.

Condition (a) is the Schinzel condition at $2$, and condition (b) is the
triviality of the Hilbert symbol at $2$. Recall that $d=d_{1}+d_{2}$. We have

$\sigma_{2}=\delta_{2}(2,{\bf d})-4^{-d-2}\\#T=1/2-1/4^{2}-4^{-d-2}\\#T,$

where $T$ is the set of pairs $P_{1}(t),P_{2}(t)\in({\mathbb{Z}}/4)[t]$ of
degrees ${\rm{deg}}(P_{1})\leq d_{1}$, ${\rm{deg}}(P_{2})\leq d_{2}$
satisfying condition (a) and taking only values $0$, $2$, and $3$ modulo $4$.

There are $2^{e-1}$ polynomials of degree at most $e$ in ${\mathbb{F}}_{2}[t]$
with given values at $0$ and $1$. Each of these can be lifted to exactly
$2^{e-3}$ polynomials of degree at most $e$ in $({\mathbb{Z}}/4)[t]$ with
given values at the elements of ${\mathbb{Z}}/4$ that are compatible with the
values at $0$ and $1$ modulo $2$. Thus there are $4^{e-2}$ polynomials
$P(t)\in({\mathbb{Z}}/4)[t]$ of degree at most $e$ with given values at the
elements of ${\mathbb{Z}}/4$ such that the values at 0 and 2 have the same
parity, and similarly for the values at 1 and 3.

For $i\in{\mathbb{Z}}/4$ we write $v_{i}=P_{1}(i)$ and
$v^{\prime}_{i}=P_{2}(i)$. Condition (a) is equivalent to the condition that
$v_{0}v^{\prime}_{0}$ is odd or $v_{1}v^{\prime}_{1}$ is odd. We obtain that
$\\#T$ is the product of $4^{d-4}$ and the number of
$(v_{j})\in\\{0,2,3\\}^{4}$, $(v^{\prime}_{j})\in\\{0,2,3\\}^{4}$ such that

$v_{0}v^{\prime}_{0}\text{ or }v_{1}v^{\prime}_{1}\,\text{is odd,
and}\,v_{0}-v_{2},v_{1}-v_{3},v^{\prime}_{0}-v^{\prime}_{2},v_{1}^{\prime}-v_{3}^{\prime}\
\text{are even}.$

Since $2\mid v_{0}-v_{2}$, the values of $(v_{0},v_{2})$ can only be

$(0,0),(0,2),(2,0),(2,2),(3,3).$ (20)

The same can be said about
$(v_{1},v_{3}),(v^{\prime}_{0},v^{\prime}_{2}),(v^{\prime}_{1},v^{\prime}_{3})$.
If $v_{0}v^{\prime}_{0}$ is odd, then
$v_{0}=v_{2}=v^{\prime}_{0}=v^{\prime}_{2}=3$. Then there are 25 possibilities
for $(v_{1},v_{3},v^{\prime}_{1},v^{\prime}_{3})$ since both $(v_{1},v_{3})$
and $(v^{\prime}_{1},v^{\prime}_{3})$ can be chosen arbitrarily from the list
in (20). A similar counting gives another $25$ cases when
$v_{1}v^{\prime}_{1}$ is odd. We have counted twice the cases where both
$v_{0}v^{\prime}_{0}$ and $v_{1}v^{\prime}_{1}$ are odd. This implies that all
$v_{i}$ and $v^{\prime}_{i}$ equal 3, so there is in fact only one such case.
We obtain $\\#T=25+25-1=49$, and hence

$\sigma_{2}=1/2-1/4^{2}-\frac{49}{4^{6}}=\frac{1743}{4096}=42.553\ldots.$

Evaluating the product $\sigma_{2}\prod_{\ell\geq 3}\delta_{2}(\ell,{\bf d})$
using Lemmas 3.1 and 3.3 gives the following bound.

###### Proposition 3.4

Let $d_{1}$ and $d_{2}$ be positive integers. The density of pairs of integer
polynomials $P_{1}(t)$, $P_{2}(t)$ of degrees ${\rm{deg}}(P_{1})=d_{1}$ and
${\rm{deg}}(P_{2})=d_{2}$ with positive leading coefficients, ordered by
height, such that the equation

$P_{1}(t)x^{2}+P_{2}(t)y^{2}=z^{2}$

is solvable in ${\mathbb{Z}}$, is at least

$\frac{1743}{4096}\prod_{3\leq\ell\leq 1+\min
d_{i}}\left(1-\left(\frac{2\ell-1}{\ell^{2}}\right)^{\ell}\right)\prod_{\ell>1+\min
d_{i}}\left(\left(1-\frac{1}{\ell^{1+d_{1}}}\right)\left(1-\frac{1}{\ell^{1+d_{2}}}\right)+O\left(\frac{2^{\ell}}{\ell^{\ell}}\right)\right).$

When $\min\\{d_{i}\\}\to\infty$ this product becomes asymptotic to

$\mathcal{P}:=\frac{1743}{4096}\prod_{\text{prime }\ell\geq
3}\left(1-\left(2/\ell-1/\ell^{2}\right)^{\ell}\right).$

This is strictly exceeding $0.35$. Verifying $\mathcal{P}>0.35$ with any
computer package will lead to troubles related to accumulating rounding errors
owing to $(2/\ell-1/\ell^{2})^{\ell}$. Thus, it is better to use the bound

$\left(2/\ell-1/\ell^{2}\right)^{\ell}<(2/\ell)^{\ell}\leq
1/\ell^{4},\quad\text{for}\ \ \ell\geq 17,$

which implies that

$\mathcal{P}_{0}\prod_{\text{prime }\ell\geq
17}\left(1-1/\ell^{4}\right)<\mathcal{P}<\mathcal{P}_{0},$

where

$\mathcal{P}_{0}=\frac{1743}{4096}\prod_{\text{prime
}\ell\in[3,13]}\left(1-\left(2/\ell-1/\ell^{2}\right)^{\ell}\right).$

We can easily calculate that

$\prod_{\ell\geq 17}(1-1/\ell^{3})=0.9999723\ldots\text{ and
}\mathcal{P}_{0}=0.3504\ldots,$

therefore,

$0.3503<\mathcal{P}<0.3504.$

## References

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Department of Mathematics, South Kensington Campus, Imperial College London,
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Problems, Russian Academy of Sciences, Moscow, 127994 Russia

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