Generic diagonal conic bundles revisited

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 [SS23, Thm.1.2].Here we give some improvements, complements and further applications of the results of [SS23] relevant to diagonal conic and quadric bundles over the projective line.
In the first section we prove a stronger form of [SS23, 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 [SS23, 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 1.5.(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 [SS23, Remark 6.2] and proved in the second section of this note, see Theorem 2.1 and Corollary 2.3.
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 is soluble in 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).We are very grateful to the referees for their thorough reading of the paper and helpful comments.
1 Schinzel hypothesis on average with quadratic residue conditions Non-constant polynomials P 1 (t), . . ., P n (t) ∈ Z[t] are called a Schinzel n-tuple if the leading coefficient of each P i (t) is positive, and for every prime ℓ the product n i=1 P i (t) is not divisible by t ℓ − t modulo ℓ.The height of a polynomial P (t) ∈ 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 P = (P 1 (t), . . ., P n (t)) ∈ (Z[t]) n is defined as |P| = max i=1,...,n (|P i |) The following result is [SS23, Thm.1.2] with additional properties (2), (3) and (6).The proof of (6) uses [SS23, Prop.6.5] based on Heath-Brown's bound for character sums [HB95,Cor. 4].
Theorem 1.1 Fix any (d 1 , . . ., d n ) ∈ N n , ε > 0, M ∈ N, m 0 ∈ Z/M and Q ∈ ((Z/M)[t]) n such that deg(Q i ) ≤ d i and gcd(Q i (m 0 ), M) = 1 for all i = 1, . . ., n.For every (i, j) ∈ (N∩[1, n]) 2 with i < j let ǫ ij ∈ {1, −1}.Then for 100% of Schinzel n-tuples P ∈ (Z[t]) n of respective degrees d 1 , . . ., d n such that P ≡ Q mod M, there exists a natural number m with the following properties: the Legendre symbol P i (m) P j (m) equals ǫ ij for all i < j. Proof.Define Let us write Ω = {(i, j) : 1 ≤ i < j ≤ n}.For any S ⊂ Ω define 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 ∅,P (x) = θ P (x).Assuming (5), the following function takes the value 1 if (6) holds and the value 0 otherwise: which leads to For the rest of the proof we restrict attention only to the range where A 1 , A 2 are arbitrary fixed constants satisfying n < A 1 < A 2 .By [SS23, Eq. (6.7)] the term corresponding to S = ∅ equals where θ P (x) is defined similarly to θ P (x) by dropping the condition that the primes P 1 (m), . . ., P n (m) are necessarily distinct.Let us define We next show that uniformly in the range (8) and for all S = ∅ one has Letting makes it plain that T S,P takes the form of the function η P from [SS23,Def. 6.4].This allows us to apply [SS23, Prop.6.5] to verify (10).Feeding (9)-( 10) into (7) yields Hence the number of P in Poly(H) Therefore, for 100% of P in Poly(H) one has which implies where we used [SS23, Eq. (4.10)] for the last deduction.Hence, there exists m ≤ x satisfying (4)-(6).To verify (3), we take which proves (3).To prove (2) we note that if the largest integer m ≤ x that satisfies (4)-( 6) is m 1 then For almost all P with |P| ≤ H we have shown that C P (x) > x log x .Combined with the upper bound ).Note that for almost all P with |P| ≤ H one has min i,j |c ij | ≥ H log log H , hence, for all i we have Hence, for all i one has In particular, min i P i (m 1 ) > |P|(log |P|) ε/2 , which concludes the proof of (2).

Generic diagonal conic bundles
As an application we give a simplified proof of [SS23, 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 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 3 i=1 f i x 2 i represents zero in Q, then there exists a solution (x 1 , x 2 , x 3 ) ∈ N 3 such that For m = (m 0 , m 1 , . . ., m r ) ∈ Z r we write P (t, m) for the polynomial r k=0 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 ≥ 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, . . ., n i .Define such that the n-tuple (P ij (t, m ij )) is Schinzel.Let M be the set of m ∈ P such that for each p|2a 1 a 2 a 3 the equation has a solution in Z p for which the value of each polynomial P ij (t) is a p-adic unit.
Then there is a subset M ′ ⊂ M of density 1 such that for every m ∈ M ′ the equation (11) has a solution The set M ′ has positive density in 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 M is a finite disjoint union of subsets given by the condition for which there exists an m 0 ∈ Z such that (11) with t = m 0 has a solution in Z p for each p|M, and gcd(Q ij (m 0 ), M) = 1.Let us fix such a Q and such an m 0 .Then for any P ≡ Q mod M and any m ≡ m 0 mod M the equation ( 11) with t = m is soluble in Z p for each p|M.Suppose that p ij := P ij (m), where i = 1, 2, 3 and j = 1, . . ., n i , are distinct primes, where 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 ≡ m 0 mod M so that ( 11) is solvable at each of the primes p ij = P ij (m).
For i = 1, 2, 3 define Here the middle term is the Legendre symbol and the right hand term is the Hilbert symbol.By global reciprocity we obtain Define λ ij ∈ F 2 as follows: • λ 1,j = λ 1j , for j = 1, . . ., n 1 ; ) mod M, we see from (12) that the λ ij depend only on Q and m 0 .Thus the same is true for the λ ij .
Lemma 1.4 We have i=1,2,3 Proof.By assumption, a 1 , a 2 , a 3 are not of the same sign.Since M = 8a 1 a 2 a 3 , from global reciprocity we obtain Local solubility of the conic (11) at a prime p is equivalent to so this holds for all p|M.Thus the product of From the definition of λ ij it is immediate that this product equals the product of (−1) λ ij over all i and j.
We continue the proof of Corollary 1.3.
Local solubility of the conic (11) at p ij is equivalent to the condition where If i > i ′ , we define x ij,i ′ j ′ to be equal to x i ′ j ′ ,ij .Thus we always have x ab,cd = x cd,ab .We observe that (13) is equivalent to the equation This is clear for i = 1.For i = 2 and i = 3 we prove (14) using which immediately follows from global reciprocity.Consider the system of n = n 1 +n 2 +n 3 linear equations (14) in n 1 n 2 +n 2 n 3 +n 1 n 3 variables x ij,i ′ j ′ .By assumption we have n 1 > 0 and n 2 > 0, so we have at least one variable, namely x 11,21 , and n ≥ 2 equations.The sum of the left hand sides of all equations ( 14) is zero, and it is easy to see that the matrix of this linear system has rank n − 1.Thus the linear map given by this matrix is surjective onto the subspace of vectors with zero sum of coordinates in (F 2 ) n .We conclude that the system ( 14) is solvable for arbitrary λ ij with zero sum.In our case this holds by Lemma 1.4, so (14) has a solution, say ε ij,i ′ j ′ ∈ F 2 , where (ij) and (i ′ j ′ ) are pairs as above.
Applying Theorem 1.1, for P ≡ Q mod M in a subset of density 1 we find a natural number m ≡ m 0 mod M, m ≤ (log |P|) n+1/2 , such that the numbers P ij (m) are distinct primes satisfying Then the conic (11) with t = m is everywhere locally solvable, and so is solvable in Z. Furthermore, Proposition 1.2 ensures the existence of a solution where

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 , . . ., d n be positive integers and let d = j=0 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, . . ., n}.The equation defines a family of quadrics Q t parametrised by the affine line with coordinate t over the field Q(m), where m = (m 1 , . . ., m n ).The generic fibre Q η is a quadric of dimension 2 over Q(t, m).We can multiply (15) by a non-zero element of Q(m) and absorb squares into coordinates x i without affecting the isomorphism class of Q η .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, . . ., n} belongs to more than two of the sets S i .Define Corollary 1.5 In the above notation assume that δ is not a square in Q(t, m).Let P be the set of m = (m i ) ∈ Z d+n such that the n-tuple (P i (t, m i )) is Schinzel.Let M be the set of m ∈ P such that for each p|2a the equation (15) has a solution in Z p for which the value of each polynomial P i (t) is a p-adic unit.Then there is a subset M ′ ⊂ M of density 1 such that for every m ∈ M ′ the equation (15) has a solution in Z.The set M ′ has positive density in 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 M is a finite disjoint union of subsets given by the condition P ≡ Q mod M, where Q is an n-tuple of polynomials in (Z/M)[t] for which there exists an m 0 ∈ Z such that (15) with t = m 0 has a solution in Z p for each p|M, and gcd(Q ij (m 0 ), M) = 1.Let us fix such a Q and such an m 0 .Then for any P ≡ Q mod M and any m ≡ m 0 mod M the equation (15) with t = m is solvable in Z p for each p|M.
Suppose that p i := P i (m), where i = 1, . . ., n, are distinct primes, where P i (t) ≡ Q i (t) mod M and m ≡ m 0 mod M. Thus p i ≡ Q i (m 0 ) mod 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, . . ., r} belongs to exactly one of the sets S i , and every element of {r + 1, . . ., n} belongs to two of these sets.The condition on δ implies r ≥ 1, so the prime p 1 belongs to exactly one of the sets S i .The local solubility of (15) with t = m at the primes p i , where i = 1, . . ., r, is automatic: this follows from the fact that the conic obtained by setting x j = 0 in (15), where i ∈ S j , has good reduction modulo p i .It remains to show that for P ≡ Q mod M in a subset of density 1 we can choose m ≡ m 0 mod M so that ( 15) is solvable at each of the primes p r+1 , . . ., p n .
To apply Theorem 1.1 we define the values ǫ ij = p i p j for i < j, as follows.The values of ǫ 1,i = 1 for i = 2, . . ., r are of no importance and can be chosen arbitrary.For i = r + 1, . . ., n we define Since p i ≡ Q i (m 0 ) mod M, we see that this depends only on Q and m 0 .For k ≥ 2 we define ǫ k,l = 1 for all l > k.Then (15) with t = m is solvable in Z p i for i = r + 1, . . ., n, because up to multiplication by a square the product of all four coefficients is ap 1 . . .p r , which is a non-square modulo p i .This implies solubility in Z p i by [Cas78, Ch. 4, Lemma 2.6].An application of Theorem 1.1 finishes the proof since the resulting quadric Q m is everywhere locally solvable, hence solvable in Z.
The case when δ is a square in Q(m) can be reduced to the case of conic bundles.Indeed, let C → 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 × 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 δ is a square in Q(m).
When δ is a square in Q(t, m) but not a square in Q(m), prime values of polynomials seem to be insufficient to prove the analogue of Corollary 1.5.Indeed, consider the following particular case of (15): 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 [SS23, 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, . . .n i .Let m ijk be independent variables, where i = 1, 2, 3, j = 1, . . .n i , and k = 0, . . ., d ij .Write m ij = (m ijk ).Consider the generic polynomials of degree d 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, . . .n i , and k = 0, . . ., d ij .For a 1 , a 2 , a 3 ∈ K × consider the subvariety X ′ ⊂ P 2 F × A 1 F given by (11).It is easy to see that X ′ is smooth and geometrically integral, and the projection to A 1 F is a proper morphism whose fibres are conics.There is a natural compactification of X ′ → A 1 F to a smooth projective surface X with a conic bundle structure X → P 1 K .The main result of this section is Theorem 2.1 Let n 1 > 0, n 2 > 0, n 3 ≥ 0. Then the natural map Br(F ) → Br(X) is an isomorphism.
We need to introduce some notation.Let F be an algebraic closure of F .Each polynomial P ij (t) is irreducible over F , thus is a field extension of F of degree d ij .We write N ij : F ij → F for the norm map.The norm map N ij gives rise to a homomorphism , which we shall denote also by N ij .
For distinct pairs (ij) and (rs) define R ij,rs ∈ F as the resultant of P ij (t) and P rs (t) considered as polynomials in F [t]. Since P ij (t) and P rs (t) have no common root in F , we have R ij,rs = 0. We note that each R ij,rs is an (absolutely) irreducible polynomial in the variables m ijk over K, see [GKZ94,p. 398].The polynomials R ij,rs and R i ′ j ′ ,r ′ s ′ differ by an element of K × if and only if either i = i ′ , j = j ′ , r = r ′ , s = s ′ or i = r ′ , j = s ′ , r = i ′ , s = j ′ .In the second case we have R ij,rs = (−1) d ij drs R rs,ij .
To simplify notation, we write p ij = m i,j,d ij for the leading coefficient of P ij (t).Write N ij (P rs ) for the norm of the image of P rs (t) in F ij .Then we have For i = 1, 2, 3 we define P i (t) = n i j=1 P ij (t) and write p i = n i j=1 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 = n i j=1 d ij and let n = n 1 + n 2 + n 3 .Proof of Theorem 2.1.Let A ∈ Br(F (t)) be the class of the conic (11) over F (t). Equivalently, A is the class of the quaternion algebra associated to this conic.Each closed point M = Spec(F M ) of P 1 F gives rise to the residue of A at this point; this is an element of For any j ′ ≤ n i ′ this is a product of the irreducible polynomial R ij,i ′ j ′ and a rational function coprime to R ij,i ′ j ′ .In particular, the norm N ij (α ij ) is non-trivial, hence α ij is non-trivial.By assumption n 1 ≥ 1 and n 2 ≥ 1, thus each α ij is non-trivial.In particular, α 1,1 and α 2,1 are two non-trivial residues of A. It follows that A does not belong to the image of the natural map Br(F ) → Br(F (t)), hence the map Br(F ) → Br(X) is injective by [CTS21,Lemma 11.3.3].
The above calculation also shows that all fibres of X → P 1 F over the points of 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 first assume that this is the case.By [CTS21, Prop.11.3.4], the cokernel of Br(F ) → Br(X) is the homology group of the following complex: where the first map sends the generator 1 ∈ Z/2 to (1, . . ., 1), and the second map sends the generator of the (i, j)-summand Z/2 to N ij (α ij ) ∈ F × /(F × ) 2 .This sequence is a complex by Faddeev's reciprocity law [CTS21, Thm.1.5.2].
Let ε ij ∈ {0, 1}, for i = 1, 2, 3 and j = 1, . . ., n i , not all of them zero, be such that i=1,2,3 The factors given by pairs (ij) and (rs) contribute the irreducible element R ij,rs to the left hand side and no other factor does.Thus if ε ij = 1 for some i and j, then we must have ε rs = 1 for all r = i and all s = 1, . . ., n r .Repeating the argument, we obtain that ε ij = 1 for all possible values of i and j.This proves 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 ′ , i ′′ }, where d i and d i ′ have the same parity.Let α ∞ be the residue of A at infinity.We calculate that then this is simply divisible by p 2,1 .In all cases we conclude that α ∞ is non-trivial.Thus the fibre of X → P 1 F at infinity is singular, reduced and irreducible.Now [CTS21, Prop.11.3.4] says that the map Br(F ) → Br(X) is injective and its cokernel is the homology group of the following complex: Here the first map sends the generator 1 ∈ Z/2 to (1, . . ., 1), and the second map sends the generator of the (i, j)-summand Z/2 to N ij (α ij ) ∈ F × /(F × ) 2 and sends the generator of the extra copy of Z/2 (given by the point at infinity) to α ∞ .
Proof.Since the smooth proper surface X F is rational, we have Br(X F ) = 0 by the birational invariance of the Brauer group [CTS21, Cor.6.2.11].Thus we obtain a functorial exact sequence be the Zariski open subset given by the condition that the discriminants and the leading coefficients of all the polynomials P ij (t) and the resultants of all pairs of the polynomials P ij (t) are non-zero.We have U = ∅, e.g., because given by ( 11) is smooth and geometrically integral when m ∈ U(Q), because no polynomial in t divides more than one coefficient and if it divides some coefficient then it simply divides it.For m ∈ U(Q) we denote by X m → P 1 Q the conic bundle surface over Q which is a natural smooth compactification of X ′ m .Specialisation at m ∈ U(Q) preserves the degrees of the polynomials P ij (t), thus X m can be obtained as the specialisation of the conic bundle surface X → P 1 F considered above.
Corollary 2.3 For 100% of points m ∈ Z d+n ordered by height the natural map Br(Q) → Br(X m ) is an isomorphism.
Proof.100% of points m ∈ Z d+n are contained in U, and for such points the surface X m and thus the map Br(Q) → Br(X m ) are well defined.For m ∈ U(Q) we have compatible specialisation maps sp m : Pic(X) → Pic(X m ) and sp m,Q : Pic(X F ) → Pic(X The exact sequence (18) and Theorem 2.1 imply that the top horizontal map is an isomorphism.It follows that for m ∈ H the bottom map is also an isomorphism.This map fits into an exact sequence Proof.The probability for a given polynomial of positive degree to have a nonzero value at a given point of F ℓ is 1 − 1/ℓ.The product of n polynomials of positive degrees does not vanish at a given point of F ℓ with probability (1 − 1/ℓ) n , so it vanishes with probability 1 − (1 − 1/ℓ) n .If ℓ ≤ d i for all i, these events are independent, so the product of polynomials vanishes everywhere on F ℓ with probability Remark 3.2 For ℓ = 2 the assumptions of the lemma are always met, so we have δ n (2, 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 → ∞.

The case n = 2
Expression [SS23, Eq. (2.6)] is complicated to evaluate if 1 + min{d i } < ℓ ≤ d.We now give a more practical lower bound in the case n = 2.
Proof.In the counting function in the definition of δ 2 (ℓ, d) we can assume that neither polynomial is identically zero, thus Since P 1 (t)P 2 (t) vanishes everywhere on F ℓ , we see that which gives the desired bound.
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 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 ℓ = 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 mod 4. Thus we need to calculate the probability σ 2 that a pair of polynomials P 1 (t), P 2 (t) ∈ (Z/4)[t] of degrees deg(P 1 ) ≤ d 1 , deg(P 2 ) ≤ 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) ≡ 1 mod 4 for some i ∈ {1, 2} and some n ∈ 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 where T is the set of pairs P 1 (t), P 2 (t) ∈ (Z/4)[t] of degrees deg(P 1 ) ≤ d 1 , deg(P 2 ) ≤ 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 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 (Z/4) [t] with given values at the elements of Z/4 that are compatible with the values at 0 and 1 modulo 2. Thus there are 4 e−2 polynomials P (t) ∈ (Z/4)[t] of degree at most e with given values at the elements of Z/4 such that the values at 0 and 2 have the same parity, and similarly for the values at 1 and 3.
so by [CTS21, Remark 5.4.3 (2)] the last map in (18) is surjective.It remains to apply Theorem 2.1.Let K = Q.We can think of m = (m ijk ) as the coordinates of the affine space A d+n Q , where n = n 1 + n 2 + n 3 and d = ij d ij .Let U ⊂ A d+n Q Evaluating the product σ 2 ℓ≥3 δ 2 (ℓ, 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 deg(P 1 ) = d 1 and 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 from which we conclude that for m ∈ H the natural map Br(Q) → Br(X m ) is an isomorphism.The complement Q d+n \ 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 H contains 100% of points of Z d+n , when they are ordered by height.Let ℓ be a prime.For P (t) ∈ F ℓ [t] we denote by Z P (ℓ) the number of zeros ofP (t) in F ℓ .For n ∈ N and d = (d 1 , . .., d n ) ∈ N n , let δ n (ℓ, d) := #{P ∈ F ℓ [t] n : deg(P i ) ≤ d i , Z P 1 •••Pn (ℓ) = ℓ} ℓ n+d 1 +...+dn be the density of Schinzel n-tuples modulo ℓ.Write d = d 1 + . . .+ d n .An explicit expression for δ n (ℓ, d) is given in [SS23, Eq. (2.6)], but when ℓ is large or small compared to the degrees d i it is easy to calculate δ n (ℓ, d) directly.For example, if ℓ > d, then Z P 1 •••Pn (ℓ) = ℓ if and only if each P i (t) is a non-zero polynomial, thus