Brauer–Manin obstructions for sums of two squares and a power

# Integral Brauer–Manin obstructions for sums of two squares and a power

Fabian Gundlach Mathematisches Institut, Ludwig-Maximilians-Universität München, Theresienstr. 39, 80333 München, Germany
###### Abstract.

We use Brauer–Manin obstructions to explain failures of the integral Hasse principle and strong approximation away from for the equation with fixed integers and . Under Schinzel’s hypothesis (H), we prove that Brauer–Manin obstructions corresponding to specific Azumaya algebras explain all failures of strong approximation away from at the variable . Finally, we present an algorithm that, again under Schinzel’s hypothesis (H), finds out whether the equation has any integral solutions.

###### 2010 Mathematics Subject Classification:
11P05 (14F22, 11G35)

## 1. Introduction

For integers and we consider the equation

 (1) x2+y2+zk=m.

For the famous theorem of Gauß about sums of three squares says that (1) has an integral solution if and only if and is not of the form for non-negative integers and . The non-existence of integral solutions can in this case always be explained by the non-existence of real or 2-adic solutions.

Vaughan conjectured in [Vau81, Chapter 8] that for sufficiently large there is an integral solution to (1) satisfying whenever for each prime there is some solution to the above equation such that . For odd such local solutions always exist. His conjecture would then imply the integral Hasse principle for sufficiently large .

This was however disproved by Jagy and Kaplanski in [JK95]. They gave an elementary proof using quadratic reciprocity that there is no integral solution if and for some prime . The remark following their theorem mentions that if is an odd composite integer, then for infinitely many equation 1 has no solution.

Dietmann and Elsholtz gave examples of failures of strong approximation in [DE08b] for and more general ones in [DE08a] for arbitrary .

Brauer–Manin obstructions were originally introduced by Manin to explain failures of the Hasse principle for rational points on certain cubic surfaces (see for example [Man70]). For an overview of further developments of Brauer–Manin obstructions for the Hasse principle and weak approximation for rational points see [Pey05].

This method was adapted to integral points and applied to quadratic forms such as by Colliot-Thélène and Xu in [CTX09]. Further examples of failures of the integral Hasse principle and strong approximation explained by Brauer–Manin obstructions are given in [KT08], [CTW12], [CTX13] and [CTH12].

We show that the counterexample to the integral Hasse principle given in [JK95] can be explained by a Brauer–Manin obstruction (see Theorem 4.7).

Furthermore, we systematically find new counterexamples to the integral Hasse principle and strong approximation:

###### Theorem 1.

The following equations do not fulfill strong approximation away from due to Brauer–Manin obstructions:

 x2+y2+zk=nk, k≥3 odd, n≡1mod4 x2+y2+zk=nk, k≥2 even, n>0
###### Proof.

See Corollary 4.4. ∎

Our second goal is to show the fulfillment of the integral Hasse principle and strong approximation away from at the variable in case there is no Brauer–Manin obstruction via certain elements of the Brauer group. Unfortunately, we can only do this under assumption of Schinzel’s hypothesis (H), a generalization of Dirichlet’s theorem on primes within arithmetic progressions to prime values of polynomials.

Schinzel’s hypothesis (H) has been employed by Colliot-Thélène and Sansuc in [CTS82] to prove the Hasse principle and weak approximation for rational solutions of equations similar to (1). This technique has subsequently been used for example in [CTSD94], [CTSSD98a], [CTSSD98b], [Wit07] and [Wei12].

However, as far as we know, for integral points the potential use of Schinzel’s hypothesis (H) was so far only briefly mentioned in Remark (v) on pages 618–619 of [CTSSD98a].

###### Theorem 2.

Let be an odd integer. Under Schinzel’s hypothesis (H) each solution to equation 1 without any Brauer–Manin obstruction generated by Azumaya algebras of the form described in Section 3.1 can be approximated with respect to the variable by integral solutions to equation 1.

###### Proof.

See Theorem 5.4. ∎

Jagy and Kaplanski conjectured in [JK95] that (1) has an integral solution whenever is an odd prime.

###### Theorem 3.

Let be an odd prime. Under Schinzel’s hypothesis (H) every integer is of the form for integral .

###### Proof.

See Corollary 5.9. ∎

For each prime and let .

###### Theorem 4.

Let be the product of two primes and let .

For the existence of integral solutions to equation 1, it is necessary and under Schinzel’s hypothesis (H) also sufficient that the following two statements are both true.

• There is no such that and for each prime dividing :

or or there is no such that

 p∣rp(n)a−1+⋯+z′(a−1)b.
• There is no such that and for each prime dividing :

or or there is no such that

 p∣rp(n)b−1+⋯+z′(b−1)a.
###### Proof.

See Theorem 5.11. ∎

For and odd an algorithm is given in Section 6, which, using Schinzel’s hypothesis (H), determines whether is of the form .

Finally, lists of small positive integers not of the form are given for small odd .

Acknowledgements. We thank Jean-Louis Colliot-Thélène, Christian Elsholtz, Dasheng Wei and the referee for their comments.

## 2. Preliminaries

From now on, let be a number field, the set of places of and the set of archimedian places of . Let be the completion of with respect to for each . Let be the corresponding valuation ring for each and let for each . The valuation associated to is called . The ring is called the ring of integers of .

In this section, let be a variety over .

For topological rings over , the set of -rational points obtains the induced topology.

Given a class of varieties, one often wants to know whether the existence of local solutions implies the existence of global solutions, or, even better, whether the existence of local integral solutions implies the existence of integral solutions. This leads to

###### Definition 2.1.

Let be a subset of . The set of -adeles

 AS:={(xv)v∈Ω∖S∈∏v∈Ω∖SKv∣∣xv∈Ov for almost every v∈Ω∖S}

is a ring by coordinatewise addition and multiplication. The ring is called the adele ring of . The sets

 {∏v∈Ω∖SAv∣∣Av=Ov for almost all v∈Ω∖S and Av open in Kv% for all v∈Ω∖S}

define a basis for the topology on .

For the field may be diagonally embedded into as for every there are only finitely many such that . Below, the images of these embeddings are identified with .

Given a variety and some , obviously . It is of interest how relates to .

###### Definition 2.2.

We say that the variety satisfies strong approximation away from if (where the closure is taken inside ), i.e., if is dense in .

An introduction to Brauer–Manin obstructions can be found in [Sko01].

###### Definition 2.3 ([Mil80, Chapter IV]).

An -algebra is called an Azumaya algebra over if it is coherent (i.e., there is some open covering by affine schemes , such that for some finitely generated -module for each ) and if is a central simple algebra over the residue field for every .

If furthermore is a field extension of , then for each (i.e., each morphism of -schemes) let .

###### Remark 2.4.

If is an Azumaya algebra over , then is a central simple -algebra for each , as is a central simple -algebra and is a -algebra by the morphism , such that

 A(x)≅(Ax(η)⊗OX,x(η)κ(x(η)))⊗κ(x(η))k.
###### Definition 2.5.

For let be the invariant map from local class field theory. For simplicity, we will refer to the class of in by , too.

###### Theorem 2.6 (Brauer–Manin obstruction, [Sko01, Chapter 5.2] and [Ctx13, Section 2]).

For every Azumaya algebra over the set

 X(A)A:={(xv)v∈X(A)∣∣∣∑vinvv(Av(xv))=0}

contains .

Hence, for each the set defined as

 {(xv)v∈Ω∖S∈X(AS)∣∣∣∃(xv)v∈S∈X(AΩ∖S) such % that ∑vinvv(Av(xv))=0}

contains .

## 3. Azumaya algebra

In this section, we will define an Azumaya algebra over the scheme defined by equation 1. We will then compute its local invariants.

### 3.1. Construction

Let . Recall, that for each prime and we defined .

For each let denote the Hilbert symbol of degree 2 (i.e., if and only if there exist such that ).

For each ring of characteristic different from 2 and let denote the quaternion algebra over with parameters (i.e., it is a free -module with basis such that , and ).

###### Lemma 3.1.

For all , we have:

 (a,−1)=1 ⇔∃x,y∈Qv:a=x2+y2 ⇔invv(a,−1Qv)=0 ⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a>0,v=∞,r2(a)≡1mod4,v=2,0=0,v≡1mod4,2∣vv(a),v≡3mod4.

For -adic integers , we even have:

 (a,−1)=1⇔∃x,y∈Zv:a=x2+y2.
###### Proof.

See [Ser73, III.1, Theorem 1] and [GS06, Proposition 1.1.7].

The last equivalence is trivial if , so let be prime. The implication from right to left is obvious. Conversely, remark that there are at least such that . Let . If , then , so assume . Then we have with . Therefore, . As and are not both divisible by , this implies that is a quadratic residue modulo . Hence, , so there are such that . According to the pigeonhole principle, is the sum of two quadratic residues, which we can lift to satisfying using Hensel’s Lemma (as and ). Finally, repeated application of Brahmagupta’s identity

 (xx′′−yy′′)2+(xy′′+yx′′)2=(x2+y2)(x′′2+y′′2)=p(x2+y2)

yields such that . ∎

Let and be integers such that or . Consider the equation

 (2) x2+y2+zab=na.

Let

 X:=SpecZ[X,Y,Z]/(X2+Y2+Zab−na)

and

 XQ:=X⊗ZQ=SpecQ[X,Y,Z]/(X2+Y2+Zab−na).

The variety is covered by the principal open subsets and . Indeed, as , we have

 V(n−Zb)∩V(na−1+⋯+Z(a−1)b) =V(n−Zb,na−1+⋯+Z(a−1)b) =V(n−Zb,ana−1)=∅.

Consider the -algebra

 A1:=(n−Zb,−1OXQ(U1))∼

and the -algebra

 A2:=(na−1+⋯+Z(a−1)b,−1OXQ(U2))∼.

There is an -algebra isomorphism induced by

 i ↦Xi′+Yi′j′na−1+⋯+Z(a−1)b j ↦j′

where and are the canonical generators of and , respectively (this is an -algebra isomorphism as ). Hence and can be glued along to obtain an -algebra such that and . Quaternion algebras over fields (with nonzero arguments) are central simple algebras, so is an Azumaya algebra.

In the following, we are interested in strong approximation “at ” away from . To this end, we choose a suitable topology: In we equip the first two components (i.e., those belonging to the variables and ) with the trivial topology (sometimes called indiscrete topology) and the last one (i.e., that belonging to the variable ) with the usual topology on . Accordingly, the sets , , etc. obtain the induced topologies.

Strong approximation with respect to the usual topology (i.e., at , and ) seems more difficult, as for fixed the equation does not fulfill the integral Hasse principle (unlike the equation ).

If strong approximation “at ” away from is not fulfilled, then strong approximation away from with respect to the usual topology is not fulfilled, either.

###### Lemma 3.2.

Let

 U:=U1∩U2=D(na−Zab)

and

 Iv:=invv(A(X(Zv))).

for any . Then and

 Iv=invv(A(U(Qv)∩X(Zv))).

For we have

 invv(A(x,y,z))=0 ⇔(n−zb,−1)=1 if n−zb≠0, invv(A(x,y,z))=0 ⇔(na−1+⋯+z(a−1)b,−1)=1 if na−1+⋯+z(a−1)b≠0,
 w∈Iv⇔∃z∈Zv such that (na−zab,−1)=1 and (n−zb,−1)={1,w=0,−1,w=1/2.
###### Proof.

The inclusion follows from the fact that quaternion algebras over fields have order 2 in the Brauer group .

The first two equivalences follow straight from the definition of and Lemma 3.1.

It is easy to see that is dense in . Hence, is dense in because is an open subset of . As is locally constant (even in the topology chosen above!), this implies that .

For each satisfying , there exist such that if and only if . Together with the first equivalence and this proves the final one. ∎

### 3.2. Place ∞

We have .

###### Proof.

A real solution is (as or ), so .

If , then and or , so . As we have , i.e., , so . Hence . ∎

### 3.3. Place 2

###### Lemma 3.4.

If and is odd, then .

###### Proof.

Due to for it is obvious that . The set is nonempty as there is some odd such that and this fulfills (as and are odd), so . ∎

If , then .

###### Proof.

Let . Then , so . ∎

###### Lemma 3.6.

If and and , then .

###### Proof.

Let and . Now and , so .

Furthermore , so . ∎

###### Lemma 3.7.

If and are odd and , then .

###### Proof.

Let . Then and , so . ∎

###### Lemma 3.8.

If are odd and , then .

###### Proof.

We know from the previous lemma.

Let .

If is odd, then , so .

If is even, then

 1≡r2(na−zab)≡r2((n/2)a−2a(b−1)(z/2)ab)≡r2(n/2)a≡3mod4

###### Lemma 3.9.

If are odd and , then .

###### Proof.

The values of in the following table fulfill :

0 1 2 3 4 5 7
-1 0 0 1 3 5

### 3.4. Odd places

###### Lemma 3.10.

We have for all odd primes .

###### Proof.

One of the numbers or is not divisible by . Set or , respectively. Then and hence are not divisible by , so . ∎

Hence it is only interesting whether .

###### Lemma 3.11.

We have for all primes .

###### Proof.

The previous lemma implies . Furthermore, we have for all . ∎

Hence only the case is interesting, so let be prime for the rest of Section 3.4.

If and , then .

###### Proof.

Take and . Then

 (na−zab1,−1) =−1 (as 2∣0=vp(na−zab1)) (na−zab2,−1) =−1 (as 2∣avp(n)=vp(na−zab2)) (n−zb1,−1) =−1 (as 2∣0=vp(n−zb1)) (n−zb2,−1) =−1 (as 2∤vp(n)=vp(n−zb2)).

Let be odd for the rest of Section 3.4. Then the following lemma simplifies the analysis of .

###### Lemma 3.13.

Let such that . Then the following statements are equivalent:

1. There are such that and (hence ).

2. and

 1 ≡vp(rp(n)a−1+⋯+rp(z)(a−1)b) mod2 1+vp(n) ≡vp(rp(n)−rp(z)b) mod2 vp(n) ≡vp(rp(n)a−rp(z)ab) mod2 .
###### Remark 3.14.

The sum of the first two congruences in statement b) is the third one, so only two of them have to be proved.

###### Proof of the lemma.

Assume a). Then (as ):

 2∤vp(na−1+⋯+z(a−1)b)

If , then yields a contradiction.

If , then yields a contradiction.

Hence .

Then

 1 ≡vp(na−1+⋯+z(a−1)b) ≡(a−1)vp(n)+vp(rp(n)a−1+⋯+rp(z)(a−1)b) ≡vp(rp(n)a−1+⋯+rp(z)(a−1)b)mod2

and (as )

 1+vp(n) ≡vp(n)+vp(n−zb) ≡2vp(n)+vp(rp(n)−rp(z)b) ≡vp(rp(n)−rp(z)b)mod2.

Conversely, b) implies

 vp(na−zab) ≡avp(n)+vp(rp(n)a−rp(z)ab) ≡(a+1)vp(n)≡0mod2,

so there are such that .

Furthermore

 vp(n−zb)≡vp(n)+vp(rp(n)−rp(z)b)≡1mod2,

so . ∎

Assume . Then .

###### Proof.

Suppose and . Then and are odd. Hence and have to be divisible by , so together . Therefore . ∎

Assume . Then .

###### Proof.

Suppose and . According to Lemma 3.13 we have , so . ∎

###### Lemma 3.17.

Let . Then the following two statements are equivalent:

1. .

2. and and there is some such that .

###### Proof.

Assume a). Let such that .

Lemma 3.13 shows that (so ). It also shows that , so .

If , then according to Lemma 3.13. Therefore and . Together , which is obviously impossible. Therefore .

Conversely, assume b). As and obviously , we have

 rp(n)a−z′ab≢rp(n)a−z′ab−abz′ab−1p≡rp(n)a−(z′+p)abmodp2.

Hence or . Let or , respectively.

Therefore (as )

 1≤vp(rp(n)a−1+⋯+rp(z)(a−1)b)≤