Sums of squares - EPFL · 2018. 10. 2. · CHAPTER 1 Sums of squares One of the great theorems in arithmetic in the 19th century is Lagrange’s four square theorem: Theorem (Lagrange,

CHAPTER 1

Sums of squares

One of the great theorems in arithmetic in the 19th century is Lagrange’sfour square theorem:

Theorem (Lagrange, 1770). A non-zero integer n is representable as asum of four squares, that is, there exists (a, b, c, d) 2 Z4 such that

(0.1) n = a2 + b2 + c2 + d2.

if and only if n > 0.

A quadruple (a, b, c, d) 2 Z4 satisfying the above equation is called anintegral representation of n as a sum of four squares. Alternatively, let q4denote the Euclidean quadratic form on Q4

q4(a, b, c, d) = a2 + b2 + c2 + d2,

the quadruple (a, b, c, d) is called an (integral) representation of n by q4.Let R4(n) be the set of all such representations:

R4(n) = {(a, b, c, d) 2 Z4, a2 + b2 + c2 + d2 = n};for n 6= 0, that set is non-empty if an only if n is positive.

1. A proof of Lagrange’s theorem after A. Venkov

We shall now give a proof of Lagrange’s theorem following ideas of A:Venkov [Ven22,Ven29]: this involve the introduction into the picture ofanother object: the algebra of Hamilton1 quaternions.

1.1. The Hamilton quaternion. This is the 4-dimensional associa-tive, non commutative, Q-algebra

B(Q) = Q+Q.i+Q.j +Q.k

with i, j, k satisfying the relations

i2 = j2 = k2 = �1, ij = �ji = k.

The center of B is the algebra of scalars

ZB(Q) = Q = Q.1.

1Discovered by Hamilton in 1843

7

8 1. SUMS OF SQUARES

The algebra B may be explicitely realized as a Q-algebra of 2⇥ 2 (com-plex) matrices via

(1.1) 1 7! Id, i 7!✓

0 �11 0

◆, j 7!

✓0 ii 0

◆, k 7!

✓�i 00 i

◆.

Alternatively, letting B(Q) act on itself by left multiplications (viewingB(Q) ' Q4 as a Q-vector space), one also may realize B as an algebraof 4⇥ 4 matrices: the map defined by

(1.2) a+ bi+ cj + dk 7!

0BB@a �b �c �db a �d cc d a �bd �c b a

1CCAis an isomorphism of Q-algebras.

1.2. Reduced Norm, Reduced trace. The algebra B is endowedwith an involutive Q-linear anti-automorphism: the canonical involution

z = a+ bi+ cj + dk 7! z = a� bi� cj � dk

(ie. satisfyingz1z2 = z2.z1.)

Remark 1.1. When B is identified with the algebra of 2 ⇥ 2 complexmatrices generated by the four matrices given in (1.1), the canonical involu-tion is the ”conjugate-transpose” map. When B is realized as an algebra of4⇥ 4 rational matrices, the canonical involution is just the ”transpose”map

Out of the canonical involution, one defines the Q-valued reduced traceand reduced norm maps

(1.3) tr(z) = z + z = 2a,

(1.4) Nr(z) = zz = a2 + b2 + c2 + d2.

The trace is Q-linear and satisfies

tr(z1z2) = tr(z2z1).

From its concrete expression (1.4), the norm defines a quadratic form, whopolarization is the bilinear, symetric function of two variables given by

(z1, z2) ! Nr(z1 + z2)�Nr(z1)�Nr(z2) = z1z2 + z1z2 = tr(z1z2).

In particular the linear map

(1.5)B(Q) 7! Q4

a+ bi+ cj + dk 7! (a, b, c, d),

is an isometry between the quadratic spaces (B(Q),Nr) and (Q4, q4) sendingthe basis {1, i, j, k} to the canonical basis {e1, e2, e3, e4}.

Another quite remarkable property of the norm is that it ismultiplicative:

Nr(z1z2) = Nr(z1)Nr(z2);

1. A PROOF OF LAGRANGE’S THEOREM AFTER A. VENKOV 9

indeed

Nr(z1z2) = z1z2z1z2 = z1z2z2z1 = z1Nr(z2)z1

= Nr(z2)z1z1 = Nr(z1)Nr(z2).

(in that way one retrieves the so-called Lagrange identity in dimension 4)The quadratic form Nr is obviously anisotropic: given z 2 B(Q), one has

Nr(z) = 0 if and only if z = 0.A consequence is that a quaternion z is invertible in B(Q) if and only if

it is non zero, its inverse being:

z�1 = z/Nr(z).

In other terms B(Q) is a division algebra (or a skew field): the multiplicativegroup of invertible elements of B(Q) equals

B⇥(Q) = B(Q)� {0}.

Remark. The adjective reduced, for the trace or the norm come fromthe fact that, if we view B(Q) as a subalgebra of the algebra, EndQ(B(Q)),of linear endomorphism of the Q-vector space (cf. the matrix realization1.2), the trace and the determinant of a quaternion z are equal respectively,to twice the reduced trace and to the square of the reduced norm of z.

1.2.1. The characteristic polynomial. Any z 2 B(Q) is annihilated bythe quadratic polynomial with rational coe�cients:

(1.6) Pz(X) = X2 � tr(z)X +Nr(z) = (X � z)(X � z);

We have evidently

| tr(z)|2 4Nr(z)

and moreover this inequality is strict if and only if z is not a scalar (z 62Q); in such a case, Pz(X) is an irreducible polynomial (even over R) andthe commutative Q-algebra generated by z, Q[z] = Q(z) is a (imaginary)quadratic field embedded into B(Q) and one has the equalities between tracesand norms

tr(z) = trQ(z)/Q(z), Nr(z) = NQ(z)/Q(z).

1.2.2. Trace 0-quaternions. The kernel of the trace map is a linear sub-space of B(Q), called the space of trace zero or pure quaternions

B0(Q) = {z 2 B(Q), tr(z) = 0} = {bi+ cj + dk, b, c, d 2 Q}.

1.2.3. Norm-1 quaternions. Similarly we denote by B(1)(Q) the kernelof the norm homomorphism on B⇥(Q):

B(1)(Q) = {z 2 B(Q), Nr(z) = 1} = {a+bi+cj+dk, a2+b2+c2+d2 = 1}.


1.3. The integral quaternions and the Hurwitz quaternions.From the shape of the reduced norm (1.4), we see that the question ofdetermining the representations of n as a sum of four squares is equivalentto determining the quaternions of norm n with having integral entries: inother terms, let

B(Z) = Z+ Zi+ Zj + Zkbe the set of integral quaternion and let

B(n)(Z) = {z = a+ bi+ cj + dk, a, b, c, d 2 Z, Nr(z) = n}

be the subset of integral quaternion of norm n; the map

(1.7) (a, b, c, d) 7! a+ bj + cj + dk

yields a bijection between R4(n) and B(n)(Z).The Z-module B(Z) is a subring of B(Q) (ie. it contain the identity and

is stable under multiplication) and is free rank 4: in other terms B(Z) is anorder in B(Q). As it turns out, B(Z) is contained is a slightly bigger order(with index 2): the ring of Hurwitz quaternions

OB = Zi+ Zj + Zk + Z1 + i+ j + k

2= Z[i, j, k,

1 + i+ j + k

2].

The proof of Lagrange’s theorem rests of the following:

Theorem 1.1. The ring of Hurwitz quaternions satisfies the followingproperties

(1) The elements of OB have integral reduced norm and trace.(2) The group of units of OB and B(Z) are

B(Z)⇥ = {±1,±i,±j,±k},

O⇥B = {±1,±i,±j,±k,

±1± i± j ± k

2}.

(3) For any z 2 OB there is a unit w 2 O⇥B such that wz 2 B(Z).

(4) The ring OB is principal: any left (resp. right) OB-ideal I ⇢ B(Q)is of the form OB.z (resp. z.OB) for some z 2 B⇥

Proof. (1) is a general property of elements of finitely generated (asZ-module) subrings of B(Q): if z is contained into some finitely generatedsubring, the commutative ring Z[z] is finitely generated and hence the monicpolynomial Pz(X) (cf. (1.6)) which annihilates z has integral coe�cients.

Property (2) follows immediately from the fact, a consequence of (1),that z 2 OB is a unit if and only if Nr(z) = 1.

(3) is a simple computation.As for (4): recall that a left (resp. right) OB-ideal I ⇢ B is a finitely

generated Z-module of maximal rank (4) such that OB.I ⇢ I (resp. I.OB ⇢I). The principality is a consequence of the following


Claim. the order OB is euclidean w.r.t the norm Nr: for any q, z 2 OB,q 6= 0, there is h, r 2 OB such that

z = hq + r, Nr(r) < Nr(q).

Proof. (of the Claim) Indeed, considering zq�1 2 B there is l 2 B(Z)such that z.q�1�l = a+bi+cj+dk has its coordinates contained in the cubeof radius 1/2 [�1/2, 1/2]4 = {|a|, |b|, |c|, |d| 1/2} so that k(a, b, c, d)k =(a2 + b2 + c2 + d2)1/2 = Nr(z.q�1 � l)1/2 1; moreover equality hold ifand only if |a| = |b| = |c| = |d| = 1/2 but in that case replacing l bysome h = l + ±1±i±j±k

2 2 OB we can always insure that k(a, b, c, d)k =

Nr(z.q�1 � h)1/2 < 1 hence Nr(z � hq) < Nr(q). ⇤Now given I a OB-left ideal; up to multiplying I by a scalar a 2 Q⇥

we may assume that I ⇢ OB. Let q 2 I � {0} of minimal norm (it existsbecause I is discrete in B(R)); then for any z 2 I, there is h 2 OB such thatr = z � hq 2 I has norm < Nr(q), but the minimality of Nr(q) guarantiesthat r = 0. ⇤

We have the following consequence of (4)

Corollary 1.1. The ring of Hurwitz quaternions is a maximal orderof B(Q). Any order, and more generally, any finitely generated subringO ⇢ B(Q) is conjugate to a subring of OB: there is q 2 B⇥(Q) such that

q�1Oq ⇢ OB.

Proof. The maximality property follows from the second part of thecorollary. Let O ⇢ B(Q) be any finitely generated subring, then O.OB isfinitely generated hence is a right OB-ideal; it i therefore principal:

O.OB = qOB

for some q 2 B⇥(Q) and then

q�1Oq ⇢ q�1OqOB = q�1O.O.OB ⇢ q�1O.OB = OB.

⇤Remark. The ring B(Z) is not principal.

1.4. Proof of Lagrange theorem. Lagrange’s theorem is equivalentto showing for any n � 1 the existence of an integral quaternion z 2 B(Z)of norm n.

1. If n = 1, take for z any element of B(Z)⇥, so we may assume thatn � 2.

2. Since Nr(z1z2) = Nr(z1)Nr(z2), we see, by decomposing n into a prod-uct of primes, that it is su�cient to show the existence of such a quaternionwhen n = p a prime.

3. If p = 2, z = 1+i has norm 2 and clearly the set of integral quaternionof norm 2 is precisely B(Z)⇥.(1 + i) (so has order 8).

4. Let now p be an odd prime, we claim the following


Lemma. The equation

a2 + b2 + c2 = 0

has a solution in Fp3 � {(0, 0, 0)}.

Proof. There are precisely p+12 squares in Fp ((Fp

⇥)2 [ {0}), hence

the sets {1 + b2, b 2 Fp} and {�c2, c 2 Fp} have cardinality p+12 ; since

p+12 + p+1

2 > p these sets have non empty intersection ⇤From the above lemma, there exists (a, b, c) 2 Z3 such that a ⌘ 1(mod p)

anda2 + b2 + c2 ⌘ 0(mod p).

Set w = ai + bj + ck 2 OB and let I be the left ideal I = OBw + OBp; onehas

OB.p ( I ( OB.

Indeed the first inclusion is strict since w 62 OBp and the second becauseall the elements of I have norm divisible by p. By the above lemma I isprincipal; let z be a generatorof I, then z 62 O⇥

B there is z0 2 OB such thatp = z0z. We have Nr(p) = p2 = Nr(z)Nr(z0) and since z0 is not a unit (sinceI contains OB.p properly) we must have Nr(z) = Nr(z0) = p. A priori zbelong to OB but from (3) of the previous proposition we may multiply itby a unit w 2 O⇥

B such that wz 2 B(Z) and Nr(wz) = Nr(z) = p. ⇤1.5. Quaternions and quadratic spaces. As we have seen, one of

the main virtue of introducing the Hamilton quaternions has been to endowthe quadratic space (Q4, a2 + b2 + c2 + d4), with an extra multiplicativestructure, via the linear isomorphism,

(1.8)B(Q) 7! Q4

a+ bi+ cj + dk 7! (a, b, c, d),

so that the four squares quadratic form becomes multiplicative with respectto that structure2.

In that section, we discuss in greater details the relationships betweenthe quadratic and the quaternionic structures. The norm form is a quadraticform on B(Q) whose polarization is the inner product

(1.9) hz, z0i = Nr(z + z0)�Nr(z)�Nr(z0) = tr(zz0)

and the map (1.8) is an isometry of quadratic spaces (Q4, q4) ' (B(Q),Nr),hence

SO4 ' SOB .

We now express SOB in terms of quaternions; we refer to Chapter 10for the details and for more general statements. Observe that given z, z0 2B⇥(Q), and w 2 B(Q) one has

Nr(zwz0�1) = Nr(z/z0)Nr(w),

2 For general quadratic forms a related construction exists: it yields the Cli↵ordalgebra (or algebra of spinors).


therefore the linear map

⇢z,z0 :B(Q) 7! B(Q)w 7! zwz0�1

is an orthogonal similitude of the quadratic space (B(Q),Nr) with similitudefactor

µ(⇢z,z0) = Nr(z/z0).

Moreover a computation shows that

det(⇢z,z0) = µ2(⇢z,z0),

so the above maps defines a group homomorphism

⇢ :B⇥(Q)⇥ B⇥(Q) 7! GSOB(Q) ' GSO4(Q)

(z, z0) 7! ⇢z,z0.

where GSOB denote the connected component of GOB the group of orthog-onal similitudes of (B,Nr) (the kernel of the character

⇢ 2 GOB 7! det(⇢)/µ(⇢)2).

This map is surjective and its kernel is the diagonal scalar subgroup

�Z⇥B (Q) = {(�,�), � 2 Q⇥}.

In fact, we obtain an isomorphism of Q-algebraic groups

(1.10) B⇥ ⇥ B⇥/�Z⇥B ' GSOB .

The kernel of the similitude factor µ is the group of special orthgonal isome-tries SOB and restricting to it we obtain a isomorphism ofQ-algebraic groups

(1.11) (B⇥ ⇥ B⇥)Nr/�Z⇥B ' SOB

where(B⇥ ⇥ B⇥)Nr = {(z, z0) 2 B⇥ ⇥ B⇥, Nr(z) = Nr(z0)}.

1.5.1. The trace zero quaternions as quadratic space. We consider nowthe three dimensional space of pure quaternion

B0(Q) = {z 2 B(Q), tr(z) = 0}with its quadratic space structure given by the norm form. From (1.9) we seethat B0(Q) is precisely the subspace of vectors orthogonal to the quaternion1,

B0(Q) = Q.1? = {z 2 B(Q), hz, 1i = tr(z.1) = tr(z) = 0}.It follows that its special orthogonal group SOB0 is the stabilizer of thevector 1 in GSOB. In term of quaternions, this subgroup is given by theisometries of the shape ⇢z,z for z 2 B⇥; in other terms

SOB0 = StabGSOB(1) ' {(z, z), z 2 B⇥}/�Z⇥B ' B⇥/Z⇥

B = PB⇥

the projective group of invertible quaternions.The map (1.8) induces an isomorphism between the quadratic spaces

(Q3, b2 + c2 + d2) ' (B0(Q),Nr)


and therefore we have an isomorphism of Q-algebraic groups

(1.12) ⇢ : PB⇥ ' SOB0 ' SO3 .

1.5.2. The group of quaternions of norm one. Let

B(1) = {z 2 B, Nr(z) = 1} ⇢ B⇥

denote the variety of quaternions of norm 1; being the kernel of the normhomomorphism, this is a normal algebraic subgroup of B⇥. By the inclusionB(1) ,! B⇥, B(1) acts on B0 and one has a short exact sequence of algebraicgroups (over C)

(1.13) 1 ! {±1} ! B(1) ! PB⇥ ! 1.

In fact relative to the identification PB⇥ ' SOB0 ' SO3, B(1) getsnaturally identified with the spin group (the simply connected 2-coveringgroup of the special orthogonal group)

B(1) ' SpinB0 ' Spin3.

1.5.3. The Hopf fibration. The previous section is especially useful todiscuss the Hopf filtration: over the reals the sequence (1.13) remains exact:

1 ! {±1} ! B(1)(R) ! PB⇥(R) ! 1.

By Witt’s theorem B(1)(R) acts transitively on B0,(1)(R) the set of vectorsof norm 1 (by conjugation) making it a principal homogeneous space: themap

(1.14)B(1)(R) 7! B0,(1)(R)

z 7! ⇢z,z(k) = zkz�1

is surjective, its fiber above any w 2 B0,(1)(R) is the stabilizer of w in B(1)(R),

B(1)(R)w = StabB(1)(R)(w) = {a+ d.w, a, d 2 R, a2 + d2 = 1} ' S1

and yields to the homeomorphism

B0,(1)(R) ' B(1)(R)/B(1)(R)k;

(indeed the set of quaternions z commuting with w (such that zw = wz) isthe set of quaternions of the shape a+ d.w, a, d 2 R.)

Now, via (1.8), B(1)(R) is identified with the 3-sphere

S3 = {(a, b, c, d) 2 R4, a2 + b2 + c2 + d2 = 1},B0,(1)(R) with the 2-sphere

S2 = {(0, b, c, d) 2 R4, b2 + c2 + d2 = 1},and from the above discusion we obtain a realization of S3 as a (topological)S1-bundle over the 2-sphere

(1.15) S1 ,! S3 ⇣ S2.

This is the so-called Hopf fibration.

2. THE NUMBER OF SUMS OF FOUR SQUARES REPRESENTATIONS 15

2. The number of sums of four squares representations

After having established the existence of representations of and integeras a sum of four squares, the next step is: how many representations arethere ?

Thirty years after Lagrange, Jacobi, using theta series, gave a closedformula of the number of such representations:

Theorem (Jacobi, 1828). For n � 1,

|R4(n)| = |B(n)(Z)| = 8Xd|n4-d

d = 8(1 + 2�2|n)Y

p↵kn,p 6=2

p↵+1 � 1

p� 1.

The factor 8 is the order of the group of units of the ring B(Z)B⇥(Z) = {±1,±i,±j,±k}

which acts faithfully on B(n)(Z) by left multiplication. So Jacobi’s formulacould be rewritten

(2.1) |B⇥(Z)\B(n)(Z)| =Xd|n4-d

d = (1 + 2�2|n)Y

p↵||n,p 6=2

p↵+1 � 1

p� 1.

It follows from these formulas and from the estimate for the number ofdivisors

d(n) =Xd|n

1 = no(1)

that, as n ! +1, 4 - n

(2.2) |R4(n)| = n1+o(1).

We will not discuss the proof here; for now let us merely say that themain ingredient of Jacobi’s proof is the introduction of an analytic object:Jacobi’s theta function, defined for z 2 C, =(z) > 0 by

✓4(z) =X

(a,b,c,d)2Z4

exp(2⇡i(a2 + b2 + c2 + d2)z) =Xn�0

|R4(n)| exp(2⇡inz).

Using the Poisson summation formulation, one shows that ✓4(z) is an holo-morphic modular form of weight 2 relative to the subgroup

�0(4) = {✓

a bc d

◆2 SL2(Z), c ⌘ 0(4)};

that is for � =

✓a bc d

◆2 �0(4)

✓4(az + b

cz + d) = (cz + d)2✓4(z),

and in fact is an Eisenstein series. Using this information one can thenobtain Jacobi’s formula (cf. [Iwa97,Sar91]).


3. The distribution of four squares representations

As we have seen, as n ! +1, there are “more and more” representationof n as a sum of four squares. Our next question then is

How are these representations distributed as n ! 1 ?One of the simplest way to investigate this question is the following: any

representation (a, b, c, d) is a vector in R4 of euclidean norm n1/2. Projectingsuch points radially to the 3-sphere we obtain a sequence of finite (essentiallygrowing) sets of points

n�1/2.R4(n) ⇢ S3.

3.1. Equidistribution of points on the 3-sphere. As it turn out,these points become distributed in the most natural possible way: let µS3

denote the ”Lebesgue”, rotationally invariant, probability measure on the3-sphere

Theorem (Malyshev). As n ! +1 n 6⌘ 0(4), the set n�1/2.R4(n)becomes equidistributed on S3 with respect to µS3: for any ' 2 C(S3),

1

|R4(n)|X

x2R4(n)

'(x

n1/2) !

ZS3

'(x)dµS3(x)

Below we give several equivalent formulations of Malyshev’s equidistri-bution theorem.

3.2. Equidistribution of quaternions. As we have seen R4 is isomet-ric to B(R), and under this isometry, R4(n) is identified with the of integralquaternions of norm n, B(n)(Z), and the 3-sphere S3 with the group realquaternion of norm one B(1)(R). The later a compact group and underthis identification, the Lebesgue measure on S3 correspond the Haar prob-ability measure µB(1)(R). Therefore an equivalent formulation of Malyshev’s

theorem is: for any ' 2 C(B1(R)),

(3.1)1

|B(n)(Z)|X

z2B(n)(Z)

'(z

n1/2) !

ZB1(R)

'(z)dµB1(z), n ! +1.

3.3. Equidistribution of rotations. The Hopf map discussed in §1.5.3makes it possible to represent graphically this equidistribution property: toeach non-zero quaternion z is associated a rotation rz 2 SO3(R), namelythe rotation corresponding to the rotation ⇢z,z 2 SOB0(R) via the isometryB0(R) ' Re2 + Re3 + Re4 ' R3. This yields an homeomorphism

B(1)(R)/{±1} ' SO3(R)

and the Haar probability measure on SO3(R) correspond to the Haar proba-bility measure on B(1)(R) (restricted to the continuous functions invariantunder multiplication by ±1). Therefore we obtain

3. THE DISTRIBUTION OF FOUR SQUARES REPRESENTATIONS 17

Theorem 3.1. As n ! +1 n 6⌘ 0(4), the set of rotations

{rz/n1/2 , z 2 B(n)(Z)} ⇢ SO3(R)

becomes equidistributed on SO3(R) with respect to the Haar probability mea-sure: for any f 2 C(SO3(R)),

1

|R4(n)|X

z2R4(n)

f(rz/n1/2) !ZSO3(R)

f(⇢)dµSO3(⇢).

3.3.1. Equidistribution on the 2-sphere. By Witt’s theorem the action ofSO3(R) on S2 is transitive; so any v0 2 S2, the map

r 2 SO3(R) 7! r(v0)

induces and homeomorphism

SO3(R)/ SO3(R)v0 ' S2.

Under this map, the rotation invariant Lebesgue probability measure µS2

on the 2-sphere corresponds to the Haar probability measure on SO3(R)(restricted to the subspace of continuous function on SO3(R) which areSO3(R)v0 . Therefore we deduce

Corollary 3.1. Given any v0 2 S2, as n ! +1 n 6⌘ 0(4), the multiset{rz/n1/2(v0), z 2 B(n)(Z)} becomes equidistributed on S2 w.r.t. µS2: for any

' 2 C(S2),

1

|B(n)|X

z2B(n)

'(⇢z/n1/2(v0)) !ZS2

'(v)dµS2(v).

In fact this Corollary (the equidistribution of the translates of v0 by the⇢z for any v0 is equivalent to Theorem 3.1.

3.4. Sketch of a proof. Malyshev’s theorem is to be proven in thecourse of this book in a much more general form but let us sketch theprinciple of the proof as it is exposed in [Iwa97, Chap. 11].

It is known ([Far08]) that any continuous function on S3 may be approx-imated uniformly on S3 by a linear combination of harmonic homogeneouspolynomials: let us recall that an harmonic homogeneous polynomial, is therestriction to S3 of a polynomial �⌫ (⌫ 2 N an integer) on R4 which ishomogeneous of degree ⌫, that is such that for � 2 R

�⌫(�x) = �⌫�⌫(x), and such that �R4�⌫ = 0

were

�R4 =@2

@2x+

@2

@2y+

@2

@2z+

@2

@2t


Figure 1. The distribution of {⇢z(v0), z 2 R4(53)}, v0 = (1, 0, 0).

Figure 2. The distribution of {⇢z(v0), z 2 R4(54)}, v0 = (1, 0, 0).

is the Laplace operators on R4. This is a consequence of the Peter-Weyltheorem3 for the compact group SO4(R). Explicitely, one can show thatthe space of harmonic homogeneous polynomial of degree ⌫ is generated by

3A continuous function on a compact group may be approximated uniformly by linearcombination of the matrix coe�cients of irreductible continuous representations the group

3. THE DISTRIBUTION OF FOUR SQUARES REPRESENTATIONS 19

polynomials of the shape

�⌫(x, y, z, t) = ax+ by + ct+ dt

for (a, b, c, d) 2 C4, varying over the set of q4 isotropic vectors: ie. such that

q4(a, b, c, d) = a2 + b2 + c2 + d2 = 0.

In view of this, it is su�cient (by Weyl’s equidistribution criterion) toshow that for any such polynomial

(3.2) W (�⌫ ;n) =1

|R4(n)|X

x2R4(n)

�⌫(x

|n|1/2 ) !ZS3

�⌫(x)dµS3(x)

The quantitiesW (�⌫ ;n) are called theWeyl sum associated with this equidis-tribution problem.

– If ⌫ = 0, �⌫ is a constant polynomial and

W (�⌫ ;n) = �⌫ =

ZS3

�⌫(x)dµS3(x).

– If ⌫ > 0, thenRS3 �⌫(x)dµS3(x) = 0 (�⌫ is orthogonal to the constants)

and we have to show that W (�⌫ ;n) ! 0.– If ⌫ is odd, then since R4(n) = �R4(n),

W (�⌫ ;n) = (�1)⌫W (�⌫ ;n) = 0

and we are done.– Let us now suppose that ⌫ is even. We consider, for

z 2 H = {z 2 C, =(z) > 0}( the upper half-plane), the theta function

✓(�⌫ , z) =X

(a,b,c,d)2Z4

�⌫(a, b, c, d) exp(2⇡i(a2 + b2 + c2 + d2)z)

=Xn�1

� Xx2R4(n)

�⌫(x)�exp(2⇡inz)

=Xn�1

|R4(n)|W (�⌫ ;n)n⌫/2 exp(2⇡inz).

Using the Poisson summation formula, one can show that ✓(�⌫ , z) is a holo-morphic modular cusp form of weight k = 2+ ⌫ with respect to the congru-ence subgroup

�0(4) = {✓

a bc d

◆2 SL2(Z), c ⌘ 0(4)}.

In other terms, ✓(�⌫ , z) this is an holomorphic function on H, such that for

any � =

✓a bc d

◆2 �0(4),

✓(�⌫ , �z) = ✓(�⌫ ,az + b

cz + d) = (cz + d)2+⌫✓(�⌫ , z)


such that the �0(4) invariant function z 7! (=z)k/2|✓(�⌫ , z)| is bounded onH.

The sumsXx2R4(n)

�⌫(x) =

Z 1

0✓(�⌫ , x+ iy)e(�2⇡in(x+ iy))dx

are then the Fourier coe�cients of this holomorphic cusp form; by the Pe-tersson formula, the square of their modulus are bounded by

n�(k�1)|X

x2R4(n)

�⌫(x)|2 ⌧�⌫ 1 +X

c⌘0(4)

|S(n, n; c)|c

|Jk�1(4⇡n

c)|

where Jk�1(x) is the Bessel function of order k � 1 and

S(n, n; c) =X

x(mod c)(x,c)=1

exp(2⇡inx+ nx�1

c)

is a Kloosterman sum. Such sums were bounded non-trivially for the firsttime by Kloosterman (for c prime which is the fundamental case) and Salie(for general c): there is an absolute constant � > 0 such that

(3.3) |S(n, n; c)| c1��+o(1)(n, c)�.

It follows from such a bound that (see [Iwa97, Chap. 11])Xx2R4(n)

�⌫(x) ⌧�⌫ n⌫2+1� �

2 ,

hence (using the estimate |R4(n)| = n1+o(1)) we obtain

W (�⌫ ;n) ⌧�⌫ n��/2+o(1) ! 0, n ! +1

which conclude the proof of equidistribution. ⇤

Remark 3.1. Kloosterman proved the bound (3.3) for � = 1/3 and Weilusing his proof of the Riemann hypothesis for curves over finite fields im-proved the bound to � = 1/2; 30 years later, Deligne [?DelBour, ?WeilI]proved the Weil conjectures and as a consequence of it the Ramanujan-Petterson conjecture on the size of Fourier coe�cients of holomorphic mod-ular cuspforms of integral weight. This implies the following bound

W (�⌫ ;n) ⌧�⌫ n�1/2+o(1).

This bound is up to the term no(1) optimal; moreover since |R4(n)| = n1+o(1)

the Weyl sum is essentially bounded by the inverse of squareroot of thenumber of terms occuring the sum; this is a manifestation of the square-rootcancellation phenomenon.

4. SUMS OF THREE SQUARES 21

4. Sums of three squares

We consider now the question of the representability of a given integern as a sum of three squares: the existence of solutions (a, b, c) 2 Z3 to theequation

(4.1) n = a2 + b2 + c2.

This question is significantly harder than for than for the sums of foursquares. Its solution came 30 years after the proof Lagrange’s theorem, whenLegendre gave a classification integers which are sums of three squares; Le-gendre’s argument however was not complete4 and a di↵erent unconditionalproof was given by Gauss

Theorem (Legendre, 1798; Gauss 1801). Every non-negative integer nnot of the form 4k(8l � 1), k, l 2 N may be represented as a sum of threesquares

n = a2 + b2 + c2, a, b, c 2 Z.

4.1. An idea of the proof of the Gauss-Legendre Theorem. Wewill not give the complete argument here but at least we discuss the arith-metical part of the proof using the Hamilton quaternions.

Let us make first a trivial reduction: let a2+b2+c2 = n be a representa-tion of n as a sum of three squares; if 4|n then a2+b2+c2 ⌘ 0(mod 4) and byinspecting the sums of three squares in Z/4Z, we see that 2 must divide a, band c so (a/2, b/2, c/2) is a representation of n/4; in other terms, if n = 4kn0

with 4 - n0 then the representations of n are of the form 2k(a0, b0, c0) for(a0, b0, c0) ranging over the representations of n0. Considering these, we seethat since 7 is not a sum of three squares modulo 8, n0 is not representableas a sum of three squares unless n 6⌘ 7(mod 8). This proves the necessity ofthe condition n 6= 4k(8l � 1).

The hard part of the Gauss-Legendre Theorem is that this condition issu�cient. The starting and key point is the fact that, under that condition,n admit at least a rational representation: there is (a, b, c) 2 Q3 such thatn = a2 + b2 + c2. This is special case of the Hasse-Minkowski principle (seeChap. 3 below for a discussion and [Ser73][Thm. 8, p. 41] for a proof.)

From this, one can easily conclude the proof: we denote by q3 the “threesquare” quadratic form (the euclidean quadratic form on Q3)

q3(a, b, c) = a2 + b2 + c2.

As noted before, the euclidian 3 space is isometric to the space of pureHamilton quaternions

(Q3, q3) ' (B0(Q),Nr).

4Legendre’s proof assumed the inifiniteness of the set primes in a given arithmeticprogression which was only proven about 50 years later by Dirichlet


The problem of representing n as a sum of three squares is equivalent to theproblem of finding a pure integral quaternion of that norm. Consider thepure (rational) quaternion

z = ai+ bj + ck 2 B(0)(Q).

It has norm n, and so, by (1.6), satisfies the quadratic equation

z2 + n = 0.

The commutative ring Z[z] ⇢ OB is isomorphic to the quadratic ring Z[p�n]

hence is finitely generated and by the Corollary 1.1 there is q 2 B⇥(Q) suchthat z0 = q�1zq = ⇢q(z) 2 OB. Since Nr(z0) = Nr(z) = n, tr(z0) = tr(z) = 0and OB \ B(0)(Q) = B(0)(Z), z0 = a0i + b0j + c0k has integral entries and(a0, b0, c0) is a representation on n. ⇤

5. The number of three squares representations

Let

R3(n) = {(a, b, c) 2 Z4, a2 + b2 + c2 = n}be the set of integral representations of n by q3. We know precisely whenR3(n) is non empty and, we will discuss now its finer structure and propertiesas n varies. The next question coming is the size of that set. On that matterwe have the following

Theorem 5.1. For n 6⌘ 0, 4, 7(mod 8) one has, as n ! +1

|R3(n)| = n1/2+o(1).

Remark. From the previous discussion, we may and will assume that4 - n.

The proof of this theorem will be discussed in the next sections. Againits proof is significantly harder than the corresponding question for R4(n)and builds on three principal ingredients:

The first ingredient is due to Gauss: the set of representations R3(n) isclosely related to another arithmetic set which we will discuss again in thenext chapter: set d = �n if n ⌘ 3 (mod 8) or �4d if n ⌘ 1, 2, 5, 6 (mod 8),and let

Rdisc(d) = {(a0, b0, c0) 2 Z3, (b0)2 � 4a0c0 = d},be the set of integral representations of d by the discriminant quadratic form

disc(a0, b0, c0) = (b0)2 � 4a0c0.

To describe the relationship more precisely we need the notion of primitiverepresentation.

5. THE NUMBER OF THREE SQUARES REPRESENTATIONS 23

5.0.1. Relation with the class number. Let (a0, b0, c0) be an integral rep-resentation of some integer n0 by q3, then for any integer f 6= 0, f.(a0, b0, c0) =(fa0, fb0, fc0) is a representation of n = n0f2. A representation (a, b, c) of nis primitive if it does not come from a representation of some proper divisorn0 = n/f2, or in other terms if and only if a, b, c are coprime5. The set ofprimitive representations is noted

R⇤3(n) = {(a, b, c) 2 Z3, a2 + b2 + c2 = n, a, b, c coprime}.

One has

(5.1) R3(n) =Gf2|n

f.R⇤3(n/f

2).

Similarly, one consider the set of primitive representations of d by the dis-criminant

R⇤disc(d) = {(a0, b0, c0) 2 Z3, (b0)2 � 4a0c0 = d, a0, b0, c0 coprimes}.

It is easy to see that the set R⇤disc(d) is non-empty if and only d ⌘ 0, 1 (mod 4)

(search a representation with a0 = 1 or b0 = 1) and in this case R⇤disc(d).

The reason is that R⇤disc(d) is acted by the infinite group SL2(Z) and the

stabilizers of the points R⇤disc(d) in are finite (since d < 0). This action is

defined by noting that R⇤disc(d) correspond bijectively with the set of integral,

primitive, binary quadratic forms of discriminant d via the map

(a0, b0, c0) ! qa0,b0,c0(X,Y ) = a0X2 + b0XY + c0Y 2,

and SL2(Z) acts on the set of such quadratic forms via the linear transfor-

mation in Z2: for

✓a bc d

◆2 SL2(Z)

(X,Y ).

✓a bc d

◆= (aX + cY, bX + dY ).

A fundamental result of Gauss is

Theorem 5.2 (Gauss). The set of orbits SL2(Z)\R⇤disc(d) is finite. Its

cardinality is called the class number of the discriminant d and is noted

h(d) = |SL2(Z)\R⇤disc(d)|.

The set SL2(Z)\R⇤disc(d) has the structure of an abelian group which

nowaday is expressed as the ideal class group of the imaginary quadraticorder of discriminant d and Gauss proved that this group acts on transitivelythe quotient SO3(Z)\R⇤(n) and computed the size of the stabilizers for thisaction. From this, it follows that

Theorem (Gauss). For n � 1, n 6⌘ 0, 4, 7 (mod 8). One has

|R⇤3(n)| = (48/w)h(d), with d = �n if n ⌘ 3(8),

|R⇤3(n)| = (24/w)h(d), with d = �4n if n ⌘ 1, 2, 5, 6(4);

here w = 6 for d = �3, w = 4 for d = �4, w = 2 for d < �4.

5one also says that (a, b, c) is a primitive vector of Z3


Remark. If n = 1, 2 or 3 , the above theorem remains valid with,the 12 replaced by 6, the 12 remains unchanged and the 24 replaced by 8respectively.

5.0.2. The class number formula. The second ingredient is the Dirich-let’s class number formula which express the order of the ideal class groupof imaginary quadratic orders in terms of L-functions: for d as above, let

�d(.) = (d

.) : (Z/dZ)⇥ 7! {±1}

be the Kronecker symbol of d. This is a (not-necessarily primitive) realcharacter of modulus d and conductor dK = d/f2 the unique fundamentaldiscriminant in the square class of d (ie. the discriminant of the quadraticfield K = Q(

pd)). Let

L(�d, s) =Xn�1

(d

n)1

ns=

Yp

(1� (d/p)

ps)�1

be the Dirichlet L-function associated with �d. One has

Theorem (Dirichlet). The class number h(d) has the following expres-sion

h(d) =w

2

L(�d, 1)

⇡|d|1/2.

Combining Gauss and Dirichlet formulae we obtain

|R⇤3(n)| = 12

L(�d, 1)

⇡n1/2.

Remark. We are grateful to Gergely Harcos for pointing out that theformula above is valid for any n not divisible by 4 and that it is not restrictedto squarefree n.

5.0.3. Siegel’s theorem. By Dirichlet’s formula (and the fact that a grouphas order at least 1) L(�d, 1) � 1. Landau and subsequently Siegel showedthat this lower bound is far from optimal: (see [Dav00, Chap. 21]):

Theorem (Siegel). As d ! 1L(�d, 1) = |d|o(1).

Combining these three results we obtain

|R⇤3(n)| = n1/2+o(1)

and using (5.1) we obtain Theorem (5.1).

6. Sums of three squares and quadratic fields

In this section, we discuss the arguments leading to the proof of GaussTheorem above, in the language of quaternion; this lead us naturally toconsider ideal classes of imaginary quadratic order instead classes of binaryquadratic forms.

6. SUMS OF THREE SQUARES AND QUADRATIC FIELDS 25

6.1. Embedding of quadratic field. As pointed out before, the map

(a, b, c) 7! z = ai+ bj + ck

identifies the quadratic space (Q3, q3) with (B0(Q),Nr); under that mapR3(n) is identified with the pure integral quaternions of reduced norm n,

B0,(n)(Z) = O0,nB , R⇤

3(n) with the primitive ones noted B0,(n)(Z)⇤ = O0,(n)⇤B .

In the sequel we will use the notations R3(n), R⇤3(n), B

0,(n)(Z), O0,nB B0,(n)(Z),

O0,(n)⇤B etc... interchangeably.Given z = ai+bj+ck 2 B(0)(Q) a pure quaternion of norm n, z satisfies

the equationz2 + n = 0

The polynomial X2+n is irreducible and therefore z yield an embedding ◆zof the imaginary quadratic field K = Q(

p�n) into B(Q)

◆z : u+ vp�n 7! u+ vz.

To such an embedding, one associate an order in K in the following way

Definition 6.1. Let z = ai + bj + ck 2 B(0)(Q) be a non-zero purequaternion. The order associated with z is the order

Oz = Q[z] \ OB.

We would like now to determine the preimage of Oz in K

◆�1z (Oz) ⇢ K.

Obviously the order Oz depends on z only up to multiplication by scalars:for any � 2 Q⇥

Oz = O�z,

we may assume that z is an integral primitive pure quaternions (that is z hasintegral coprime coordinates). If is remarkable that after such a reduction,the preimage in K depends only on the norm of z, n:

Proposition 6.1. Let z = ai + bj + ck be a pure integral primitivequaternion of norm n. We have

◆�1z (Oz) =

(Z[p�n] if �n ⌘ 2, 3(4)

Z[1+p�n

2 ] if �n ⌘ 1(4).

In other terms,

◆�1z (Oz) = Od = Z[

d+pd

2]

is the quadratic order of discriminant d = d(n) where d = �n if �n ⌘1(mod 4) and d = �4n if �n ⌘ 2, 3(mod 4).

Proof. Let u+p�nv 2 ◆�1

z (Oz), u, v 2 Q, ◆z(u+p�nv) = u+ vai+

vbj + vck 2 OB thus either (u, va, vb, vc) 2 Z4 which impose u, v 2 Z sincea, b, c are coprime or (u, va, vb, vc) 2 (12 + Z)4 which can occur if and only


if a, b, c ⌘ 1(2) which is equivalent to a2 + b2 + c2 ⌘ 3(4) and which imposethat u, v 2 1

2 + Z. Evidently if a, b, c ⌘ 1(2), then 1+z2 2 OB. ⇤

Definition 6.2. Given K a quadratic field, O ⇢ K an order and ◆ :K 7! B an embedding of Q-algebras. ◆ is an optimal embedding of O intoOB if ◆(O) ⇢ OB and ◆�1(OB) = O is maximal for this property.

In view of this definition, ◆z defines an optimal embedding of Od intoOB.

6.2. The ideal class group of a quadratic order. Let O ⇢ K bean order in a quadratic field, an ideal I ⇢ K is an O-module which is ofrank 2 as a Z-module. An O-ideal is proper if its stabilizer in K is preciselyO (and not a bigger order)

OI = {z 2 K, zI ⇢ I} = O.

Let Prop(O) be the set of propers O-ideals: the group K⇥ acts on the setof propers O-ideals by multiplication

(�, I) 2 K⇥ ⇥ Prop(O) ! �.I 2 Prop(O)

and the set of proper ideal classes is the set of orbits under this action, it isnoted

Pic(O) = Prop(O)/K⇥

and is finite.Moreover Prop(O) has the structure of commutative group for the mul-

tiplication of ideals

I, J ! I.J = Z-module generated by the products xy, x 2 I, y 2 J ;

this gives Pic(O) the structure of a finite commutative group: this is thePicard group of O.

7. The action of the class group on sums of three squares

Let z 2 R⇤3(n) be a primitive pure quaternion of norm n and let Oz be its

corresponding order and O = ◆�1z (Oz) ⇢ K by the corresponding “abstract

order”. The Theorem 5.0.1 of Gauss is a consequence of the existence of atransitive action of the Picard group Pic(O) on a quotient of R⇤

3(n). In thissection we will describe this action.

We have seen that B⇥(Q) act isometrically on B(0)(Q) by conjugationand that the group of projective units PB⇥(Q) = B⇥(Q)/Q⇥ is isomorphicto the orthogonal group SOB(0)(Q); one can also show thst the stabilizer ofthe lattice B(0)(Z) ' Z3, SOB(0)(Z) ' PB⇥(Z) say, is of order 24 and equalsto the image of the group generated by O⇥

B and the image of any integralquaternion of norm 2. Clearly PB⇥(Z) acts on R⇤

3(n). Denote the quotientby fR3(n) := PB⇥(Z)\R⇤

3(n).

The generalization of Theorem 5.0.1 is

8. THE DISTRIBUTION OF THE REPRESENTATIONS 27

Theorem 7.1. The quotient fR3(n) is endowed with a transitive actionof the Picard group Pic(O) and the stabilizer of an element under this actionhas order at most 2. Consequently

1

2|Pic(O)| |R⇤

3(n)| 24|Pic(O)|

Proof. We will first show how, from an integral representation z ofn and a proper6 O-ideal I ⇢ K, one can obtain a new representation ofn. The Z-module ◆z(I).OB is a left OB-ideal; since OB is principal, one has◆z(I)OB = qOB for some q 2 B⇥(Q). Then we claim that z0 = q�1zq 2 R⇤

3(n).Indeed Nr(z0) = n;

z0 = q�1zq 2 q�1zqOB = q�1z◆z(I).OB ⇢ q�1◆z(I).OB = OB,

since z◆z(I) = ◆z(p�nI) ⇢ ◆z(I); it is easy to check that z0 is primitive.

Note that the quaternion q is defined only up to multiplication on theright by an element of O⇥

B ; if we replace q by qu for u 2 O⇥B we see that z0

is replaced by the conjugate of z0 ⇢u�1(z0) 2 R⇤3(n). Thus the ideal I defines

a map between orbits

[I] :fR3(n) 7! fR3(n)z 7! [I].z := z0

.

If we replace I by an homothetic lattice �.I � 2 K⇥, then q is replaced byq0 = ◆z(�)q and q0�1zq0 = q�1◆z(��1

p�n�)q = q�1zq so that [�.I] = [I];

finally it is not di�cult to check that [I.J ] = [I] � [J ] and that [O] = Id

so that we obtain a well defined action of Pic(O) on fR3(n). We postponethe proof of the transitivity of this action and of the size of the stabilizer toChapter 6.1 were these results are obtained in greater generality.

To prove (??), we need first to to evaluate the size of fR3(n): for anyz 2 R⇤

3(n), the stabilizer of z for the action of O⇥B is the set of u 2 O⇥

Bcommuting with z. It follows from Chap ?? that u 2 ◆z(K)\O⇥

B = ◆z(O⇥).The later has order 4, 6 or 2 for n = 1, 3, 6= 1, 3 respectively. Hence

|R⇤3(n)|

⇤

8. The distribution of the representations

As for the case of four squares, since |R3(n)| (when non-zero) growsrelatively rapidly with n, we may wish toconsider the distribution of thedirections in R3 defined by the points of R3(n) ⇢ R3 or equivalently we mayconsider the distribution of their projection on the sphere S2. Again, thesepoints turn out to be well distributed

6Since K is a quadratic field, the notions of proper and locally principal ideal coincide


Figure 3. The distribution of R3(101).

Theorem 8.1 (Duke). As n ! +1, n 6⌘ 0, 4, 7(8), the set

n�1/2.R3(n) ⇢ S2,

become equidistributed on S2 w.r.t. the Lebesgue measure: for any f 2C(S2),

1

|R3(n)|X

x2R3(n)

'(x

n1/2) !

ZS2

'(v)dµS2(v).

As we will see below, this beautiful theorem is significantly deeper thanMalyshev’s Theorem. Until the work of Duke [Duk88] the best result wasthe work of Linnik [Lin68] who, by his“ergodic method”proved this equidis-tribution of n�1/2R3(n) for n satisfying an extra congruence condition (atsome fixed odd prime). One can find in [EMV10] an exposition of Linnik’sproof in a modern language.

8.1. Principle of the proof. Duke’s original approach follow closelythe principles of the proof of the equidistribution of representations as sumsof four squares sketched previously. In the course of this book we will discussan alternative approach but for now, and for completeness, let us presentDuke’s original approach as in [Duk88].

As for the 3-sphere, continuous function on S2 may be uniformly ap-proximated by a linear combinations of harmonic homogeneous polynomials:

8. THE DISTRIBUTION OF THE REPRESENTATIONS 29

Figure 4. The distribution of R3(78540).

the restriction to S2 of a polynomial �⌫ , of degree ⌫, on R3 such that for� 2 R

�⌫(�x) = �⌫�⌫(x), and �R3�⌫ = 0

were

�R3 =@2

@2y+

@2

@2z+

@2

@2tis the Laplace operator. Such polynomial may be taken of the form

(8.1) �⌫(y, z, t) = (by + cz + dt)⌫ .

Where (b, c, d) 2 C3, is an isotropic complex vector: ie. such that q3(b, c, d) =b2 + c2 + d2 = 0.

By Weyl’s equidistribution criterion is it su�cient to show that for suchnon-constant polynomials, the corresponding Weyl sums decay:

(8.2) W (�⌫ ;n) =1

|R3(n)|X

z2R3(n)

�⌫(z

|n|1/2 ) !ZS2

�⌫(x)dµS2(x) = 0.

Again, we may restrict to even degree ⌫ (otherwise the Weyl sum iszero). In view of Theorem 5.1 and by homogeneity, it will be su�cient toshow that

(8.3)X

x2R3(n)

�⌫(x) ⌧� n12+

⌫2��

for some absolute constant � > 0.


As before, the theta function

✓(�⌫ ,x) =X

(a,b,c)2Z3

�⌫(a, b, c) exp(2⇡i(a2 + b2 + c2)z)

=Xn�1

� Xx2R3(n)

�⌫(x)�exp(2⇡inz).

is a holomorphic modular cusp form of weight k = 3/2 + ⌫ with respect tothe congruence subgroup

�0(4) = {✓

a bc d

◆2 SL2(Z), c ⌘ 0(4)}.

That means that ✓(�⌫ , z) is bounded on the upper-half plane H and that for

any � =

✓a bc d

◆,

✓(�⌫ , �z) = ✓(�⌫ ,az + b

cz + d) = j(�, z)3+2⌫✓(�⌫ , z),

withj(�, z) = (

c

d)"�1

d (cz + d)1/2

with ( cd) the extended Legendre symbol, "d = 1, ◆ for d ⌘ 1, 3(4) and (.)1/2

is the usual branch of the square root which is positive on R>0.The sums

Px2R3(n)

�⌫(x) which we seek to bound are the Fourier co-e�cients of ✓(�⌫ , z) and by the Petersson formula, their square may bebounded by sums of Salie sums. Salie sums may be evaluated by elementarymeans and bounded individually in an optimal way; from this one obtainsthe following boundX

x2R3(n)

�⌫(x) ⌧�⌫ nk�12 + 1

4+o(1) = n12+

⌫2+o(1)

which is just not su�cient. In the remarkable paper, Iwaniec [Iwa87] wasable to take into account the summation over the Salie sums and to improvethe constant 1

4 above to 14 � 1

22 replacing the final exponent by 12 + ⌫

2 � 122

at least for weights k � 7/2. Duke generalized Iwaniec’s method to boundFourier coe�cients to forms of weight � 3/2 as well as to more general –notnecessarily holomorphic– forms of 1/2-integral weight and used this boundto prove the equidistribution of integral three squares representations as wellas for other (possibly indefinite) quadratic forms. We discuss these in thenext chapter.

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Sums of squares - EPFL · 2018. 10. 2. · CHAPTER 1 Sums of squares One of the great theorems in arithmetic in the 19th century is Lagrange’s four square theorem: Theorem (Lagrange,