Journal of Algebra 239, 174᎐190 Ž2001.
doi:10.1006rjabr.2000.8681, available online at http:rrwww.idealibrary.com on
A Six Generalized Squares Theorem, with Applications
to Polynomial Identity Algebras
Paula B. Cohen1
School of Mathematics, Institute for Ad¨ anced Study, Princeton, New Jersey 08540 and
UMR AGAT au CNRS, Uni¨ ersite´ des Sciences et Technologies de Lille,
59655 Villeneu¨ e d’Ascq cedex, France
E-mail: pcohen@math.ias.edu, Paula.Cohen@univ-lille1.fr
and
Amitai Regev 2
Department of Mathematics, The Weizmann Institute of Science, Reho¨ ot 76100, Israel
E-mail: regev@wisdom.weizmann.ac.il
Communicated by Susan Montgomery
Received March 3, 2000
The theories of superalgebras and of P.I. algebras lead to a natural ⺪ 2-graded
extension of the integers. For these generalized integers, a ‘‘six generalized
squares’’ theorem is proved, which can be considered as a ⺪ 2-graded analogue of
the classical ‘‘four squares’’ theorem for the natural numbers. This theorem was
conjectured by A. Berele and A. Regev Ž‘‘Exponential Growth of Some P.I.
Algebras,’’ wBR2x. and has applications to p.i. algebras.
䊚 2001 Academic Press
Key Words: additive number theory; four squares theorem; exponents of P.I.
algebras; Amitsur᎐Cappelli polynomials.
1. INTRODUCTION
The celebrated four squares theorem from number theory says that every
positive integer is a sum of four squares. The study of superalgebras leads
to a natural generalization B s ⺪w t x, t 2 s 1, of the integers, for which we
1
The first author thanks the Weizmann Institute of Science where this research was
initiated. She also thanks the Institute for Advanced Study, Princeton, NJ, where this
research was completed with the support of the Ellentuck Fund.
2
The second author is partially supported by ISF Grant 6629r1.
174
0021-8693r01 $35.00
Copyright 䊚 2001 by Academic Press
All rights of reproduction in any form reserved.
SIX GENERALIZED SQUARES THEOREM
175
prove a six generalized squares theorem. Writing t s Ž0, 1., we can identify
B with ⺪ = ⺪, endowed with the addition
Ž a, b . q Ž c, d . s Ž a q c, b q d . ,
Ž a, b . , Ž c, d . g ⺪ = ⺪
and the multiplication
Ž a, b . Ž c, d . s Ž ac q bd, ad q bd . ,
Ž a, b . , Ž c, d . g ⺪ = ⺪.
We adopt throughout the convention that 0 g ⺞. We call
P s Ž r 2 , r 2 . , Ž r 2 q s 2 , 2 rs . N r , s g ⺞ 4
the set of the generalized squares in ⺪w t x. Justification of this terminology
is given in the sequel.
In this paper we prove,
THE SIX GENERALIZED SQUARES THEOREM. Gi¨ en r G s G 0 in ⺞, the
pair Ž r, s . is always the sum of at most six elements in P.
Note that, for example, Ž10, 3. is not a sum of any five elements of P.
It is possible to formulate the six generalized squares theorem in terms
of ⺞ = ⺞ Žendowed with the usual componentwise addition and multiplication.. Under the map Ž a, b . ¬ Ž a q b, a y b . the generalized squares P
are mapped bijectively onto the set
S s Ž 2 n2 , 0 . , Ž n2 , m2 . , n ' m mod 2, n G m4 ,
and the six generalized squares theorem can be restated as follows.
SIX GENERALIZED SQUARES THEOREM RESTATED. Gi¨ en u G ¨ G 0 in ⺞
with u ' ¨ mod 2, the pair Ž u, ¨ . is always the sum of at most six elements
in S .
We now explain how representation theory and Young diagrams suggest
that B is the right ⺪ 2 Žor ‘‘super’’. generalization of ⺪ and that P is the
right generalization of the squares in ⺪. Applications of the six generalized
squares theorem, which appear in wBR2x, are described below.
Squares of integers are the dimensions of the matrix algebras. In the
classical ŽJacobson. ring theory, arbitrary algebras are, in a sense, approximated by simple Ži.e., matrix. algebras. Thus, for example, the classical
four square theorem in number theory implies that the dimension of any
finite dimensional algebra is the sum of the dimensions of four or less
matrix algebras.
The theory of algebras satisfying polynomial identities ŽP.I. algebras.
suggests a natural generalisation of the above. In Kemer’s structure theory
wKx, the passage from the general to the finitely generated case uses
176
COHEN AND REGEV
superalgebras, and it introduces new classes of ‘‘simple’’ superalgebras,
called ¨ erbally prime. Through the theory of the cocharacters of P.I.
algebrasᎏand their corresponding Young diagramsᎏthe dimension dim A
is replaced here by a certain pair hookŽ A. s Ž k, l . g ⺞ = ⺞. When A is a
matrix algebra, hookŽ A. s Ž r 2 , 0. where r 2 s dim A. The hook pairs of
the verbally prime superalgebras are the above set
P s hook Ž A . N A verbally prime4 s Ž r 2 , r 2 . , Ž r 2 q s 2 , 2 rs . N r , s g ⺞ 4 .
We call P the set of the generalized squares. The approximation of a
superalgebra by verbally prime superalgebras leads to expressing a given
Ž k, l . g ⺞ = ⺞ as a sum of such generalized squares, where addition in
⺞ = ⺞ is coordinatewise. Conversely, expressing Ž k, l . as a sum of such
generalized squares leads to the corresponding approximation of a given
P.I. algebra by verbally prime algebras wBR2, Sect. 4x. It is natural to ask
what the minimal necessary number of such generalized squares is in such
a presentation. Moreover, as explained below, the elements of P are
indeed squares under a natural new multiplication in ⺞ = ⺞, once we
admit Ž1, 1. as a square.
We describe this now in some more detail. In the character theory of Sn
Žthe nth symmetric group., the theory of Lie superalgebras, and the theory
of P.I. algebras, an integer k g ⺞ corresponds to an infinite strip of the
Young diagrams of height at most k. Such strips of Young diagrams arise
in the representation theory of classical Lie algebras, while the more
general Ž k, l . hooks of Young diagrams arise in the study of Lie superalgebras.
The irreducible Sn characters are given by ParŽ n., the partitions of n.
An Sn character ␹n is supported by K : ParŽ n. if ␹n s Ý ␭ g K m␭ ␹␭ ,
where ␹␭’s are the Sn irreducible characters, and the m␭ g ⺞ are their
multiplicities in ␹n .
DEFINITION.
Let Par s D n ParŽ n.. Denote
H Ž k, l . s ␭ s Ž ␭1 , ␭2 , . . . . g Par N ␭ kq1 F l 4 ,
k, l g ⺞.
We shall often identify Ž k, l . with H Ž k, l ..
In P.I. theory, one associates with a given P.I. algebra A a certain
sequence ␹nŽ A., n s 0, 1, 2, . . . , of Sn characters: these are the so-called
cocharacters of A, and c nŽ A. s degŽ ␹nŽ A.. are its codimensions.
THEOREM 1 wARx. Gi¨ en a P. I. algebra A, there exist k, l such that for all
n the cocharacter ␹nŽ A. is supported by H Ž k, l ..
Here, the ‘‘strip’’ case H Ž k, 0. is characterized by the so-called Capelli
identities wR1x. In the above theorem, denote: hookŽ A. : H Ž k, l .. If in
SIX GENERALIZED SQUARES THEOREM
177
addition k, l are minimal and unique, denote: hookŽ A. s H Ž k, l . Žalso,
the notation hookŽ ␹n . : H Ž k, l . is clear..
In Kemer’s structure theory for P.I. algebras over a field F of characteristic zero wKx, the basic algebras are three classes of superalgebras, called
¨ erbally prime:
Mr Ž F . , Mr Ž E . , Mr , s .
Here Mr Ž F . are the r = r matrices over F, and E is the infinite Grassmann Žexterior. algebra, while Mr, s is a certain subalgebra of Mrqs Ž E ..
The algebra F w t x s F ⭈ 1 [ F ⭈ t, t 2 s 1, is instrumental in this classification of the verbally prime algebras wK, p. 21x.
PROPOSITION.
The following ha¨ e been pro¨ ed in wBr1, R2x:
Ž1. hookŽ Mr Ž F .. s H Ž r 2 , 0. Ž the ‘‘strip’’ case.,
Ž2. hookŽ Mr Ž E .. s H Ž r 2 , r 2 .,
Ž3. hookŽ Mr, s . s H Ž r 2 q s 2 , 2 rs . ŽŽ2. and Ž3. are the ‘‘hook’’ case..
We call the abo¨ e Ž1., Ž2., and Ž3. ‘‘the ¨ erbally prime hooks.’’
The set of the verbally prime hooks is now identified with
P s Ž r 2 , r 2 . , Ž r 2 q s 2 , 2 rs . N r , s g ⺞ 4 .
These are the hook generalizations of the strips H Ž r 2 , 0. ' Ž r 2 , 0. of the
matrix algebras. For that reason we call P the set of generalized squares. In
fact, we show below that with a natural multiplication of pairs that is
induced by the Kronecker product, these are indeed squares, once the pair
Ž1, 1. s hookŽ E . is admitted as a generalized square.
Two natural operations in the character theory of Sn wJKx that frequently arise in the study of cocharacters of P.I. algebras are:
ˆ if hookŽ ␹i . : H Ž k i , l i ., i s 1, 2, then
Ža. The outer product m:
ˆ ␹ 2 . : H Ž k 1 q k 2 , l1 q l 2 . .
hook Ž ␹ 1 m
Žb. The inner ŽKronecker. product m: if hookŽ ␹ i . : H Ž k i , l i ., i s
1, 2, then
hook Ž ␹ 1 m ␹ 2 . : H Ž k 1 k 2 q l 1 l 2 , k 1 l 2 q k 2 l 1 . .
ˆ induces the
Identifying H Ž k, l . with Ž k, l ., we see that the operation m
coordinatewise addition
Ž a, b . q Ž c, d . s Ž a q c, b q d . ,
178
COHEN AND REGEV
while m induces the following multiplication:
Ž a, b . Ž c, d . s Ž ac q bd, ad q bc . .
This gives a commutative ring structure on B s ⺪ = ⺪ s ⺪w t x, where
t s Ž0, 1.. We remark that on B there is a natural conjugation Ž a, b . s
Ž a, yb . with corresponding ‘‘hyperbolic’’ norm
< Ž a, b . < 2 s Ž a, b . Ž a, yb . s Ž a2 y b 2 , 0 . ' a2 y b 2 .
Therefore B is also called the hyperbolic integers. In a sense, the elements
of B are the Žgeneralized. integers in F w t x, where ⺪ : F is a field.
Note that Ž r 2 q s 2 , 2 rs . s Ž r, s . 2 , and also Ž r 2 , r 2 . s Ž1, 1.Ž r, 0. 2 . Thus
the elements of P are indeed squares, provided we agree to call Ž1, 1. '
H Ž1, 1. s hookŽ E . a Žgeneralized. square. Note also that Ž1, 1. is almost a
square: let ␮ s 12 Ž1, 1.; then ␮2 s ␮. Trivially, if Ž r, s . g B is a sum of
elements of P then r G s G 0.
THEOREM 2 wGZx. Gi¨ en any P. I. algebra A in characteristic zero, with
the corresponding sequence of codimensions c nŽ A., the limit
exp Ž A . [ lim Ž c n Ž A . .
1rn
nª⬁
exists and is always an integer. It is called the exponent of A.
Given A, it is natural to wish to compute the invariant expŽ A.. For
certain polynomials Ckq 1 w x; y x, called Capelli polynomials, A satisfies
Ckq 1 w x; y x s 0 if and only if ␹nŽ A. : H Ž k, 0. for all n wR1x. For these
Capelli identities Ži.e., the ‘‘strip’’ case., expŽ A. is calculated in wMRZx.
Denote by Uk the universal algebra of the kth Capelli identity. We have
0 F k y exp Ž Uk . F 3.
The proof utilizes a construction involving the verbally prime algebras,
which in this ‘‘strip’’ case are just the matrix algebras Mr Ž F .. Consequently
it depends on expressing integers as sums of squares r 2 Ž' Ž r 2 , 0... The
four square theorem from number theory implies the above upper bound 3
Žwhich cannot be improved..
In the Ž k, l . hook case there are analogous ‘‘Amitsur᎐Capelli’’ type
polynomials Ckq 1, lq1w x; y x s 0 wARx, with the corresponding universal
algebra Uk, l replacing Uk . The computation of expŽ A. is done in wBR2x,
where it is shown that when k G l G 0,
0 F Ž k q l . y exp Ž Uk , l . F 4.
SIX GENERALIZED SQUARES THEOREM
179
In this case, the proof depends on expressing elements Ž k, l . g B, k G l G
0, as sums of generalized squares. For these generalized squares, a seven
Žgeneralized. squares theorem was proved, and a six Ž generalized. squares
theorem is conjectured in wBR2x. As a consequence of the proof of this
conjecture given in this paper we have the sharp bound
0 F Ž k q l . y exp Ž Uk , l . F 3.
We express our special thanks to Umberto Zannier for his interest in
our work, especially in connection with the remarks of Section 3.
2. PROOF OF THE SIX GENERALIZED
SQUARES THEOREM
In this section we prove the six generalized squares theorem by a series
of lemmata. We shall often use the result of Legendre wDi, Vol. II, p. 261x,
which states that a number is a sum of three squares with no common
factor unless it is of the form 4 u Ž8l q 7., u, l g ⺞, in which case it is a sum
of four squares. We also use a number of corollaries to this result, like the
fact that every odd integer can be written in the form x 2 q y 2 q 2 z 2 . A
full treatment of such results can be found in wDix and when our statements are not found in that reference we give details.
DEFINITION. Call any of the following presentations of s g ⺞ a
quadratic-ternary presentation:
Ž1.
Ž2.
s s ␧ x 2 q ␩ y 2 q ␥ z 2 , ␧ , ␩ , ␥ g 0, 1, 24 ,
s s ␧ x 2 q ␩ y 2 q 2 z Ž z q l ., ␧ , ␩ g 0, 1, 24 ,
Ž3.
s s ␧ x 2 q 2 y Ž y q k . q 2 z Ž z q l ., ␧ g 0, 1, 24 .
We define the length of such a presentation to be the number of its
non-zero summands and its shift to be 0 in case Ž1., l 2 in case Ž2., and
k 2 q l 2 in case Ž3..
LEMMA 1. In ⺞, let r G s G 0. Assume s has a quadratic-ternary presentation of length F 2 with shift F r y s. Then Ž r, s . g B can be written as a
sum of six or fewer elements in P ; that is, Ž r, s . is a sum of six generalized
squares.
Similarly, if s admits any quadratic-ternary presentation with shift s d F
r y s, such that r y s y d is a sum of three squares in ⺞, then Ž r, s . is a sum
of six generalized squares.
180
COHEN AND REGEV
Proof. The proof follows from the four squares theorem in ⺞, applied
to r y s y shift G 0. Here are the cases.
Ž1. s s ␧ x 2 q ␩ y 2 , ␧ , ␩ g 0, 1, 24 Žhence shift s 0.. In ⺞ let r y s
s q ⭈⭈⭈ qq42 ; then
q12
Ž r , s . s Ž ␧ x 2 , ␧ x 2 . q Ž ␩ y 2 , ␩ y 2 . q Ž q12 , 0 . q ⭈⭈⭈ q Ž q42 , 0 . ,
a sum of six terms from P.
Ž2. s s ␧ x 2 q 2 y Ž y q k ., ␧ g 0, 1, 24 . Let r y s y shift s r y s y
2
k s q12 q ⭈⭈⭈ qq42 ; then
2
Ž r , s. s Ž ␧ x2 , ␧ x2 . q Ž y2 q Ž y q k . , 2 yŽ y q k . .
q Ž q12 , 0 . q ⭈⭈⭈ q Ž q42 , 0 .
Žsince y 2 q Ž y q k . 2 s 2 y Ž y q k . q k 2 ..
Ž3. s s 2 y Ž y q k . q 2 z Ž z q l .. Now r y s y shift s r y s y Ž k 2 q
2.
l s q12 q ⭈⭈⭈ qq42 , and
2
2
Ž r , s. s Ž y2 q Ž y q k . , 2 yŽ y q k . . q Ž z2 q Ž z q l . , 2 z Ž z q l . .
q Ž q12 , 0 . q ⭈⭈⭈ q Ž q42 , 0 . .
LEMMA 2. Let 5 F M g ⺞ and assume that none of M, M y 1, and
M y 4 is a sum of three squares.
Then M s 4 u Ž8l q 7., with 0 F l and 3 F u. In particular, 448 s
3
4 7 F M.
Proof. By assumption, M s 4 u Ž8l q 7. s Ž M y 1. q 1 s 4¨ Ž8 k q 7.
q 1. If 1 F ¨ then u s 0 so l.h.s.' 3 mod 4 while r.h.s.' 1 mod 4, a
contradiction. Thus ¨ s 0, so M s 4 u Ž8l q 7. s 8Ž k q 1., implying 2 F u.
Similarly M s Ž M y 4. q 4 s 4 w Ž8 m q 7. q 4, so 4 u Ž8l q 7. s 4 w Ž8 m
q 7. q 4; hence 1 F w. This implies 4 uy 1 Ž8l q 7. s 4 wy 1 Ž8 m q 7. q 1
which implies that w y 1 s 0 since 2 F u. Deduce that 4 uy 1 Ž8l q 7. s
8Ž m q 1., which implies that 2 F u y 1.
LEMMA 3. As in Lemma 2, let 5 F M g ⺞, and assume that none of M,
M y 1 and M y 4 is a sum of three squares. Let t g ⺞ be an integer that
modulo 8 is congruent to either 2 or 5. Then M y t is a sum of three squares.
Proof. Assume this is not the case. With the above notations, deduce
that M s 4 u Ž8l q 7. s Ž M y t . q t s 4 q Ž8 r q 7. q t, 0 F l, 3 F u. If q s
0, deduce that t ' 1 mod 8, a contradiction. If q s 1, then 4 divides t, so
mod 8, t is congruent to either 0 or 4, a contradiction. Finally, if 2 F q,
deduce that t ' 0 mod 8, again a contradiction, and the proof follows.
SIX GENERALIZED SQUARES THEOREM
181
LEMMA 4. As in Lemma 2, let 9 F M be an integer and assume that none
of M, M y 1, and M y 4 is a sum of three squares. Then M y 8 is a sum of
three squares.
Proof. Assume the contrary. Deduce that M is of the form
M s 4 u Ž 8l q 7 . s Ž M y 8 . q 8 s 4 y Ž 8 s q 7 . q 8
for certain integers u G 3, l G 0, y G 0, s G 0. Clearly, by parity, y G 1, so
that on dividing both sides of the above equation by 4 we have
4 uy 1 Ž 8l q 7 . s 4 yy 1 Ž 8 s q 7 . q 2.
Again by parity, as u G 3, we must have y G 2. Reducing the above
equation modulo 4 gives a contradiction and the proof follows.
LEMMA 5. E¨ ery s g ⺞ can be represented as s s x 2 q y 2 q ␧ z 2 , x, y, z
g ⺞, ␧ g 0, 1, 24 .
Proof. Every odd integer s can be written in the form
s s x2 q y2 q 2 z2.
Suppose now that s is even. If s ' 2 mod 4, then s is a sum of three
squares. If s ' 0 mod 4 then either s is a sum of three squares or it is of
the form s s 4 u Ž8l q 7. with u G 1, so that we can write it as
s s 2 ⭈ 4 uy 1 Ž 8l q 7 . q 2 ⭈ 4 uy 1 Ž 8l q 7 . .
Clearly 2 ⭈ 4 uy 1 Ž8l q 7. can be written as a sum of three squares x 2 q y 2
q z 2 so that
2
2
s s 2 x2 q 2 y2 q 2 z2 s Ž x q y. q Ž x y y. q 2 z2.
This completes the proof.
LEMMA 6. E¨ ery s g ⺞ can be represented as s s ␧ x 2 q y 2 q 2 z Ž z q 1.,
where ␧ g 0, 1, 24 , x, y, z g ⺞.
Proof. We consider first the case where s is a positive even integer.
Now, every odd integer is the sum of four squares of which two are
consecutive wDi, Vol. II, p. 293x. Hence,
s q 1 s p 2 q q 2 q z 2 q Ž z q 1.
and so
s s p 2 q q 2 q 2 z Ž z q 1. .
This deals with the case where s is even.
2
182
COHEN AND REGEV
If s is odd, then s is congruent to either 1 or 3 mod 4. Now suppose that
s is congruent to 1 mod 4, and write s s 4 k q 1. We know that for any
integer k G 0 the integer 8 k q 3 is a sum of three squares:
8 k q 3 s a2 q b 2 q c 2 .
Reducing mod 4 we see that a, b, c are all odd. Therefore we may rewrite
8 k q 4 s 1 q a2 q b 2 q c 2
as
8k q 4 s
1
2
Ž Ž a y 1. 2 q Ž a q 1. 2 q Ž b y c . 2 q Ž b q c . 2 . .
Thus,
4k q 2 s
ž
ay1
2
aq1
2
q
/ ž
2
q
/ ž
2
byc
2
2
q
/ ž
bqc
2
2
/
and
s s 4k q 1 s 2
ž
ay1
2
/ž
aq1
2
q
/ ž
byc
2
2
q
/ ž
bqc
2
2
/
.
If s is congruent to 3 mod 4, we write s s 4 k q 3 and argue in a similar
way as above but with the representation, for any integer k G 0, of 8 k q 7
as
8 k q 7 s a2 q b 2 q 2 c 2 ,
where we have a G 1 odd and b G 0 even. We deduce from this that
s s 4k q 3 s 2
ž
ay1
2
/ž
aq1
2
q2
b
/ ž /
2
2
q c2 .
LEMMA 7. E¨ ery s g ⺞ can be represented as s s ␧ x 2 q y 2 q 2 z Ž z q 2.,
where ␧ g 0, 1, 24 , x, y, z g ⺞.
Proof. Assume first that s is odd. Then, we may write
s q 2 s a2 q b 2 q 2 c 2 .
Clearly, if c / 0, then
s s a2 q b 2 q 2 Ž c y 1 . Ž c q 1 .
and we are done. So assume that c s 0. If a, b / 0 then we can suppose
that a is even, as a and b have opposite parity. Write a s 2 k, k G 1. Then
2
s q 2 s Ž 2 k . q b2 s 2 k 2 q 2 k 2 q b2
183
SIX GENERALIZED SQUARES THEOREM
so that
s s b 2 q 2 k 2 q 2 Ž k y 1. Ž k q 1. ,
as required. We can use previous arguments to see that we can suppose
ab / 0. Alternatively, argue as follows. Suppose that
s q 2 s b2
with b odd. Then b 2 is congruent to 1 mod 8 and 2 b 2 is congruent to
2 mod 8 and so is not of the form 8 n q 7 or 4 n. We can therefore
represent 2 b 2 as a sum of three squares with no common factor,
2 b2 s x 2 q y 2 q z 2 .
We may suppose that x, y are odd and that z is even. If z / 0 we write
ps
1
2
Ž x q y. ,
qs
1
2
Ž x y y. ,
rs
1
2
z
so that
b2 s p2 q q 2 q 2 r 2 ,
rG1
and we conclude that
s s b 2 y 2 s p 2 q q 2 q 2 Ž r y 1. Ž r q 1. .
If z s 0, then x and y are coprime so that we may assume pq / 0 and
b2 s p2 q q 2 ,
with p s 2 k even and q odd. This enables us to write
b2 s 2 k 2 q 2 k 2 q q 2
and
s s b 2 y 2 s q 2 q 2 k 2 q 2 Ž k y 1. Ž k q 1. .
Now suppose that s is even. Consider first the case where s q 2 s
2 2 eq1 ¨ , some e G 0, and ¨ odd. As ¨ is odd, it is of the form
¨ s p2 q q 2 q 2 r 2 ,
where one of p or q is non-zero, say p / 0. Therefore
2
2
2
2 2 e ¨ s Ž 2 e p . q Ž 2 eq . q 2 Ž 2 e r . s A2 q B 2 q 2C 2
184
COHEN AND REGEV
for A s 2 e p, B s 2 eq, C s 2 e r, and
2
s q 2 s 2 ⭈ 2 2 e ¨ s 2 A2 q 2 B 2 q Ž 2C . ,
with A / 0 as p / 0. Therefore,
2
s s 2 Ž A y 1 . Ž A q 1 . q 2 B 2 q Ž 2C . .
Now, consider the case where s q 2 s 2 2 e ¨ , some e G 1 and ¨ odd. We
write again
¨ s p2 q q 2 q 2 r 2 .
We have
2
2
s q 2 s Ž 2 e p . q Ž 2 eq . q 2 Ž 2 e r .
2
and if r / 0 then
2
2
s s Ž 2 e p . q Ž 2 eq . q 2 Ž 2 e r y 1 . Ž 2 e r q 1 . .
If r s 0, then ¨ s p 2 q q 2 and we can suppose that p / 0. Then
2
s q 2 s Ž 2 e p . q Ž 2 eq .
2
and as e G 1, with A s 2 ey 1 p / 0 and B s 2 eq, we can write
s q 2 s 2 A2 q 2 A2 q B 2
and hence
s s 2 Ž A y 1 . Ž A q 1 . q 2 A2 q B 2 ,
as required.
COROLLARY 1. Let r G s G 0 and r y s - 448. Then Ž r, s . is a sum of
six hyperbolic squares.
Proof. Let M s r y s. If M F 4 then M is a sum of three squares in
⺞. If 5 F M then by Lemma 2, either M or M y 1 or M y 4 is a sum of
three squares in ⺞. If M is a sum of three squares, apply Lemma 5; if
M y 1 is a sum of three squares, apply Lemma 6; if M y 4 is a sum of
three squares, apply Lemma 7.
For example, assume M y 4 s q12 q q22 q q32 . By Lemma 7, we have
s s ␧ x 2 q y 2 q 2 z Ž z q 2., ␧ g 0, 1, 24 , and we have
2
Ž r , s . s Ž ␧ x 2 , ␧ x 2 . q Ž y 2 , y 2 . q Ž z 2 q Ž z q 2. , 2 z Ž z q 2. .
q Ž q12 , 0 . q Ž q22 , 0 . q Ž q32 , 0 . .
185
SIX GENERALIZED SQUARES THEOREM
Remark. Lemma 8 below will show that most s g ⺞ admit the presentation
Ž 2.1.
s s ␧ x 2 q 2 y Ž y q 1. q 2 z Ž z q 2. ,
␧ g 0, 1, 2 4 .
We comment in Section 3 on the difficulty of establishing this fact for all
integers / 3, 23. We have
COROLLARY 2. Let s g ⺞ admit the presentation Ž2.1., and let r G s G 0;
then Ž r, s . is a sum of six hyperbolic squares.
Proof. This follows by similar arguments to those of Corollary 1, using
Lemmas 2 and 3.
LEMMA 8. E¨ ery s g ⺞ for which 2 s q 5 is not of the form a2 q 2 c 2 ,
with a odd and c odd, can be written in the form s s ␧ x 2 q 2 y Ž y q 1. q
2 z Ž z q 2., ␧ g 0, 1, 24 , x, y, z g ⺞.
Proof. Write
2 s q 5 s a2 q b 2 q 2 c 2 ,
with a G 1 odd and b G 0 even.
Step 1. We show we can assume that either b ) 0 or c ) 0.
This is because we can represent 2Ž2 s q 5. as a sum of three squares
with no common factor:
2 Ž 2 s q 5. s x 2 q y 2 q z 2 .
We can assume that z is even and that x and y are odd, x G y.
Substituting
as
1
2
Ž x q y. ,
bs
1
2
Ž x y y. ,
cs
1
2
z,
we recover
2 s q 5 s a2 q b 2 q 2 c 2 .
We can assume that a is odd and non-zero and b is even. If b s c s 0 this
would mean that z s 0 and x s y, contradicting the fact that x, y and z
have no common factor.
Step 2. Assume b / 0, so b G 2 since it is even. From
2 s q 5 s a2 q b 2 q 2 c 2
deduce that
2 s q 6 s 1 q a2 q b 2 q 2 c 2 s
1
2
2
2
Ž a y 1. q Ž a q 1. q 2 b 2 q Ž 2 c . .
Ž
2
186
COHEN AND REGEV
and so
sq3s
ž
ay1
2
q
/ ž
2
aq1
2
b
q2
2
q c2 .
/ ž /
2
2
We conclude that
ss2
ž
ay1
2
/ž
aq1
2
q2
/ ž
b
2
y1
/ž
b
2
q 1 q c2 ,
/
as required.
Step 3. In this case b s 0. Then c ) 0 and is even by the assumptions.
So let
2 s q 5 s a2 q 2 c 2 ,
with a G 1 odd and c G 2 even Žthus s is congruent to 2 modulo 4.. We
then have
2s q 6 s
1
2
Ž Ž a y 1. 2 q Ž a q 1. 2 q 4 c 2 . .
It follows that
ay1
sq3s
ž
ay1
aq1
2
2
q
/ ž
aq1
2
2
q4
c
/ ž /
2
2
.
We deduce that
ss2
ž
2
/ž
2
q2
/ ž
c
2
y1
/ž
c
2
q1 q2
c
/ ž /
2
2
,
as required.
This completes the proof. Note that in the remaining cases where
2 s q 5 s a2 q 2 c 2 ,
with a, c G 1 odd, s is congruent to 3 modulo 4.
Remark. Notice that the integers s s 3, 23 cannot be written in the
form ␧ x 2 q 2 y Ž y q 1. q 2 z Ž z q 2., ␧ g 0, 1, 24 .
It follows from Corollary 2 that for r G s G 0, if s satisfies the assumptions of Lemma 8, then Ž r, s . is a sum of six elements in P.
We now complete the proof of the six generalized squares theorem.
SIX GENERALIZED SQUARES THEOREM
187
By Corollary 2 and Lemma 8, it remains to prove this theorem in the
case where 2 s q 5 s a2 q 2 c 2 , where both a, c G 1 are odd. By Corollary
1, we can assume that r y s G 448. There remain a number of cases to
treat.
Case 1. Assume a G 7. Then
2 s y 44 s a2 y 49 q 2 c 2 s Ž a y 7 . Ž a q 7 . q 2 c 2 ,
so that on dividing by 2 we have
s y 22 s 2
ž
ay7
2
/ž
aq7
2
/
q c2 ,
and finally we deduce the representation
s s 2 ⭈ 1 ⭈ 11 q 2
ž
ay7
2
/ž
aq7
2
/
q c2 .
This representation has the shift 10 2 q 7 2 s 149 which is congruent to
5 modulo 8 and is less than 448. This enables us to appeal to Lemmas 3
and 1.
Notice that this gives the representation
23 s 2 ⭈ 1 ⭈ 11 q 1,
using
51 s 7 2 q 2 ⭈ 12 .
Case 2. Assume a s 5. Then
2 s q 5 s 25 q 2 c 2 s 3 2 q 4 2 q 2 c 2
and we have
s s 2 ⭈ 1 ⭈ 2 q 2 ⭈ 1 ⭈ 3 q c2 ,
giving rise to a shift of 5; hence we are done by Lemmas 3 and 1.
Case 3. Assume a s 3. Then
2 s q 5 s 9 q 2 c2 ,
so that
2 s s 4 q 2 c2
and
s s 2 ⭈ 1 q c2 ,
which is a representation of length 2 and hence sufficesᎏby Lemma 1.
188
COHEN AND REGEV
For example, when s s 3 we have c s 1.
Case 4. Assume a s 1. Then
2 s q 5 s 1 q 2 c2 .
As s G 1 we may assume that c G 3. We have
2 s q 4 s 2 c2
and
s s c 2 y 2.
Recalling that c is odd, we may write
s s c2 y 2 s 1 q 2
ž
cy1
2
/ž
cq3
2
q2
/ ž
cq1
2
/ž
cy3
2
/
.
This is a representation of the form
s s 1 q 2 y Ž y q 2. q 2 z Ž z q 2. ,
which gives rise to a shift of 8, which will enable us to appeal to Lemmas 4
and 1.
This completes the proof of the six generalized squares theorem.
3. CONCLUDING REMARKS
During the 19th and earth 20th centuries, various mathematicians obtained results containing additional information about the four squares
theorem for natural numbers Žsee wDi, Vols. II and III., and this motivated
our present treatment of the six generalized squares theorem which relies
on some of these very old classical observations.
The discussion of Section 2 makes it clear that we would have a shorter
proof of the six generalized squares theorem if we knew that Lemma 8
held for all integers except an explicit finite list. A computer search leads
us to ask the following.
QUESTION.
Is it true that e¨ ery integer s / 3, 23 can be written in the form
s s ␧ x 2 q 2 y Ž y q 1. q 2 z Ž z q 2. ,
␧ g 0, 1, 2 4 ?
By the treatment of Section 2, it is clear that the answer to the question
is ‘‘yes’’ given that we can settle it for integers s congruent to 3 mod 4, of
which s s 3, 23 are special cases. In Section 2 we reduced this question to
the study of representations of 2 s q 5 in the form a2 q b 2 q 2 c 2 with a
SIX GENERALIZED SQUARES THEOREM
189
and c odd Žwe can assume a, c G 1. and b G 0 even. We relied on being
able to take b / 0: for example for s s 23 2 ⭈ 23 q 5 s 51 s 12 q 2 ⭈ 5 2 s
7 2 q 2 ⭈ 12 are the only representations of this form for 51 and both have
b s 0. As recalled at the beginning of Section 2, the representation of an
odd integer t in the form a2 q b 2 q 2 c 2 follows from the representation
of 2 t as a sum of three squares x 2 q y 2 q z 2 , with x, y odd and z even.
We have b s 0 precisely when x s y. Therefore, we are faced with
comparing the number of representations of 2 t as x 2 q y 2 q z 2 with
x / y, to the number of such representations with x s y. We have the
following result wDi, Vol. II, p. 262x.
PROPOSITION 1. The number ␸ Ž m. of proper representations x 2 q y 2 q z 2
of m as a sum of three squares with no common factor is gi¨ en by
␸ Ž m . s 12 ⭈ 2 ␮ ⭈ h⬘ Ž y4m . ,
s 2 ␮q2 ⭈ h⬘ Ž y4m . ,
m ' 1, 2, 5, 6 mod 8
m ' 3 mod 8.
Here h⬘Žy4m. is the number of classes in the principal genus of the properly
primiti¨ e binary quadratic forms of discriminant y4m and ␮ is the number
of distinct odd prime factors of m Ž see w Cax for definitions..
On the other hand we have wDi, Vol. III, p. 38x
PROPOSITION 2. Let m be a natural number which we factorize into its
primes as m s 2 ␣ Ł p p r Ł q q s, where the primes p are those congruent to 1
or 3 mod 8 and the primes q are those congruent to 5 or 7 mod 8. The number
␺ Ž m. of proper representations z 2 q 2 x 2 of m, where x and z ha¨ e no
common factor, is gi¨ en by ␺ Ž m. s 2Ł p Ž r q 1.Ł q ŽŽ1 q Žy1. s .r2..
Consider the case where m s 2 t and t is a product of ␮ distinct prime
factors p congruent to either 1 or 3 mod 8. Then, by Proposition 1 we have
␸ Ž2 t . s 12 ⭈ 2 ␮ ⭈ h⬘Žy8t . and by Proposition 2 we have ␺ Ž2 t . s 2 ⭈ 2 ␮.
The number of automorphs wCa, Chap. 9, p. 127x of the form x 2 q y 2 q z 2
is 6 times the number of automorphs of z 2 q 2 x 2 . On the other hand, for
the t under consideration we have ␸ Ž2 t .r␺ Ž2 t . s 6 ⭈ h⬘Žy8t . and therefore, taking into account the automorphs, there is a representation 2 t s x 2
q y 2 q z 2 with x / y precisely when h⬘Žy8t . ) 1. It is an open problem
to list explicitly those m for which h⬘Žy4m. s 1, although it is known that
there are 65 such m with at most 1 extra m about which there is
effectively no information. These m are the numeri idonei of Euler Žalso
called ‘‘convenient’’ or ‘‘suitable’’ numbers, listed, for example, in wBoSh, p.
427x.. From the Riemann hypothesis, it would follow that the list of the 65
numeri idonei of Euler is complete. Inspection of Euler’s list shows that
m s 22, 102 are its only entries which are of the form m s 2Ž2 s q 5., with
2 s q 5 a product of distinct prime factors congruent to either 1 or 3 mod 8,
190
COHEN AND REGEV
so this checks with the fact m s 2Ž2 s q 5. s 22, 102 cannot be written as
a sum of three distinct squares and hence the cases s s 3, 23 cannot be
treated by the arguments of Lemma 8. As we can verify the question above
by computer for all numbers s / 3, 23 well beyond those of the numeri
idonei list, the only obstacle to answering the question in the affirmative is
the knowledge that this list is complete.
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