Pub
.
Ma
.
UAB
Vol
.
30
Ns
1
Maig
1986
In oduc ion
.
THE
CONCEPT
OF
k-LEVEL
POR
POSITIVE
INTEGERS
Angela
A enas
I is
said
(c
.
[4])
ha
a
posi i e
in ege n
sa is ies
p ope y
(N)
i
he e
exis s
a
ep esen a ion
o
n
as
a
sum
o
3
squa es,
n =
x1+x2+x3
,
wi h
(x
1
,n)
=
1
and
x~
<n
3
1 .
I
has
been
checked
ha
e e yposi i e
in ege
n
<
600000,
n -
3(mod
8),
e i ies
p ope y
(N)
.
Such
p ope y
appea s
in
connec ion
wi h
he
esolu ion
o
a
Ga1odd
embedd¿ng
publem
in
he
ollowing
sense
[4]

:
e e y
cen al
ex ension
o
he
al e na ing
g oup
A
n
can
be
ealised
as
a
Galois
g oup
o e
i
n-
3(mod
8)
and
n
sa is ies
p ope y
(N)
.
In hispape ,
we
in oduce,
o
a
posi i e
in ege
n,
he
concep
o
k-Ie el
ela ed
o
he
ep esen a ions
o
n
as
a
sum
o
k
squa es
.
By
conside ing
he
case
k =
3
we
exhibi a
class
o
posi i ein ege s
sa is ying
p ope y
(N)
.
We
ecall
Lemma
1
o
[11
since
i willbe used
wice
in
his
pape
:
16
n =
x3+x2
+x3
í6
a
pA,ú ú í
. e
hepheJSen a i
.on
ob
n
ab
a
búm
oj
he e
pob
.í c
: e
6quaAez
and
p
.ía
a
pxime
jac on
06
n
whi
.c
h
du
.ídee
one
ob
he
dummandb,
- hen
p
-
1
oh
2(mod
4)
.
De ini ion
.
Po
a
posi i e
in ege n
we
de ine
he
k-le el,
k(n,k),
o
n
as
he
maX
.Úllun
alue
o
R
Such ha
he e
exis s
a
ep esen a ion
o
k
n
as
a
sum
o
k
squa es,
n
=

x
i

,
x
i
eZ
,
wi h
k
summands
p ime
o
n
.
i=1
41
I is
wellknown
ha
e e y
posi i e
in ege
is
a
sum
o
ou
squa es
.
I
n
is
no
a
sum
o
k squa es
(k<3),
hen
we
ag ee
ha ,
.
R(n,k)
=
-1
.
Ob iously,
o
e e y
posi i e
in ege n
is -1
<
£(n,k)
< k
.
I
k<k'
,
hen
R(n,k)
<
k(n,k')
.
And
o
e e y
k>1
is
k(1,k)
=
k
.
The
de e mina ion
o
R(n,2)
is
ai ly
easy
and
i is
gi en
in
P
oposi ion
1 .
Le
n>1
be
a
poa
.í
í
.
e
ín egeh
.
.
Then
.í)
16
4~n
and
e e yy
odd
pníme
d¿
.í
.6on
06
n
.í
.6
congnuen
o
1
modulo
4, hen
R(n,2)
=
2
.
ii)
E
.í heA í6
41n
and
n
.ce
a
eum
ob
a oo
equane6
ox íl each pxí
.me
d¿
.í-
zoh
06
n
congnuen
o
3
modulo
4appean6 ín
he
jac oníza í
.on
ob
n
ín o
pnímeÁ
wí h
a po4
.í í e
e en
exponen ,
hen
R(n,2)
= 0
.
iii)
In
a
.Ql
he
o i1
.Fh
cabeb
.L6
R(n,2)
=
-1
.
The
ollowing
p oposi ion
cha ac e izes
he
posi i e
in ege s
n
ha ing
s ic lyposi i e
4-le el
P oposi ion
2
.
R(n,4)
>

1í6
and
only
íg
n
9
0(mod
8)
.
P oo
.
I
n =
0(mod
8),
hen
e e y
ep esen a ion
o
n
as
a
sum
o
4
squa es,
n = x2
+y
2
+z
2
+
2
,
e i ies
ha
g
.c
.d
.(x,y,z, )
> 2
,
and
so
£(n,4)
= 0
.
Fu he mo e,
i
n
=
2,3,4,6,7(mod
8),
hen
ob iously
n-1
=
1,1,3,5,6(mod
8)
and,
hus,
n-1
is
a
sum
o
3
squa es,so we ha e
£(n,4)
>
1
.
Finally,
i
n = 1,5(mod
8),
hen
n-4
=
5,1(mod
8)
and,
consequen ly,
n-4
is
aleo
a
sum
o
h ee
squa es
so
ha
k(n,4)
>
1 ,
because
2h
.
Rema
k
.
Fo
k>4
,
we ha e
£(n,k)
>
1
o all
n,
jus
because
n-1
is
a
sum
o ou
squa es
.
Le
us
concen a e
om
now
on in
he
case
k=3
.
I
is
wellknown
ha
a
posi i e
in ege n
.is
exp essible
as
a
sum
o
h ee
in ege
squa es
i
and
only
i
n
is
no
o
he
o m
4
a
(8m+7)
.
Gauss
([2],
A
.
291)
p o ed,
mo eo e ,
ha
a
posi i e
in ege
admi s
a
p imi i e
ep esen a ion
as
a
sum
o
h ee
squa es
i
and
only
i
n
;!
0,4,7(mod
8)
.
Fo
£(h,3)
we ha e
he
ollowing
elemen a y
P oposi ion
3
.
Le
ne2Z
+
,
. hen
.ib
n -
o(mod
4),
i)
J,,(n,3)
< 0
ii)
9,(n,3)
<
3
íb
n -
o(mod
2)
oiL
(mod
5)
.
The
p oo
is
immedia e
by
passing
o
2Z
/m
a
wi h
m
=
4,2,5
.
We
nex p o e ha gi en
an
odd
posi i e
in ege
wi h
R(n,3)
>
1
,
i ,we
inc ease,
p ese ing
hei
pa i y,
he
exponen e
o
i s
p ime
ac o s
cong uen
o
1
modulo
4,
hen
one
can
ob ain
le el
g ea e
hano
equal
o
2
.
Lemma
4
.

(see

[11)

16
a,nca
+
ci e
bueh ha
a
=
a2
+a2
and
n = b2
+b2
+b
2
,
hen
a
2
n
--
c2
+c2+c3

,
wd
h
c 3
=

ab3

.
The
in e es o
he
abo e
lemma
lies
on
he
special
alues o
he
c
.
which allow
us
o ob ain
he
1
P oposi ion
5
.

Le n =
2
p
1

""
'
"
p
q1

"
*
q
s

'
W~h
pi
-
1(mod
4),
1

<i<
and
qj

=-
3(mod 4),

1

<
j

<
s

,

a=
o
on
1,

a
l
>
o
.

Then
1
a
y1

y
~1

Ss
(n,
3)

>

1,

and
m =

2 p1

. .
.p

q1

.
.'q
s
(mod
2),
.í
- UAnb
ou
ha
i)
16
a
=
o,
hen
£(m,3)
> 2
,
ii)
16
a =
1,
hen
k(m,3)
>
1
.
P oo
.
W i e
m = á2n
,
wi h
c
1
=
ab
1
-
2(a1b1+a2b2)al
c
2
=
ab
2-
2(a
1
b
1
+a
2
b2)a
2
,
a
=pa
.
.p
a
,
so
ha
y
.
=
2a
.+o
.,
i=1,
.
.
.,
;
ó
.
>
1 .
1
'


i

i
i

i -
>
a
i
and
y
i
-
a
.
i
Then
a
is
a
sum
o
wo
squa es
:
a = a
2
+a2
wi h
(ai
,a)
=
1 ; 1
<
i
<
2
.
As
£(n,3)

>
1
we
can
w i e
n =
b
2
+b2+b3
wi h
(b
3
,n)
=
1
and
(b
1
,b
2,b3
)
=
1
.
hen
44
Now
apply
lemma
4
o
w i e
m
=
a
2
n
=
c2+c2+c2
.
Le
p
-
1(mod
4)
be
a
p ime
di iding
m
such
ha
po
1
andpib
2
;
and
i
c
1
=
-2a
1
b
1
a
1
jZ
0(mod
p),
c
2 =
-2a
1
b
1
a2
~!
0(mod
p)
,
because
pla
.
In e changing
he
oles
o b
1
and
b
2
Le
p
=
1(mod
4)
be
a
p ime di iding
ci =
0(mod
p)
o
some
iE{1,2},
hen
As
p~b
1
we
a e
allowed
o
w i e
and
as
pla
we
ge
since
p
di ides
n
bu no
b3
.
a
l
b
1
+a
2
b2
=
0(mod
p)

,
.
a
l
= -
abb
2
(mod
p)
1
he
same esul
is
ob ained
.
m
wi h
p1y1
and~p~b
2
now
,
2 2

2
0 = a
b22+
a2
=
b2

(b2
+b~)(mod
p)

,
1

1
whence
b2+b2
=
0(mod
p)
.
Thus
n =
b(mod
p),
which
is a
con adic ion
We
ha e
hus
p o ed
ha
bo h
c
1
1
0(mod
p)
and
c
2
j!
O(mod
p), o
e e yp ime
ac o
p
=
1(mod
4)
o m
.
On
he
o he
hand,
i
q
=
3(mod
4)
is a
p ime
ac o
o m,
,
we
necessa ily
ha e ha q~c3
,
and
as
bo h
c
1
and
c
2
a e
nonze o,
by
lemma
1
o
[1]
we
ha e
ha
q~c
1
c2
.

So,
in
he
case
(i)
we ha e
k(n,3)
>
2
and
in
he case
(ii),
as
2~c
3
and
4~m,
we
ge
(c
l
,2)
=
1
o
(c2
,2)
=
1
om
which
we
in e ha
!G(n,3)
>
1
.
Theo em
6
.
Le n be a poad
í
. e
.Ln egeA,
and
wxí
e
í
6
6ac oAu
:zaUon
ín
o
.
pxc
:me
bac
om
aa

'
Nex
we
s a e
he
ollowing
1
(mod
4)
,
q
7
-
3
(mod
4)
.
W
.í
h
hí,s
no a íon
we
ha e
1á
np1
.
.
.p

,
. hen
R(n,3)
>
2
.
a
a
a

a
ii)
Ib
n
=
2
5
1
p
2
2
. .
.p
9,(n,3)
=2
.
aa
l
a
S
1

os
n
=
2
p
1 . .
.p
q
1 . .
.q
s

,
a+a
1
>
0
0<a<
1 ,
o<a
l ,
hen
a
l
a
üi)
I~
n= p
1
.
.
.P
and
n
.íz
a
num~
ídoneua
o6
EuleA,
hen
(n,
3)

=

2
.
i )
I5
n =
q
11.
.,
.gssand n Y
7(mod
8),
xhen9,(n,3)
=
3
.
aa
s
s
)
Ib
n=
2
5
1
g2
2
. .
.q
s
s
and
n Y
7(mod
8)
S+S
1
>
o,
o
<S
<
1
hen
R(n,3)
=
2í6 Sac S
1
= o
,
and
k(n,3)
>
1
o henwíae
.
i)
11
n=
p
1 1
g
l l
.
.
.q
S
and
n
;z
7(mod
8),
. h
.en
(n,3)
>
2
.
a a
S
i¡)
Ib
n=
p
1p 2q
1..
.q
s
and
n Y
7(mod
8),
hen
k(n,3)
>
1 .
1
21s
1l
os
i¡¡)
16
n= 2p
1
g
l...
q,
,
hen
2(n,3)
>
1 .
P oo
.
i)
In
his
case
n
admi s
a
p imi i e
ep esen a ion
as
a
sum
o
wo
squa es
and
he e o e
2(n,3)
>
2
.
ii) I
su ices o
apply
i)
and
p oposi ion
3
.
iii)
These
in ege s
admi
a
p imi i e ep esen a ion
as a
sum
o
wo
squa es
bu
do
no
ha e
any
ep esen a ion
as a
sum
o
3
posi i e
squa es
(c
.
[31)
.
In ege s o
his
ype
a e
13
and
37,
and
hese
a e
up o
now he
only
known examples
no
g ea e
han
5
.10
10
(see
[5])
.
i ), i), i¡)
and
i¡¡)
a e
immedia e
consequences o
lemma
1
o
[11-
)
Unde
hese
condi ions
n
admi s
a
p imi i e
ep esen a ion
as a
sum
o
hese posi i e
squa es
and
i
su ices o
apply
lenuna
1
o
111
oge he
wi h
p oposi ion
3
.
Now
we
gi e
an
applica ion
o
he
abo e
heo em
o
he
Galois
embedding
p oblem
(c
.
[41,
Th
.
5
.1)
.
Theo em
7
.
Le
n = g
1
1 .
.
.q S
s
Wi h
q
i
-
3(mod
4),
1
< i
<
s,
and
n
=_
3(mod
8)
xhen
e eAy
cen Aa2
ex ene
.íon
ob
he
aQ exna c
:ng
gnoup
A
n
can
be
ea
.F
.íded
ae
a
Ga1o
.í s
gnoup
o eA
Q(T)
and,
4o,
o en
Q
.
Bibliog áphy
A enas
Sola,
A
.
:
On
a
cen aín
ype
o4
pnímí c
: e
nepneeen ia í,onó
o6
na íonal
.ín egena
ae
sum
o6
squaAe s
.
Pub
.
Sec
.
Ma
.
Uni
.
Au
b
noma
de
Ba celona
.
Vol
.
28
;
Núm
.
2-3
(1984),
75-80
.
[21
Gauss,
C
.F
. :
U
.í,dqu,ca
.í
c
:onu
A~e í
.Cae
.
Lipsiae,
1801
.

English
asla ion
:
A hu
A
.
Cla ke,
1966,
New
Ha en
:
Yale
Uni
.
P ess
.
[31
S,chinzel,
A
.
:
SuJC
Ieb
somme6
de
A0í6
CWVCU
.
Bull
.
Acad
.
Pol
.
de
s
Sciences
.
Vol
.
11,
6
(1959),
22-25
.
[4]
Vila,
N
. :
On
een a
.2
exxenaíone
ob
A
n
ab
a
Galo
.í .e
guup
o elc
Q
.
A ch
.
Ma h
.,
Vol
.
44,
(1985),
424-437
.
[5]
weinhe ge ,
P
.J
. :
Exponen e
ob
he
~
gnoup
o6
complex
quadnaUe
6
.ebdb
.
Ac a
A i h
.
22
(1973),
118-124
.
The
au ho hanks
o
he
e e ee
o
some
use ul
sugges ions
.
Rebux
el
15
d'oc ubne
dei
.
1985
Depa amen o
de
Algeb a
y
Fundamen os
Facul ad
de
Ma emá icas
Uni e sidad
de
Ba celona
C/ G an
Via,
585
08007Ba celona
SPAIN