The concept of K-level for positive integers

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