The concept of K-level for positive integers

Pub . Mat . UAB Vol . 30 Ns 1 Maig 1986 Introduction . THE CONCEPT OF k-LEVEL POR POSITIVE INTEGERS Angela Arenas It is said (cf . [4]) that a positive integern satisfies property (N) if there exists a representation of n as a sum of 3 squares, n = x1+x2+x3 , with (x 1 ,n) = 1 and x~ <n 3 1 . It has been checked that everypositive integer n < 600000, n - 3(mod 8), verifies property (N) . Such property appears in connection with the resolution of a Ga1odd embedd¿ng publem in the following sense [4]  : every central extension of the alternating group A n can be realised as a Galois group over if n3(mod 8) and n satisfies property (N) . In thispaper, we introduce, for a positive integer n, the concept of k-Ievel related to the representations of n as a sum of k squares . By considering the case k = 3 we exhibita class of positiveintegers satisfying property (N) . We recall Lemma 1 of [11 since it willbe used twice in this paper : 16 n = x3+x2 +x3 í6 a pA,úrútí .ve hepheJSentati .on ob n ab a búm oj there pob .ítc :ve 6quaAez and p .ía a pxime jacton 06 n whi .c h duv .ídee one ob the dummandb, -then p - 1 oh 2(mod 4) . Definition . Por a positive integern we define the k-level, k(n,k), of n as the maX .Úllun value of R Suchthat there exists a representation of k n as a sum of k squares, n =  x i  , x i eZ , with k summands prime to n . i=1 41 It is wellknown that every positive integer is a sum of four squares . If n is not a sum of k squares (k<3), then we agree that, . R(n,k) = -1 . Obviously, for every positive integern is -1 < £(n,k) < k . If k<k' , then R(n,k) < k(n,k') . And for every k>1 is k(1,k) = k . The determination of R(n,2) is fairly easy and it is given in Pr oposition 1 . Let n>1 be a poa .í tí .v e íntegeh . . Then .í) 16 4~n and everyy odd pníme d¿v .í .6on 06 n .í .6 congnuent to 1 modulo 4, then R(n,2) = 2 . ii) E .ítheA í6 41n and n .ce a eum ob atoo equane6 ox íl each pxí .me d¿v .ízoh 06 n congnuent to 3 modulo 4appean6 ín the jactonízatí .on ob n ínto pnímeÁ wíth a po4 .ítíve even exponent, then R(n,2) = 0 . iii) In a .Ql the oti1 .Fh cabeb .L6 R(n,2) = -1 . The following proposition characterizes the positive integers n having strictlypositive 4-level Proposition 2 . R(n,4) >  1í6 and only íg n 9 0(mod 8) . Proof . If n = 0(mod 8), then every representation of n as a sum of 4 squares, n = x2 +y 2 +z 2 +t 2 , verifies that g .c .d .(x,y,z,t) > 2 , and so £(n,4) = 0 . Furthermore, if n = 2,3,4,6,7(mod 8), then obviously n-1 = 1,1,3,5,6(mod 8) and, thus, n-1 is a sum of 3 squares,so we have £(n,4) > 1 . Finally, if n = 1,5(mod 8), then n-4 = 5,1(mod 8) and, consequently, n-4 is aleo a sum of three squares so that k(n,4) > 1 , because 2h . Remar k . For k>4 , we have £(n,k) > 1 forall n, just because n-1 is a sum of four squares . Let us concentrate from now on in the case k=3 . It is wellknown that a positive integern .is expressible as a sum of three integer squares if and only if n is not of the form 4 a (8m+7) . Gauss ([2], Art . 291) proved, moreover, that a positive integer admits a primitive representation as a sum of three squares if and only if n ;! 0,4,7(mod 8) . For £(h,3) we have the following elementary Proposition 3 . Let ne2Z + , .then .ib n - o(mod 4), i) J,,(n,3) < 0 ii) 9,(n,3) < 3 íb n - o(mod 2) oiL (mod 5) . The proof is immediate by passing to 2Z /m a with m = 4,2,5 . We nextprovethatgiven an odd positive integer with R(n,3) > 1 , if,we increase, preserving their parity, the exponente of its prime factors congruent to 1 modulo 4, then one can obtain level greater thanor equal to 2 . Lemma 4 .  (see  [11)  16 a,nca + cite bueh that a = a2 +a2 and n = b2 +b2 +b 2 , then a 2 n -- c2 +c2+c3  , wd th c 3 =  ab3  . The interestof the above lemma lies on the special values of the c . which allow us to obtain the 1 Proposition 5 .  Let n = 2 p 1  "" ' " p rq1  " * q s  ' W~h pi - 1(mod 4), 1  <i<r and qj  =- 3(mod 4),  1  < j  < s  ,  a= o on 1,  a l > o .  Then 1 a y1  yr ~1  Ss (n, 3)  >  1,  and m =  2 p1  . . .pr  q1  . .'q s (mod 2), .í t -tUAnb out that i) 16 a = o, then £(m,3) > 2 , ii) 16 a = 1, then k(m,3) > 1 . Proof . Write m = á2n , with c 1 = ab 1 - 2(a1b1+a2b2)al c 2 = ab 22(a 1 b 1 +a 2 b2)a 2 , a =paf . .p ar , so that y . = 2a .+ot ., i=1, . . .,r ; ó . > 1 . 1 '  r  i  i i  i - > a i and y i - a . i Then a is a sum of two squares : a = a 2 +a2 with (ai ,a) = 1 ; 1 < i < 2 . As £(n,3)  > 1 we can write n = b 2 +b2+b3 with (b 3 ,n) = 1 and (b 1 ,b 2,b3 ) = 1 . then 44 Now apply lemma 4 to write m = a 2 n = c2+c2+c2 . Let p - 1(mod 4) be a prime dividing m such that po 1 andpib 2 ; and if c 1 = -2a 1 b 1 a 1 jZ 0(mod p), c 2 = -2a 1 b 1 a2 ~! 0(mod p) , because pla . Interchanging the roles of b 1 and b 2 Let p = 1(mod 4) be a prime dividing ci = 0(mod p) for some iE{1,2}, then As p~b 1 we are allowed to write and as pla we get since p divides n but not b3 . a l b 1 +a 2 b2 = 0(mod p)  , . a l = - abb 2 (mod p) 1 the same result is obtained . m with 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 contradiction We have thus proved that both c 1 1 0(mod p) and c 2 j! O(mod p), for everyprime factor p = 1(mod 4) ofm . On the other hand, if q = 3(mod 4) is a prime factor of m, , we necessarily have that q~c3 , and as both c 1 and c 2 are nonzero, by lemma 1 of [1] we have that q~c 1 c2 . So, in the case (i) we have k(n,3) > 2 and in the case (ii), as 2~c 3 and 4~m, we get (c l ,2) = 1 or (c2 ,2) = 1 from which we infer that !G(n,3) > 1 . Theorem 6 . Let n be a poad tí .ve .LntegeA, and wxí te í t6 6actoAu :zaUon ín to . pxc :me bac tom aa  ' Next we state the following 1 (mod 4) , q 7 - 3 (mod 4) . W .í th thí,s notatíon we have 1ár np1 . . .p  , .then R(n,3) > 2 . a a a  a ii) Ib n = 2 5 1 p 2 2 . . .p rr 9,(n,3) =2 . aa l ar S 1  os n = 2 p 1 . . .p r q 1 . . .q s  , a+a 1 > 0 0<a< 1 , o<a l , then a l a r üi) I~ n= p 1 . . .P r and n .íz a num~ ídoneua o6 EuleA, then (n, 3)  =  2 . iv) I5 n = q 11. ., .gssand n Y 7(mod 8), xhen9,(n,3) = 3 . aa s s v) Ib n= 2 5 1 g2 2 . . .q s s and n Y 7(mod 8) S+S 1 > o, o <S < 1 then R(n,3) = 2í6 Sac S 1 = o , and k(n,3) > 1 othenwíae . vi) 11 n= p 1 1 g l l . . .q S and n ;z 7(mod 8), .th .en t(n,3) > 2 . a a S vi¡) Ib n= p 1p 2q 1.. .q s and n Y 7(mod 8), then k(n,3) > 1 . 1 21s 1l os vi¡¡) 16 n= 2p 1 g l... q, , then 2(n,3) > 1 . Proof . i) In this case n admits a primitive representation as a sum of two squares and therefore 2(n,3) > 2 . ii) It suffices to apply i) and proposition 3 . iii) These integers admit a primitive representation as a sum of two squares but do not have any representation as a sum of 3 positive squares (cf . [31) . Integers of this type are 13 and 37, and these are up to now the only known examples not greater than 5 .10 10 (see [5]) . iv), vi), vi¡) and vi¡¡) are immediate consequences of lemma 1 of [11v) Under these conditions n admits a primitive representation as a sum of these positive squares and it suffices to apply lenuna 1 of 111 together with proposition 3 . Now we give an application of the above theorem to the Galois embedding problem (cf . [41, Th . 5 .1) . Theorem 7 . Let n = g 1 1 . . .q S s With q i - 3(mod 4), 1 < i < s, and n =_ 3(mod 8) xhen eveAy centAa2 extene .íon ob the aQtexnatc :ng gnoup A n can be tea .F .íded ae a Ga1o .írs gnoup oveA Q(T) and, 4o, oven Q . Bibliográphy Arenas Sola, A . : On a centaín type o4 pnímítc :ve nepneeenfiatí,onó o6 natíonal .íntegena ae sum o6 squaAers . Pub . Sec . Mat . Univ . Aut b noma de Barcelona . Vol . 28 ; Núm . 2-3 (1984), 75-80 . [21 Gauss, C .F . : U .í,dqu,ca .í tc :onu A~etí .Cae . Lipsiae, 1801 .  English traslation : Arthur A . Clarke, 1966, New Haven : Yale Univ . Press . [31 S,chinzel, A . : SuJC Ieb somme6 de tA0í6 CWVCU . Bull . Acad . Pol . de s Sciences . Vol . 11, 6 (1959), 22-25 . [4] Vila, N . : On eentrta .2 exxenaíone ob A n ab a Galo .í .e guup ovelc Q . Arch . Math ., Vol . 44, (1985), 424-437 . [5] weinherger, P .J . : Exponente ob the ~ gnoup o6 complex quadnaUe 6t .ebdb . Acta Arith . 22 (1973), 118-124 . The authorthanks to the referee for some useful suggestions . Rebux el 15 d'octubne dei . 1985 Departamento de Algebra y Fundamentos Facultad de Matemáticas Universidad de Barcelona C/ Gran Via, 585 08007Barcelona SPAIN