Vol. 20 (2019), No. 2, pp. 863–871 DOI: 10.18514/MMN.2019.2920
ON THE MONOID OF MONIC BINARY QUADRATIC FORMS
JEROME T. DIMABAYAO, VADIM PONOMARENKO, AND ORLAND JAMES Q. TIGAS
Received 31 March, 2019
Abstract. We consider the quadratic formx2CmxyCny2, wherejm2 4njis a prime number.
Under the assumption that a particular, small, finite set of integers is representable, we determine all integers representable by this quadratic form.
2010Mathematics Subject Classification: 11N32; 11E25; 11A41
Keywords: binary quadratic form, law of quadratic reciprocity, prime number
1. INTRODUCTION
Representation of integers by quadratic forms is a classical problem, with major contributions by Fermat, Euler, Lagrange, and Gauss. We consider those forms that are binary, quadratic, monic, and with a cross term. Specifically, given m; n2Z with associated monic quadratic form fm;n.x; y/ WDx2CmxyCny2, we define D .m; n/D jm2 4nj, the absolute value of the discriminant of this form. We will studyto determine which integersfm;n.x; y/represents.
Modern knowledge about quadratic forms comes from the genus theory of Gauss and also involves the geometry of numbers and factorization over rings of integers of number fields. But it is still interesting to ask how far elementary methods can go for the study of such forms. For instance, although it is well-known that the product of two sums of squares is a sum of squares, the fact that it extends to products of arbitrary monic binary quadratic forms seems to be less-known (cf. [1]).
This problem was revisited by Nair [4] for the form .1; 1/D3. More recently, Bahmanpour [1] determined the primes represented by forms .1; 1/ D 3 and .1; 1/D5. We extend these results to all forms with prime , provided Condi- tion P holds (as defined below). All the above-mentioned results were obtained by elementary arguments using the fact that the set of integers represented by forms a monoid under multiplication of integers (see Lemma1below).
We have verified condition P, computationally, for the following values:
D3; 5; 7; 11; 13; 17; 19; 23; 29; 37; 41; 43; 53; 61; 67; 101; 163; 173; 197:
c 2019 Miskolc University Press
It likely holds for otheras well, as only < 200were tested. It appears thatD31 is the first prime for which Condition P fails.
Our results are restricted to the case whereis prime. Note that if is prime, then mmust be odd, and hence is also odd. We classify into cases via the following.
Definition 1. Let 2N be an odd prime. We say that isof Type I if 3 .mod 4/; we say that isof Type II if1 .mod4/.
Note that by some simple case analysis, ifD4n m2(i.e. 4n > m2), then is of Type I. If instead,Dm2 4n(i.e. m2> 4n), then is of type II. Form; n2Z, let Km;nWD fx2CmxyCny2Wx; y2Zgdenote the set of representable integers. This set is a monoid under multiplication.
Lemma 1([1]). Letm; n2Z. Then.Km;n;/is a monoid.
Proof. Closure follows from the observation that.a2CmabCnb2/.c2CmcdC nd2/D.ac nbd /2Cm.ac nbd /.bcCadCmbd /Cn.bcCadCmbd /2. Iden-
tity follows from1D12Cm.1/.0/Cn.0/2.
We callKm;nof type I/II, based on whetherD jm2 4njis of type I/II. Note that the type ofKm;n is determined solely by, independently of choice ofmandn. For example,5D .1; 1/D .3; 1/.
The following lemma shows that we may assume that the coefficient of the cross term is one for the quadratic forms that we want to study.
Lemma 2. Let m; n 2Z such that D jm2 4nj is odd. We have Km;n D K1;1 .m24 4n/.
Proof. Ifis odd, thenmis also odd. Now supposetDa2CmabCnb22Km;n. If we put c Da 1 m2
b 2Z, then we see that t Dc2CcbC1 .m42 4n/b2 2 K1;1 .m24 4n/. Conversely, ifsDx2CxyC1 .m42 4n/y22K
1;1 .m24 4n/, thensD
´2Cm´yCny22Km;n, where´DxC 1 m2
y.
In view of the above result, we putKnWDK1;n D fx2CxyCny2 Wx; y 2Zg. We defineK0nWD fx2CxyCny2Wx; y2Z;gcd.x; y/D1g, a subset ofKn. Note that all nonzero squarefree elements ofKnare inK0n(in particular, all primes inKn).
Theorem3will prove thatKndepends only onand Condition P, to be defined below.
Letbe an odd prime. We define the setPusing Legendre symbols (for this and other standard notation, see [2]), as follows:
PWD (
pprimeWp r
3; p
! D1
) :
Note thatP is a finite (possibly empty) set of integers, all prime. Most of our results require the following condition. It states that all elements of P must be
representable:
PKn: (Condition P)
We determine all representable primes (in Theorems1 and2). We then find all representable integers (in Theorem3). A prime turns out to be representable in Kn
if and only if it is equal to or is a quadratic residue modulo; a positive integer turns out to be representable if and only if each quadratic nonresidue prime in its factorization appears to an even power.
Computational evidence suggests that our results hold when is not prime. How- ever we have been unable to remove either this restriction, or Condition P.
2. PRELIMINARIES
Our first result proves non-representability for roughly half of Z. If t 2Zis a quadratic nonresidue, then it is not representable.
Theorem 1. Letn; t2Zsuch thatD j4n 1jis prime and−t. Suppose that t is a quadratic nonresidue modulo. Thent…Kn.
Proof. By way of contradiction, let us assume the existence of a; b 2Z with t Da2CabCnb2. Multiplying both sides by 4 and working modulo, we have 4t4a2C4abC4nb2.2aCb/2Cb2.4n 1/.2aCb/2. Hence1D 4t
D
t
2
2
D t
D 1, a contradiction.
We next consider the special case of representing prime itself. If is of Type I, then can always be represented asf1;n. 1; 2/D. 1/2C. 1/2C4nD 1C 4nD. If is of Type II, there is no such immediate formula. For example, 37D .1; 9/2K 9as37Df1; 9.31; 12/. For the slightly larger prime97D .1; 24/2 K 24as97Df1; 24. 60797; 11208/. But the same expressionf1;n. 1; 2/as above shows that is represented when is of Type II. We thus obtain representability for in this case if we can verify representability for 1. We attain this using the following known result, whose proof will be included for completeness.
Lemma 3. Letpbe a prime withp1 .mod4/. Then there existx; y2Nsuch thatx2 py2D 1.
Proof. By a theorem of Lagrange (cf. e.g. [3, Thm. 1, p. 53]), there exist infinitely many pairs of positive integers.X; Y /¤.1; 0/that satisfies the Pell equation
X2 pY2D1: (2.1)
Consider such a pair.X0; Y0/with minimalX0. Note thatX0 andY0 are relatively prime. Sincep1 .mod 4/,X0must be odd andY0is even. WriteY0D2W. Then we may write equation (2.1) as
X0C1
2 X0 1
2 DpW2:
Since X0C1 2
X0 1
2 D1, the integers X0C1
2 and X0 1
2 must be relatively prime. Hence, there exist relatively prime positive integersaandbsuch that
8 ˆ<
ˆ: X0C1
2 Da2 X0 1
2 Dpb2
or
8 ˆ<
ˆ: X0C1
2 Dpa2 X0 1
2 Db2:
In the first case we haveX0D2a2 1D2pb2C1, which givesa2 pb2D1. But we also have
1aa22a2 1DX0:
This contradicts the minimality ofX0 and the assumption thatX0¤1. Hence we must haveX0D2pa2 1D2b2C1, which givesb2 pa2D 1.
Lemma 4. Letn2Zsuch that D1 4nis a prime of type II. Then 12Kn. Proof. Letx; y2Nsuch thatx2 y2D 1. Then we findf1;n. x y; 2y/D .xCy/2 .xCy/.2y/C4ny2Dx2 .1 4n/y2D 1.
Using Lemma4, we see that monoids of type II are nicely symmetric around0.
Lemma 5. Letm; n2Zsuch thatD1 4nis a prime of type II. Then for every t2Z,t2Knif and only if t2Kn. In particular,2Kn.
Proof. Apply Lemma1totD. 1/. t /and tD. 1/.t /.
If instead, the monoid is of type I, then it contains no negative integers at all.
Lemma 6. Letn2Zsuch that D4n 1is a prime of type I. ThenKnN0. Proof. Leta; b2Z. Since4n > 1, we must haven > 0. SetsDp1n, andcDbp
n.
We havea2CabCnb2Da2CsacCc2D2C4s.aCc/2C2 s4 .a c/2. Since4n > 1, jsj< 2 and hence both 2C4s and 2 s4 are positive. Thus a2CabCnb2 0, with
equality only foraDbD0.
3. REPRESENTING QUADRATIC RESIDUES
We turn now to the question of representing quadratic residues. This is less straightforward than Theorem1, as not all quadratic residues are representable. In several steps we prove Theorem2, which resolves representation of primes that are quadratic residues.
The next lemma, relying on the law of quadratic reciprocity, is the starting point toward Theorem2. It will be needed for all primes except2and.
Lemma 7. Letn2Zsuch that D j1 4njis a prime. Letp be any odd prime different from. Thenp is a quadratic residue modulo if and only if1 4n is a quadratic residue modulop.
Proof. Suppose first that1 4n > 0. By quadratic reciprocity, 1D 1 4n
p
! p 1 4n
!
;
since1 4n1 .mod4/. On the other hand, if1 4n < 0, then . 1/.p 1/=2D
p
! p
!
;
since .1 4n/ 1 .mod 4/. But also
p
!
D 1 p
! 1 4n p
!
D. 1/.p 1/=2 1 4n p
! :
This implies
1D 1 4n p
! p
! :
Our approach to prove that somep2Kn will be to start withpt 2Knfor some integer t. The following strong lemma shows that if p is a quadratic nonresidue modulo, then not only ispnot inK0n, but no multiple ofpis inK0neither. It also gives examples of nonrepresentible quadratic residues. Ifpandqare distinct primes which are quadratic nonresidues modulo, thenpq…K0n. It follows thatpq…Kn, even thoughpqis a quadratic residue.
Lemma 8. Letn2Zsuch that D j1 4njis a prime. Letp be any odd prime different fromand lett2Z. Ifpis a quadratic nonresidue modulo, thenpt…K0n. Proof. We assume by way of contradiction the existence ofa; b2Zwith pt D a2CabCnb2 and gcd.a; b/D1. Ifpjb, thenpj.pt ab nb2/, so pja, which contradicts gcd.a; b/D1. Hencep−b, and we can choose an integercso thatcb1 .mod p/. Working modulop, we have0a2CabCnb2b2..ac/2C.ac/Cn/.
Hence04..ac/2C.ac/Cn/.2acC1/2C4n 1, and thus.2acC1/21 4n .mod p/. Thus1 4nis a quadratic residue, modulop. By Lemma7,pis a quadratic
residue modulo; this contradicts the hypothesis.
Since every odd prime has quadratic nonresidues, we can apply Dirichlet’s the- orem on arithmetic progressions to find some odd prime p¤ that is a quadratic nonresidue modulo. Applying Lemma8to thispand totD0, implies that0…K0n.
We now present an analogue of Lemma8forpD2.
Lemma 9. Letn; t 2Z such that D j1 4njis a prime. If 2 is a quadratic nonresidue modulo, then4t…K0n.
Proof. Since 2
D 1, by the second supplement to the law of quadratic reci- procity, we must havej1 4nj D ˙3 .mod 8/. A simple case analysis shows thatnis odd. Assume by way of contradiction the existence ofa; b2Zwith4tD a2CabCnb2 and gcd.a; b/D1. In particular, a; b cannot both be even. If a; b are both odd, then a2CabCnb2 is also odd, a contradiction. If b is odd and aD2k is even, we have 0.2k/2C.2k/bCnb2b.2kCnb/ .mod4/. But now4j.2kCnb/, sonb is even and hencebis even, a contradiction. Lastly, ifais odd andbD2j is even, we have0a2Ca.2j /Cn.2j /2a.aC2j / .mod4/.
But now4j.aC2j /, soais even, a contradiction.
We now represent, not yet an arbitrary prime, but some integer multiple thereof.
The condition p >
q
3 is why most of Condition P is imposed. An improvement here would equally improve Condition P.
Lemma 10. Letn2Zsuch thatD j1 4njis a prime. Letpbe any odd prime different from. Ifpis a quadratic residue modulo, thenpt 2K0n, for somet2Z.
Ifp >q
3, then we may assume that0 <jtj< p.
Proof. By Lemma7, there is somer2Zsuch thatr21 4n .mod p/. Choose s2Zsuch that 2sC1r .modp/. We have 4s2C4sC4n0 .modp/, and hence s2CsCn0 .mod p/. Hence, for somet02Z, there is a representation t0pDf1;n.s; 1/.
We return now to the choice ofs, and try to find a different choice (but still equival- ent modulop), which will makejt0jsmall. Consider the quadratic real polynomial g.x/D.sCxp/2C.sCxp/Cn. For all x2Z, g.x/ will not only be integer- valued, but a multiple of p. The graph of g.x/ has vertex at k0 D 2s2pC1. We calculate g.k0/D 4n 14 , and g.k0C12/Dg.k0 12/D 4n 14 Cp42. Choose an in- teger k2Œk0 12; k0C12. We haveg.k/ 2Œ4n 14 ;4n 14 Cp42. Hence, jg.k/j j4n 14 j C jp42j D 4Cp42 < 3p42Cp42 Dp2. Hence, for sometwithjtj< p, we have tpDg.k/Df1;n.sCkp; 1/. We havet¤0since0…K0n. This next lemma is a generalization of a result found in [4]. It shows that if a prime pandpt, for somet2Z, are both representable, thent is also representable.
Lemma 11. Letn2Z,t2Nandpbe a prime. Iftp; p2Kn, thent2Kn. Proof. By hypothesis, there are integers a; b; c; d with tpDa2CabCnb2 and pDc2CcdCnd2. In fact we have gcd.c; d /D1, since pis prime. We calculate b2p d2tpDb2.c2CcdCnd2/ d2.a2CabCnb2/D.bc ad /.bdCbcCad /.
Hence eitherpj.bc ad /orpj.bdCbcCad /, which splits the proof into two cases.
Suppose first thatpj.bc ad /. There is somer2ZwithrpDbc ad. SetyD aCrndandxDb rc. We substitute fora; bto getrpDr.c2CcdCnd2/ rcd
ydCxc, so0D rcd ydCxcand hencec.x rd /Ddy. Since gcd.c; d /D1, there is some w2 Z with y Dcw. Substituting, we get x Dd.wCr/. Hence aDcw rnd andbDd.wCr/Crc. Since.w2CwrCnr2/.c2CcdCnd2/D .cw rnd /2C.cw rnd /.d.wCr/Crc/Cn.d.wCr/Crc/2Da2CabCnb2D tp, we find thattDw2CwrCnr22Kn.
Suppose now thatpj.bdCbcCad /. There is somer2ZwithrpDbdCbcC ad. Sety Da rnd and xDb rc. We substitute for a; b to get rpDr.c2C cdCnd2/CxdCxcCyd, so0DxdCxcCyd and henced.xCy/Dc. x/.
Since gcd.c; d /D1, there is some w 2Z with xDdw. Substituting, we get y Dw.dCc/. Hence aDw.dCc/Crnd and b D dwCrc. Since .w2C wrCnr2/.c2CcdCnd2/D.w.dCc/Crnd /2C.w.dCc/Crnd /. dwCrc/C n. dwCrc/2Da2CabCnb2Dtp, we find thattDw2CwrCnr22Kn.
We now prove that all primes that are quadratic residues are representable, subject to Condition P.
Theorem 2. Letn2Zsuch thatD j1 4njis a prime. Suppose that Condition P holds. Thenp2Knfor every primepthat is a quadratic residue modulo.
Proof. By way of contradiction, let p be the smallest prime with p
D1 and p…Kn. Ifpq
3, thenp2P, which contradicts Condition P.
We now chooset2Zto have minimal absolute value to satisfy both 0 <jtj< p andpt2K0n. Note that such atexists by Lemma10.
IftD1we contradictp…Kn. Ifis of Type I,tD 1is impossible by Lemma6.
Suppose is of Type II andtD 1. From Lemma4we have 12Kn. Thus Lemma 1givespD. 1/.tp/2Kn, a contradiction. Hence we may assume thatjtj> 1and writejtj Dp1p2 pk, a product of (not necessarily distinct) primes, each less than p.
Suppose that somepi 2Kn; we will show that this is impossible. By Lemma11, ppt
i 2Kn. Henceppt
i Da2CabCnb2for somea; b2Z. We have gcd.a; b/2jppt
i. If gcd.a; b/Dp, thenp2jpt, a contradiction. Hence gcd.a; b/2jpti. We now have pp t
igcd.a;b/2 D.gcd.a;b/a /2C.gcd.a;b/a /.gcd.a;b/b /Cn.gcd.a;b/b /22K0n. This contra- dicts our choice oft. Hence eachpi …Knand in particularpi¤.
Consequently eachpiis a quadratic nonresidue modulo, otherwise by our choice ofp we must havepi 2Kn. If anypi were odd, we would obtain a contradiction to Lemma 8. Hence t D2c for some c 2N, where 2 is a quadratic nonresidue modulo . Ifc D1, then tpD2p is the product of a quadratic nonresidue and a quadratic residue. Hencetpis a quadratic nonresidue, and by Theorem1,tp…Kn, a contradiction. Hencec2, but by Lemma9,tpD4.2c 2p/…K0n, a contradiction.
We can now reproduce the known results. For the known .1; 1/D3of Type I and . 1; 1/D .1; 1/D5of type II, we haveP3D¿andP5D¿, respectively.
Thus, condition P holds vacuously for these two examples. For another example, take 17D .3; 2/D .1; 4/. We have P17D f2g. Then condition P is verified since2Df1; 4.2; 1/. We also have2Df3; 2.1; 1/by inspection (or using the proof of Lemma2).
4. COMPLETE DETERMINATION OF REPRESENTABLE INTEGERS
We turn now to the question of characterizing irreducibles in the monoidKn. Due to Lemmas5and6, we concern ourselves only with irreducibles inKn\N, itself a monoid.
Lemma 12. Letn2Zsuch thatD j1 4njis a prime. Suppose that Condition P holds. Then the irreducibles in the monoidKn\Nare exactly those integers of the form:
(1) ,
(2) p, wherepis prime and a quadratic residue modulo; and (3) q2, whereqis prime and a quadratic nonresidue modulo.
Proof. We have 2 Kn from Section 2 and p 2 Kn by Theorem 2. We have q2Df1;n.q; 0/2Kn; it is irreducible by Theorem1. Now lett2Knbe some other irreducible. Writet Dp1p2 pk, for not necessarily distinct primespi. We must have k2 by Theorem 1 again. If any pi 2 Kn, then by Lemma 11, pt
i 2Kn, which contradicts irreducibility. In particular, by Condition P, no pi can be . If any pi is odd, then by Lemma 8, t …K0n. But then, writing t Da2CabCnb2, there is some primer dividing gcd.a; b/. We haver2Df1;n.r; 0/and rt2 D.ar/2C .ar/.br/Cn.br/2. Butrt2> 1sincetis not among the two types of irreducibles already described. Hencet is reducible, which is a contradiction. The remaining possibility is thattis a power of2, where2is a quadratic nonresidue modulo. An even power of2may be written as a product of irreducibles22, while an odd power of2is not in
Knby Theorem1. Hence no suchtcan exist.
With the irreducibles we may easily determine the full monoidKn. The statement of Theorem3is similar to a well-known theorem on representing integers as the sum of two squares, i.e. the quadratic form 4D .0; 1/. We recall also Lemmas 5and 6, which combine with Theorem3to resolve the membership question for negative integers.
Theorem 3. Letn2Zsuch thatD j1 4njis a prime. Suppose that Condition P holds. Lett2N. Thent2Knif and only if the prime decomposition oftcontains no prime, that is a quadratic nonresidue modulo, raised to an odd power.
Proof. Immediate from Lemma12.
REFERENCES
[1] K. Bahmanpour, “Prime numberspwith expressionpDa2˙ab˙b2.”J. Number Theory, vol.
166, pp. 208–218, 2016, doi:10.1016/j.jnt.2016.02.024.
[2] G. Hardy and E. Wright,An introduction to the theory of numbers. Edited and revised by D. R.
Heath-Brown and J. H. Silverman. With a foreword by Andrew Wiles. 6th ed., 6th ed. Oxford:
Oxford University Press, 2008.
[3] L. Mordell,Diophantine equations., 1969.
[4] U. P. Nair, “Elementary results on the binary quadratic forma2CabCb2,”ArXiv Mathematics e-prints, 2004.
Authors’ addresses
Jerome T. Dimabayao
Institute of Mathematics, College of Science, University of the Philippines Diliman, C.P. Garcia St., U.P.Campus, Diliman, 1101 Quezon City, Philippines
E-mail address:jdimabayao@math.upd.edu.ph
Vadim Ponomarenko
Department of Mathematics and Statistics, San Diego State University, 5500 Campanile Drive, San Diego California, USA
E-mail address:vponomarenko@mail.sdsu.edu
Orland James Q. Tigas
Institute of Mathematics, College of Science, University of the Philippines Diliman, C.P. Garcia St., U.P.Campus, Diliman, 1101 Quezon City, Philippines
E-mail address:oqtigas@upd.edu.ph
