ON MONIC BINARY QUADRATIC FORMS

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1 ON MONIC BINARY QUADRATIC FORMS JEROME T. DIMABAYAO*, VADIM PONOMARENKO**, AND ORLAND JAMES Q. TIGAS*** Abstract. We consider the quadratic form x +mxy +ny, where m n is 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. 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, n Z with associated monic quadratic form f m,n (x, y) := x + mxy + ny, we define = (m, n) = m n, the absolute value of the discriminant of this form. We will study to determine which integers f m,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 recently by Nair [] for the form (1, 1) = 3. More recently, Bahmanpour [1] determined the primes represented by forms (1, 1) = 3 and (1, 1) = 5. We extend these results to all forms with prime, provided Condition 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 Lemma 1.1 below). We have verified condition P, computationally, for the following values: = 3, 5, 7, 11, 13, 17, 19, 3, 9, 37, 1, 3, 53, 61, 67, 101, 163, 173, 197. It likely holds for other as well, as only < 00 were tested. It appears that = 31 is the first prime for which Condition P fails. Our results are restricted to the case where is prime. Note that if is prime, then m must be odd, and hence is also odd. We classify into cases via the following. Definition 1. Let N be an odd prime. We say that is of Type I if 3 (mod ); we say that is of Type II if 1 (mod ). Note that by some simple case analysis, if = n m (i.e. n > m ), then is of Type I. If instead, = m n (i.e. m > n), then is of type II. For m, n Z, let K m,n := {x +mxy+ny : x, y Z} denote the set of representable integers. This set is a monoid under multiplication. Lemma 1.1 (from [1]). Let m, n Z. Then (K m,n, ) is a monoid. 010 Mathematics Subject Classification. Primary 06B05, 06D35; Secondary 35R30, 39A10. Key words and phrases. MV-algebra, effect algebra, Navier-Stokes equation, oscillatory solution, asymptotic behaviour. 1

2 JEROME T. DIMABAYAO, VADIM PONOMARENKO, AND ORLAND JAMES Q. TIGAS Proof. Closure follows from the observation that (a + mab + nb )(c + mcd + nd ) = (ac nbd) + m(ac nbd)(bc+ad+mbd)+n(bc+ad+mbd). Identity follows from 1 = 1 +m(1)(0)+n(0). We call K m,n of type I/II, based on whether = m n is of type I/II. Note that the type of K m,n is determined solely by, independently of choice of m and n. For example, 5 = (1, 1) = (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 1.. Let m, n Z such that = m n is odd. We have K m,n = K 1 (m 1, n). Proof. If is odd, then m is also odd. Now suppose t = a + mab + nb K m,n. If we put c = a ( ) 1 m b Z, then we see that t = c + cb + 1 (m n) b K 1 (m 1, n). Conversely, if s = x +xy+ 1 (m n) y K 1 (m 1, n), then s = z +mzy+nz K m,n, where z = x+ ( ) 1 m y. In view of the above result, we put K n := K 1,n = {x + xy + ny : x, y Z}. We define K n := {x + xy + ny : x, y Z, gcd(x, y) = 1}, a subset of K n. Note that all nonzero squarefree elements of K n are in K n (in particular, all primes in K n ). Theorem.1 will prove that K n depends only on and Condition P, to be defined below. Let be an odd prime. We define the set P using Legendre symbols (for this and other standard notation, see []), as follows: P := { p prime : p 3, ( ) } p = 1. Note that P 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: P K n. (Condition P) We determine all representable primes (in Theorems.1 and 3.1). We then find all representable integers (in Theorem.1). A prime turns out to be representable in K n 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. However we have been unable to remove either this restriction, or Condition P.. Preliminaries Our first result proves non-representability for roughly half of Z. nonresidue, then it is not representable. If t Z is a quadratic Theorem.1. Let n, t Z such that = n 1 is prime and t. Suppose that t is a quadratic nonresidue modulo. Then t / K n. Proof. By way of contradiction, let us assume the existence of a, b Z with t = a + ab + nb. Multiplying both sides by and working modulo, we have t a + ab + nb (a + b) + b (n 1) (a + b). Hence 1 = ( t ) = ( t )( ) = ( t ) = 1, a contradiction. We next consider the special case of representing prime itself. If is of Type I, then can always be represented as f 1,n ( 1, ) = ( 1) + ( 1) + n = 1 + n =. If is of Type II, there is no such immediate formula. For example, 37 = (1, 9) K 9 as 37 = f 1, 9 (31, 1). For the slightly larger prime 97 = (1, ) K as 97 = f 1, ( 60797, 1108). But the same

3 ON MONIC BINARY QUADRATIC FORMS 3 expression f 1,n ( 1, ) 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.1. Let p be a prime with p 1 (mod ). Then there exist x, y N such that x py = 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 X py = 1. (.1) Consider such a pair (X 0, Y 0 ) with minimal X 0. Note that X 0 and Y 0 are relatively prime. Since p 1 (mod ), X 0 must be odd and Y 0 is even. Write Y 0 = W. Then we may write equation (.1) as X X0 1 = pw. Since X X 0 1 = 1, the integers X and X 0 1 must be relatively prime. Hence, there exist relatively prime positive integers a and b such that X = a X 0 1 or = pb X = pa X 0 1 = b. In the first case we have X 0 = a 1 = pb + 1, which gives a pb = 1. But we also have 1 a a a 1 = X 0. This contradicts the minimality of X 0 and the assumption that X 0 1. Hence we must have X 0 = pa 1 = b + 1, which gives b pa = 1. Lemma.. Let n Z such that = 1 n is a prime of type II. Then 1 K n. Proof. Let x, y N such that x y = 1. Then we find f 1,n ( x y, y) = (x + y) (x + y)(y) + ny = x (1 n)y = 1. Using Lemma., we see that monoids of type II are nicely symmetric around 0. Lemma.3. Let m, n Z such that = 1 n is a prime of type II. Then for every t Z, t K n if and only if t K n. In particular, K n. Proof. Apply Lemma 1.1 to t = ( 1)( t) and t = ( 1)(t). If instead, the monoid is of type I, then it contains no negative integers at all. Lemma.. Let n Z such that = n 1 is a prime of type I. Then K n N 0. Proof. Let a, b Z. Since n > 1, we must have n > 0. Set s = 1 n, and c = b n. We have a + ab + nb = a + sac + c = +s (a + c) + s (a c). Since n > 1, s < and hence both +s and s are positive. Thus a + ab + nb 0, with equality only for a = b = 0.

4 JEROME T. DIMABAYAO, VADIM PONOMARENKO, AND ORLAND JAMES Q. TIGAS 3. Representing Quadratic Residues We turn now to the question of representing quadratic residues. This is less straightforward than Theorem.1, as not all quadratic residues are representable. In several steps we prove Theorem 3.1, which resolves representation of primes that are quadratic residues. The next lemma, relying on the law of quadratic reciprocity, is the starting point toward Theorem 3.1. It will be needed for all primes except and. Lemma 3.1. Let n Z such that = 1 n is a prime. Let p be any odd prime different from. Then s a quadratic residue modulo if and only if 1 n is a quadratic residue modulo p. Proof. Suppose first that 1 n > 0. By quadratic reciprocity, 1 = ( )( 1 n p p 1 n), since 1 n 1 (mod ). On the other hand, if 1 n < 0, then ( 1) (p 1)/ = ( p p)( ), since (1 n) 1 (mod ). But also ( ) ( p = 1 )( 1 n ) ( ) ( p p = ( 1) (p 1)/ 1 n p. This implies 1 = 1 n )( p ) p. Our approach to prove that some p K n will be to start with pt K n for some integer t. The following strong lemma shows that if s a quadratic nonresidue modulo, then not only is p not in K n, but no multiple of s in K n either. It also gives examples of nonrepresentible quadratic residues. If p and q are distinct primes which are quadratic nonresidues modulo, then pq / K n. It follows that pq / K n, even though pq is a quadratic residue. Lemma 3.. Let n Z such that = 1 n is a prime. Let p be any odd prime different from and let t Z. If s a quadratic nonresidue modulo, then pt / K n. Proof. We assume by way of contradiction the existence of a, b Z with pt = a + ab + nb and gcd(a, b) = 1. If p b, then p (pt ab nb ), so p a, which contradicts gcd(a, b) = 1. Hence p b, and we can choose an integer c so that cb 1 (mod p). Working modulo p, we have 0 a + ab + nb b ((ac) + (ac) + n). Hence 0 ((ac) + (ac) + n) (ac + 1) + n 1, and thus (ac + 1) 1 n (mod p). Thus 1 n is a quadratic residue, modulo p. By Lemma 3.1, s a quadratic residue modulo ; this contradicts the hypothesis. Since every odd prime has quadratic nonresidues, we can apply Dirichlet s theorem on arithmetic progressions to find some odd prime p that is a quadratic nonresidue modulo. Applying Lemma 3. to this p and to t = 0, implies that 0 / K n. We now present an analogue of Lemma 3. for p =. Lemma 3.3. Let n, t Z such that = 1 n is a prime. If is a quadratic nonresidue modulo, then t / K n. Proof. Since ( ) = 1, by the second supplement to the law of quadratic reciprocity, we must have 1 n = ±3 (mod 8). A simple case analysis shows that n is odd. Assume by way of contradiction the existence of a, b Z with t = a + ab + nb and gcd(a, b) = 1. In particular, a, b cannot both be even. If a, b are both odd, then a + ab + nb is also odd, a contradiction. If b is odd and a = k is even, we have 0 (k) + (k)b + nb b(k + nb) (mod ). But now (k + nb), so nb is even and hence b is even, a contradiction. Lastly, if a is odd and b = j is even, we have 0 a + a(j) + n(j) a(a + j) (mod ). But now (a + j), so a is even, a contradiction. We now represent, not yet an arbitrary prime, but some integer multiple thereof. The condition p > 3 is why most of Condition P is imposed. An improvement here would equally improve Condition P. Lemma 3.. Let n Z such that = 1 n is a prime. Let p be any odd prime different from. If s a quadratic residue modulo, then pt K n, for some t Z. If p > 3, then we may assume that 0 < t < p.

5 ON MONIC BINARY QUADRATIC FORMS 5 Proof. By Lemma 3.1, there is some r Z such that r 1 n (mod p). Choose s Z such that s + 1 r (mod p). We have s + s + n 0 (mod p), and hence s + s + n 0 (mod p). Hence, for some t Z, there is a representation t p = f 1,n (s, 1). We return now to the choice of s, and try to find a different choice (but still equivalent modulo p), which will make t small. Consider the quadratic real polynomial g(x) = (s+xp) +(s+xp)+n. For all x Z, g(x) will not only be integer-valued, but a multiple of p. The graph of g(x) has vertex at k = s+1 p. We calculate g(k ) = n 1, and g(k + 1 ) = g(k 1 ) = n 1 + p. Choose an integer k [k 1, k p < 3p + p ]. We have g(k) [ n 1, n 1 + p ]. Hence, g(k) n 1 + p = = p. Hence, for some t with t < p, we have tp = g(k) = f 1,n (s + kp, 1). We have t 0 since 0 / K n. This next lemma is a generalization of a result found in []. It shows that if a prime p and pt, for some t Z, are both representable, then t is also representable. Lemma 3.5. Let n Z such that = 1 n is a prime. Let t N and let p be a prime. If tp, p K n, then t K n. Proof. By hypothesis, there are integers a, b, c, d with tp = a + ab + nb and p = c + cd + nd. In fact we have gcd(c, d) = 1, since s prime. We calculate b p d tp = b (c + cd + nd ) d (a + ab + nb ) = (bc ad)(bd + bc + ad). Hence either p (bc ad) or p (bd + bc + ad), which splits the proof into two cases. Suppose first that p (bc ad). There is some r Z with rp = bc ad. Set y = a + rnd and x = b rc. We substitute for a, b to get rp = r(c +cd+nd ) rcd yd+xc, so 0 = rcd yd+xc and hence c(x rd) = dy. Since gcd(c, d) = 1, there is some w Z with y = cw. Substituting, we get x = d(w +r). Hence a = cw rnd and b = d(w +r)+rc. Since (w +wr +nr )(c +cd+nd ) = (cw rnd) + (cw rnd)(d(w + r) + rc) + n(d(w + r) + rc) = a + ab + nb = tp, we find that t = w + wr + nr K n. Suppose now that p (bd+bc+ad). There is some r Z with rp = bd+bc+ad. Set y = a rnd and x = b rc. We substitute for a, b to get rp = r(c +cd+nd )+xd+xc+yd, so 0 = xd+xc+yd and hence d(x + y) = c( x). Since gcd(c, d) = 1, there is some w Z with x = dw. Substituting, we get y = w(d+c). Hence a = w(d+c)+rnd and b = dw+rc. Since (w +wr+nr )(c +cd+nd ) = (w(d + c) + rnd) + (w(d + c) + rnd)( dw + rc) + n( dw + rc) = a + ab + nb = tp, we find that t = w + wr + nr K n. We now prove that all primes that are quadratic residues are representable, subject to Condition P. Theorem 3.1. Let n Z such that = 1 n is a prime. Suppose that Condition P holds. Then p K n for every prime p that is a quadratic residue modulo. Proof. By way of contradiction, let p be the smallest prime with ( ) p = 1 and p / Kn. If p 3, then p P, which contradicts Condition P. We now choose t Z to have minimal absolute value to satisfy both 0 < t < p and pt K n. Note that such a t exists by Lemma 3.. If t = 1 we contradict p / K n. If is of Type I, t = 1 is impossible by Lemma.. Suppose is of Type II and t = 1. From Lemma. we have 1 K n. Thus Lemma 1.1 gives p = ( 1)(tp) K n, a contradiction. Hence we may assume that t > 1 and write t = p 1 p p k, a product of (not necessarily distinct) primes, each less than p. Suppose that some K n ; we will show that this is impossible. By Lemma 3.5, p t K n. Hence p t = a + ab + nb for some a, b Z. We have gcd(a, b) p t. If gcd(a, b) = p, then p pt,

6 6 JEROME T. DIMABAYAO, VADIM PONOMARENKO, AND ORLAND JAMES Q. TIGAS a contradiction. Hence gcd(a, b) t t a. We now have p gcd(a,b) = ( gcd(a,b) ) a + ( gcd(a,b) )( b gcd(a,b) ) + n n( gcd(a,b) ) K n. This contradicts our choice of t. Hence each / K n and in particular. Consequently each is a quadratic nonresidue modulo, otherwise by our choice of p we must have K n. If any were odd, we would obtain a contradiction to Lemma 3.. Hence t = c for some c N, where is a quadratic nonresidue modulo. If c = 1, then tp = s the product of a quadratic nonresidue and a quadratic residue. Hence ts a quadratic nonresidue, and by Theorem.1, tp / K n, a contradiction. Hence c, but by Lemma 3.3, tp = ( c p) / K n, a contradiction. We can now reproduce the known results. For the known (1, 1) = 3 of Type I and ( 1, 1) = (1, 1) = 5 of type II, we have P 3 = and P 5 =, respectively. Thus, condition P holds vacuously for these two examples. For another example, take 17 = (3, ) = (1, ). We have P 17 = {}. Then condition P is verified since = f 1, (, 1). We also have = f 3, (1, 1) by inspection (or using the proof of Lemma 1.).. Complete determination of representable integers We turn now to the question of characterizing irreducibles in the monoid K n. Due to Lemmas.3 and., we concern ourselves only with irreducibles in K n N, itself a monoid. Lemma.1. Let n Z such that = 1 n is a prime. Suppose that Condition P holds. Then the irreducibles in the monoid K n N are exactly those integers of the form: (1), () p, where s prime and a quadratic residue modulo ; and (3) q, where q is prime and a quadratic nonresidue modulo. Proof. We have K n from Section and p K n by Theorem 3.1. We have q = f 1,n (q, 0) K n ; it is irreducible by Theorem.1. Now let t K n be some other irreducible. Write t = p 1 p p k, for not necessarily distinct primes. We must have k by Theorem.1 again. If any K n, t then by Lemma 3.5, K n, which contradicts irreducibility. In particular, by Condition P, no can be. If any is odd, then by Lemma 3., t / K n. But then, writing t = a + ab + nb, there is some prime r dividing gcd(a, b). We have r = f 1,n (r, 0) and t r = ( a r ) + ( a r )( b r ) + n( b r ). But t r > 1 since t is not among the two types of irreducibles already described. Hence t is reducible, which is a contradiction. The remaining possibility is that t is a power of, where is a quadratic nonresidue modulo. An even power of may be written as a product of irreducibles, while an odd power of is not in K n by Theorem.1. Hence no such t can exist. With the irreducibles we may easily determine the full monoid K n. The statement of Theorem.1 is similar to a well-known theorem on representing integers as the sum of two squares, i.e. the quadratic form = (0, 1). We recall also Lemmas.3 and., which combine with Theorem.1 to resolve the membership question for negative integers. Theorem.1. Let n Z such that = 1 n is a prime. Suppose that Condition P holds. Let t N. Then t K n if and only if the prime decomposition of t contains no prime, that is a quadratic nonresidue modulo, raised to an odd power. Proof. Immediate from Lemma.1. References [1] BAHMANPOUR, K.: Prime numbers p with expression p = a ± ab ± b, J. Number Theory 166 (016),

7 ON MONIC BINARY QUADRATIC FORMS 7 [] HARDY, G. H., WRIGHT, E. M.: An introduction to the theory of numbers, Oxford University Press, Oxford, sixth edition, 008. Revised by D.R. Heath-Brown and J.H. Silverman, With a foreword by Andrew Wiles. [3] MORDELL, L. J.: Diophantine equations, Pure and Applied Mathematics, Vol. 30. Academic Press, London- New York, [] NAIR, U. P.: Elementary results on the binary quadratic form a + ab + b, * Institute of Mathematics College of Science University of the Philippines Diliman C.P. Garcia St., U.P.Campus Diliman, 1101 Quezon City PHILIPPINES address: jdimabayao@math.upd.edu.ph ** Department of Mathematics and Statistics San Diego State University 5500 Campanile Drive, San Diego California USA address: vponomarenko@mail.sdsu.edu *** Institute of Mathematics College of Science University of the Philippines Diliman C.P. Garcia St., U.P.Campus Diliman, 1101 Quezon City PHILIPPINES address: oqtigas@upd.edu.ph