ACTA ARITHMETICA LXX3 (1995) Predictive criteria for the representation of primes by binary quadratic forms by Joseph B Muskat (Ramat-Gan), Blair K Spearman (Kelowna, BC) and Kenneth S Williams (Ottawa, Ont) 1 Introduction Suppose that p is a prime which can be represented by primitive integral binary quadratic forms of distinct nonsquare discriminants d and d Let G denote the genus of forms of discriminant d to which the forms representing p belong Assume that the parameters in a representation of p by a binary quadratic form of discriminant d are known A criterion which partitions G into subsets and uses the parameters in the above representation to predict which subset contains the form(s) representing p is called a predictive criterion For example, if p is a prime for which the Legendre symbols (/p) and (p/31) both have the value 1, then p is represented by positive-definite binary quadratic forms of discriminants 14 and 48 In the case of discriminant 14, p is represented either by the form x + 31y or by the forms 5x ± 4xy + 7y As 5(5x ± 4xy + 7y ) (x 1y) + 31(x ± y), there are integers H ( 1 or 5), M and N such that H p M + 31N The (principal) genus of forms of discriminant 48 representing p consists of the four forms x + 6y, x + 31y, 7x ± xy + 9y Given M and N we can predict whether p is represented by one of x + 6y, x + 31y or by 7x ± xy + 9y, as follows: { M + N ±1 (mod 8) p is represented by x + 6y or x + 31y, M + N ±3 (mod 8) p is represented by 7x ± xy + 9y ; see [15, p 76] The primary purpose of this paper is to show that an elementary technique of Dirichlet [9] yields predictive criteria for positive-definite binary quadratic forms of discriminant D (D > 0) when the Sylow -subgroup Research of the third author supported by Natural Sciences and Engineering Research Council of Canada Grant A-733 [15]
16 J B Muskat et al H ( D) of the form class group H( D) is cyclic of order k, where k, for a suitable predicting discriminant D In Section we prove Theorem 1, which characterizes those D for which H ( D) is isomorphic to a cyclic group of order k, distinguishing the three possibilities: (a) k 0, (b) k 1, (c) k In Section 3, for each of the twelve cases listed in Theorem 1(c), a suitable discriminant D is defined and predictive criteria are given to determine whether a prime represented by a form in the principal genus of discriminant D is in fact represented by a form which is a fourth power under composition; see Theorem In Section 4, for each of the fifty discriminants D (D > 0) for which H( D) is cyclic of order 4, a specific formulation of the appropriate predictive criterion from Theorem is presented An example in which a predictive criterion is applied successively to a sequence of discriminants is given in Section 5 We then exhibit sequences of discriminants for which the process of making successive predictions requires knowing the parameters in only one representation of p The applicability of Dirichlet s technique is not limited to the situation where the Sylow -subgroup of H( D) is cyclic of order 4 In Section 6 we present a case where the Sylow -subgroup has two cyclic factors each of order 4, and Dirichlet s technique produces a pair of predictive criteria which together determine which one of four cosets of the principal genus contains a form class representing the prime p The present research was motivated by the work of the first author in [15] The focus of this earlier work was restricted to representability of primes by forms of discriminant 4qr, where q is either 1 or a prime and r is a prime The concept of exclusive prediction, defined in [15, p 66], is illustrated by each of the fifty examples in Section 4 In these examples the predictive criterion distinguishes between representability by the principal form, which is the unique fourth power in the form class group, and the other (ambiguous) form in the principal genus, which is not a fourth power By contrast, in the example given earlier in this introduction, both the forms x + 6y and x + 31y are fourth powers under composition This is associated with the concept of inclusive prediction [15, p 66] Other approaches to predictive criteria can be found for example in [1] [14] Throughout this paper we use the following notation: Z m denotes the cyclic group of order m [a, b, c] denotes the form class containing the form ax + bxy + cy H( D) denotes the form class group of discriminant D (D > 0) under composition
Representation of primes by binary quadratic forms 17 h( D) H( D) denotes the form class number of discriminant D H ( D) denotes the Sylow -subgroup of H( D) r (H( D)) denotes the -rank of H( D), that is, the number k in the decomposition H ( D) Z a 1 Z a k, a i 1 r 4 (H( D)) denotes the 4-rank of H( D), that is, the number of factors Z a i in this decomposition having a i ( m ) n (sometimes written as (m/n)) denotes the Legendre Jacobi Kronecker symbol as defined in [10, eqns (5) ( and (7)] for arbitrary integers m, n with n 0 m ) p is the quartic residue symbol modulo a prime p 1 4 (mod 4) defined for an integer m satisfying ( ) m p 1 by ( m ) p ( ) n 4 p, where n m (mod p) v p (a) is the exponent of the largest power of the prime p dividing the nonzero integer a, symbolically, p vp(a) a Finally, we recall that a discriminant d is called a fundamental discriminant if d/f is not a discriminant for any integer f > 1 Characterization of the class groups H( D) whose -part is cyclic Let D be a positive integer 0 or 3 (mod 4) We set (1) D m p m 1 1 p m s s, where m is a nonnegative integer,,, p s are s ( 0) distinct odd primes, and m 1,, m s are positive integers As D 0 or 3 (mod 4) we have () m 0, in which case m 1 ( 1)/ + + m s (p s 1)/ 1 (mod ), or m In this section we characterize those class groups H( D) whose -part is (a) trivial, (b) (cyclic) of order, and (c) cyclic of order 4 Theorem 1 (a) H ( D) Z 1 if and only if (A) D m (m, 3, 4); (B) D p m 1 1 (m 1 (odd) 1, 3 (mod 4)); (C) D 4p m 1 1 (m 1 (odd) 1, 3 (mod 4)) (b) H ( D) Z if and only if (A) D m (m 5, 6); (B) D 4p m 1 1 (m 1 1, 5 (mod 8)); (C) D 4p m 1 1 (m 1 (even), 3 (mod 8)); (D) D 8p m 1 1 (m 1 1, 3, 5 (mod 8)); (E) D 16p m 1 1 (m 1 (odd) 1, 3 (mod 4));
18 J B Muskat et al (F) D p m 1 1 pm (m 1 (odd) 1, m (odd) 1, 3 (mod 4), 1 (mod 4), ( / ) 1); (G) D p m 1 1 pm (m 1 (odd) 1, m (even), 3 (mod 4), ( / ) 1); (H) D 4p m 1 1 pm (m 1 (odd) 1, m (odd) 1, 3 (mod 4), 1 (mod 4), ( / ) 1); (I) D 4p m 1 1 pm (m 1 (odd) 1, m (even), m 1, 3 (mod 4), ( / ) 1) (c) H ( D) Z k, for some k, if and only if (A) D m (m 7); (B) D 4p m 1 1 (m 1 1, 1 (mod 8)); (C) D 4p m 1 1 (m 1 (even), 7 (mod 8)); (D) D 8p m 1 1 (m 1 1, 1 (mod 8)); (E) D 8p m 1 1 (m 1 (odd) 1, 7 (mod 8)); (F) D 8p m 1 1 (m 1 (even), 7 (mod 8)); (G) D 16p m 1 1 (m 1 1, 1 (mod 4)); (H) D 16p m 1 1 (m 1 (even), 3 (mod 4)); (I) D p m 1 1 pm (m 1 (odd) 1, m 1, 3 (mod 4), 1 (mod 4), ( / ) 1); (J) D p m 1 1 pm (m 1 (odd) 1, m (even), 3 (mod 4), ( / ) 1); (K) D 4p m 1 1 pm (m 1 (odd) 1, m 1, 3 (mod 4), 1 (mod 4), ( / ) 1); (L) D 4p m 1 1 pm (m 1 (odd) 1, m (even), 3 (mod 4), ( / ) 1) In preparation for the proof of Theorem 1 we state some well-known results as Lemmas 1 3 Lemma 1 Let D 0, 3 (mod 4) be a positive integer Let t be the number of distinct odd primes dividing D Then t 1 if D 3 (mod 4) or D 1 (mod 16); r (H( D)) t if D 4, 8 (mod 16) or D 16 (mod 3); t + 1 if D 0 (mod 3) P r o o f See for example [8, Proposition 311] This result has its origins in the work of Gauss [11, 57 58] (3) (4) Corollary 1 Let D 0, 3 (mod 4) be a positive integer Then r (H( D)) 0 D m (m, 3, 4), or D p m 1 1 or 4p m 1 1 (p m 1 1 3 (mod 4)); r (H( D)) 1 D m (m 5), or
Representation of primes by binary quadratic forms 19 P r o o f We have by Lemma 1, D 4p m 1 1 (p m 1 1 1 (mod 4), m 1 1), or D 8p m 1 1 or 16p m 1 1 (m 1 1), or D p m 1 1 pm or 4p m 1 1 pm (p m 1 1 pm 3 (mod 4), m 1 1, m 1) r (H( D)) 0 (t 0) and (D 4, 8 (mod 16) or D 16 (mod 3)) and or (t 1) and (D 3 (mod 4) or D 1 (mod 16)), r (H( D)) 1 (t 0 and D 0 (mod 3)) or (t 1) and (D 4, 8 (mod 16) or D 16 (mod 3)) or from which (3) and (4) follow (t ) and (D 3 (mod 4) or D 1 (mod 16)), Lemma Let E (E > 0) be a fundamental discriminant Let p and q denote distinct odd primes Then (5) (6) v (h( E)) 0 E 4, 8 or E p, p 3 (mod 4); v (h( E)) 1 E 4p, p 5 (mod 8), or E 8p, p ±3 (mod 8), or E pq, p 1 (mod 4), q 3 (mod 4), ( ) p 1 q P r o o f This result can be found for example in [7, Corollaries (184) and (196)], or it can be deduced from [4, p 413 and Theorem 4(1)], [5, Theorems 1 and ], [6, pp 5, 6, 6] Lemma 3 Let E (E > 0) be a fundamental discriminant and let f be an integer > 1, so that D f E is a nonfundamental discriminant Let w denote the number of distinct odd primes which divide f but not E If w 1 we let q denote the unique odd prime factor of f which does not divide E If E 4 then (7) (8) v (h( D)) 0 if v (f) 1, w 0; v (h( D)) 1 if (i) v (f), w 0 or (ii) v (f) 0, w 1, q ±3 (mod 8);
0 J B Muskat et al (9) v (h( D)) If E 4 then (10) v (h( D)) 0 if otherwise (i) v (h( E)) 0, v (f) 0, w 0 or (ii) v (h( E)) 0, v (f) 1, v (E) 0, w 0; (11) v (h( D)) 1 if (i) v (h( E)) 1, v (f) 0, w 0 or (ii) v (h( E)) 1, v (f) 1, v (E) 0, w 0 or (iii) v (h( E)) 0, v (f) 1, v (E) 1, w 0 or (iv) v (h( E)) 0, v (f), v (E) 0, w 0 or (v) v (h( E)) 0, v (f) 0, w 1, (E/q) 1 or (vi) v (h( E)) 0, v (f) 1, v (E) 0, w 1, (E/q) 1; (1) v (h( D)) otherwise P r o o f Gauss [11, 54 56] proved (see for example [6, p 17]) h( D) fh( E) ( 1 ( E/p) )/ u, p p f where p runs through the distinct primes dividing f and { 3 if E 3, u if E 4, 1 if E > 4 As f p f pvp(f) we can express h( D) in the form h( D) h( E) p / v p(f) 1 (p ( E/p)) u p f p f that is, h( E) p f (13) h( D) h( E) p f p E p v p(f) 1 p f p E p v p(f) p f p E p p f p E / (p ( E/p)) u, / p vp(f) 1 (p ( E/p)) u If E 4 (in which case h( E) 1 and u ) we have from (13), v (h( D)) v (f) + v (R) 1,
where Representation of primes by binary quadratic forms 1 R p f, p (p ( 1/p)) As { 0 (mod 8) if p ±1 (mod 8), p ( 1/p) 4 (mod 8) if p ±3 (mod 8), we deduce 0 if w 0 (this implies v (f) 1), v (R) if w 1, q ±3 (mod 8), 3 if w 1, q ±1 (mod 8) or w Hence we have v (f) 1 if w 0, v (h( D)) v (f) + 1 if w 1, q ±3 (mod 8), v (f) + if w 1, q ±1 (mod 8) or w This completes the proof of the lemma in the case E 4 If E 4 we have from (13), v (h( D)) v (h( E)) + (v (f) λ) + v (p ( E/p)), where λ { 1 if f, E, 0 otherwise p f, p E If f, E then v ( ( E/)) 0 so that v (p ( E/p)) v (p ( E/p)) p f, p E For p an odd prime we have so that p f, p E p p ( E/p) p f, p E p { 0 (mod 4) if (E/p) 1, (mod 4) if (E/p) 1, { 0 if w 0, v (p ( E/p)) 1 if w 1, (E/q) 1, otherwise Hence we have v (h( D)) 0 if v (h( E)) v (f) λ w 0, v (h( D)) 1 v (h( D)) if v (h( E)) 1, v (f) λ w 0, or v (h( E)) 0, v (f) λ 1, w 0, or v (h( E)) v (f) λ 0, w 1, (E/q) 1, otherwise
J B Muskat et al As v (f) λ 0 v (f) 0 or v (f) 1, v (E) 0, v (f) λ 1 v (f) 1, v (E) 1 or v (f), v (E) 0, we obtain the assertion of the lemma when E 4 P r o o f o f T h e o r e m 1 We have H ( D) Z 1 r (H( D)) 0 D m (m, 3, 4), or D p m 1 1 or 4p m 1 1 (m 1 (odd) 1, (prime) 3 (mod 4)), by Corollary 1, completing case (a) In order to complete the proof we must classify all discriminants D with r (H( D)) 1 according as H ( D) Z (equivalently v (h( D)) 1) or H ( D) Z k (k ) (equivalently v (h( D)) ) We examine each of the cases in the second part of Corollary 1, which we refine further for convenience Cases (b), (c) below refer to the sections of the statement of Theorem 1 An asterisk (*) indicates the first part of a case, and a double asterisk (**) indicates the second and final part of the case D m, m (even) 6: E 4, f m/ 1 v (f) m/ 1, w 0 v (h( D)) 1 if m 6, by (8)(i) v (h( D)) if m 8, by (9) D m, m (odd) 5: E 8, f (m 3)/ v (h( E)) 0, v (f) (m 3)/, v (E) 3, w 0 v (h( D)) 1 if m 5, by (11)(iii) v (h( D)) if m 7, by (1) D 4p m 1 1, m 1 (odd) 1, 1 (mod 4): case (b)(a) case (c)(a) case (b)(a) case (c)(a) E 4, f p (m 1 1)/ 1 v (f) 0, w 0 v (h( E)) 1 or according as 5 (mod 8) or 1 (mod 8) v (h( D)) 1 if 5 (mod 8), by (11)(i) case (b)(b) v (h( D)) if 1 (mod 8), by (1) case (c)(b) D 4p m 1 1, m 1 (even) : E 4, f p m 1/ 1 v (f) 0, w 1
Representation of primes by binary quadratic forms 3 v (h( D)) 1 if ±3 (mod 8), by (8)(ii) v (h( D)) if ±1 (mod 8), by (9) D 8p m 1 1, m 1 (even) : E 8, f p m 1/ 1 v (h( E)) 0, v (f) 0, w 1, q v (h( D)) 1 if ±3 (mod 8), by (11)(v) v (h( D)) if ±1 (mod 8), by (1) D 8p m 1 1, m 1 (odd) 1: case (b)(b), (C) case (c)(b), (C) case (b)(d) case (c)(d), (F) E 8, f p (m 1 1)/ 1 v (f) 0, w 0 v (h( E)) 1 or according as ±3 (mod 8) or ±1 (mod 8) v (h( D)) 1 if ±3 (mod 8), by (11) (i) case (b)(d) v (h( D)) if ±1 (mod 8), by (1) case (c)(d), (E) D 16p m 1 1, m 1 (odd) 1, 3 (mod 4): E, f 4p (m 1 1)/ 1 v (h( E)) 0, v (f), v (E) 0, w 0 v (h( D)) 1, by (11)(iv) D 16p m 1 1, m 1 (even) : E 4, f p m 1/ 1 v (f) 1, w 1 v (h( D)), by (9) D 16p m 1 1, m 1 (odd) 1, 1 (mod 4): E 4, f p (m 1 1)/ 1 v (h( E)) 1, v (f) 1, v (E), w 0 v (h( D)), by (1) D p m 1 1 pm, m 1 (odd) 1, m (odd) 1, 3 (mod 4), 1 (mod 4): E, f p (m 1 1)/ 1 p (m 1)/ { } 1, (p1 /p v (h( E)) ) 1, v, ( / ) 1 (f) 0, w 0 v (h( D)) 1 if ( / ) 1, by (11)(i) v (h( D)) if ( / ) 1, by (1) D p m 1 1 pm, m 1 (odd) 1, m (even), 3 (mod 4): E, f p (m 1 1)/ 1 p m / v (h( E)) 0, v (f) 0, w 1, q case (b)(e) case (c)(g), (H) case (c)(g) case (b)(f) case (c)(i)
4 J B Muskat et al v (h( D)) 1 if ( / ) 1, by (11)(v) v (h( D)) if ( / ) 1, by (1) D 4p m 1 1 pm, m 1 (odd) 1, m (odd) 1, 3 (mod 4), 1 (mod 4): case (b)(g) case (c)(i), (J) E, f p (m 1 1)/ 1 p (m 1)/ { } 1, (p1 /p v (h( E)) ) 1, v, ( / ) 1 (f) 1, v (E) 0, w 0 v (h( D)) 1 if ( / ) 1, by (11)(ii) case (b)(h) v (h( D)) if ( / ) 1, by (1) case (c)(k) D 4p m 1 1 pm, m 1 (odd) 1, m (even), 3 (mod 4): E, f p (m 1 1)/ 1 p m / v (h( E)) 0, v (f) 1, v (E) 0, w 1, q v (h( D)) 1 if ( / ) 1, by (11)(vi) v (h( D)) if ( / ) 1, by (1) case (b)(i) case (c)(k), (L) 3 Determination of predictive criteria Throughout this section D is a positive integer 0 or 3 (mod 4) such that H ( D) Z k for some integer k Thus D is one of the twelve types (A), (B),, (L) specified in Theorem 1(c) For the discriminant D there are two generic characters χ 1 and χ as specified below: χ 1 (r) ( 1/r), χ (r) (/r), case (A), χ 1 (r) ( 1/r), χ (r) (r/ ), cases (B), (C), (G), (H), χ 1 (r) ( /r), χ (r) (r/ ), cases (D), (F), χ 1 (r) (/r), χ (r) (r/ ), case (E), χ 1 (r) (r/ ), χ (r) (r/ ), cases (I), (J), (K), (L) Define the positive integer D as follows: D/, cases (A), (E), D/, cases (B), (C), (D), (G), (H), D D/, case (F), D/, cases (I), (K), D/, cases (J), (L) It is easily verified that D is formed by dividing D by the unique squarefree integer such that D is a discriminant whose generic characters are all included among the generic characters for discriminant D With χ 1 and χ as specified above, the generic characters for the discriminant D are as follows:
Representation of primes by binary quadratic forms 5 (A) χ 1, χ, (H) χ 1, χ, (B) χ 1 (if m 1 1), (I) χ 1 (if m 1), χ 1, χ (if m 1 ), χ 1, χ (if m ), (C) χ, (J) χ (if m 1 1), (D) χ 1 (if m 1 1), χ 1, χ (if m 1 ), χ 1, χ (if m 1 ), (K) χ 1 (if m 1), (E) χ, χ 1, χ (if m ), (F) χ, (L) χ (if m 1 1), (G) χ 1 (if m 1 1), χ 1, χ (if m 1 ) χ 1, χ (if m 1 ), Let p be an odd prime such that χ 1 (p) χ (p) 1, so that p is represented by a form class C p (and its inverse Cp 1 ) in the principal genus of the form class group H( D) Then, by a well-known theorem of Gauss, C p is the square of a form class in H( D), say, C p Sp Let K be a positive integer coprime with Dp which is represented primitively by of H( D) Then K p is represented primitively by the class (Sp 1 ) C p Sp S I principal class of H( D) Hence there exist integers A and B such that { A K p + (D/4)B if D 0 (mod 4), (31) A + AB + ((D + 1)/4)B if D 3 (mod 4), K > 0, (A, B) 1, (K, Dp) 1 the form class S 1 p Further, as the generic characters for the discriminant D are a subset of those for discriminant D, by a similar argument, there exist integers H, M, N such that { M H p + (D /4)N if D 0 (mod 4), (3) M + MN + ((D + 1)/4)N if D 3 (mod 4), H > 0, (M, N) 1, (H, Dp) 1 As H (resp K) is represented primitively by a form class of discriminant D (resp D) and (H, D) (K, D) 1, we have ( ) ( ) D D (33) 1 H K Our purpose is to determine a necessary and sufficient condition for the class C p to be a fourth power in H( D) for each of the cases (A) (L) We do this by extending the techniques employed by the first author in [15], who developed an idea of Dirichlet [9, ] Our results are formulated in terms of
6 J B Muskat et al arithmetic conditions involving the integers M and N in the representation (3) We prove Theorem With the above notation the following are necessary and sufficient conditions for the form class C p to be a fourth power in the form class group H( D), where H ( D) Z k for some k C a s e (A): ( ) m 7 1, M + 4N ( ) m 8 1 M C a s e (B): ( ) M + wn m 1 1 1, where w 1 (mod ), ( ) M m 1 1 C a s e (C): ( 1) (M 1+N)/ ( M ) 1 C a s e (D): ( ) M + wn m 1 1 1, where w (mod ), ( ) M m 1 1 C a s e (E): ( 1) N(+1)/8 C a s e (F): ( 1) N(+1)/8 ( ) 1 M + N ( M + N )( ) M 1 C a s e (G): ( ) M + wn m 1 1 1, where w 4 (mod ), ( ) M m 1 1
C a s e (H): ( )( ) 1 M 1 M + N Representation of primes by binary quadratic forms 7 C a s e (I): ( ) M + wn m 1 1, where w w + 1 p 4 (1 + pm 1 1 ) 0 (mod ), ( ) 4M + N m 1 C a s e (J): ( )( ) M + wn 4M + N m 1 1 1, m 1 3 where w w + 1 4 (1 + pm 1 ) 0 (mod ), ( ) 4M + N 1 C a s e (K): ( ) M + wn m 1 ( ) M m 1 C a s e (L): ( )( ) M + wn M m 1 1 ( ) M m 1 3 1 then 1, where w p m 1 1 (mod ), 1, where w p m 1 (mod ), Before proving Theorem we state and prove some lemmas Lemma 31 If x, y, z, m are integers with m 0 such that P r o o f We have x y + z (mod m) (x + y)(x + z) (x + y + z) (mod m) (x + y)(x + z) x + xy + yz + zx x + (x + y + z) (x + y + z ) (x y z ) + (x + y + z) (x + y + z) (mod m)
8 J B Muskat et al Not all parts of the next lemma will be used, but they are given for completeness Lemma 3 Let X, Y be nonzero integers and m an integer (mod 4), 3 (mod 4) or 5 (mod 8) Suppose that a X my, a 1 (i) If m (mod 4) then { a/ X, a/ Y if a is even, (a+1)/ X, (a 1)/ Y if a is odd (ii) If m 3 (mod 4) then { a/ X, a/+1 Y or a/+1 X, a/ Y if a is even, (a 1)/ X, (a 1)/ Y if a is odd (iii) If m 5 (mod 8) then a is even, and a/ 1 X, a/ 1 Y, or a/ X, a/+1 Y, or a/+1 X, a/ Y P r o o f (i) As the exponents of in X and my are even and odd, respectively, we must have a X, a+1 my if a is even, a+1 X, a my if a is odd (ii) If v (X) v (Y ) then as m 1 (mod 4), we have v(x)+1 X my, so that a v (X) + 1 If v (X) < v (Y ) then a v (X) If v (X) > v (Y ) then a v (Y ) (iii) If v (X) v (Y ) then, as X Y v(x) (mod v(x)+3 ), we have X my v (X)+ (mod v(x)+3 ), so that a v (X) + If v (X) < v (Y ) then a v (X) If v (X) > v (Y ) then a v (Y ) Part of the next lemma is used in the proof of Lemma 35 below Lemma 33 Let m, X, Y be nonzero integers Suppose that q is an odd prime such that ( ) m q a X my, a 0, 1 q Then a is even and q a/ X, q a/ Y if v q (X) v q (Y ), q a/ X, q a/+1 Y if v q (X) < v q (Y ), q a/+1 X, q a/ Y if v q (X) > v q (Y )
Representation of primes by binary quadratic forms 9 P r o o f If v q (X) v q (Y ) k (say), then X q k X 1, Y q k Y 1, where q X 1 Y 1 Thus q a k X1 my1 If a > k then (X 1 Y1 1 ) m (mod q), contradicting (m/q) 1 Thus a k If v q (X) < v q (Y ) then a v q (X) If v q (X) > v q (Y ) then a v q (Y ) Lemma 34 Let q be an odd prime, a a positive integer, and m, X, Y nonzero integers satisfying Then q a X mqy, q m { q a/ X, q a/ Y if a is even, q (a+1)/ X, q (a 1)/ Y if a is odd P r o o f As the exponents of q in X and qy are even and odd, respectively, we must have q a X, q a+1 qy q a+1 X, q a qy if a is even, if a is odd Lemma 35 Suppose p is a prime and r, s, H, K, M, N, A, B are nonzero integers such that If q is an odd prime satisfying then H p M + rn, (M, N) 1 or, K p A + rsb, (A, B) 1 or q r, ( ) s 1, q (i) q does not divide both of HA ± KM, (ii) both of v q (HA ± KM) are even P r o o f (i) Suppose on the contrary that q divides both HA + KM and HA KM Then, as q is odd, we have q (HA, KM) Thus one of the following must occur: (a) q H, q K, (b) q H, q M, (c) q A, q K, (d) q A, q M C a s e (a) We have M rn (mod q), A rsb (mod q), q M, q N, q A, q B, so that ( r ) ( ) rs 1, 1, q q and thus (s/q) 1, contradicting (s/q) 1
30 J B Muskat et al C a s e (b) From H p M + rn we have q rn, which is impossible as q r and (M, N) 1 or C a s e (c) From K p A +rsb we have q rsb, which is impossible as q r, q s, (A, B) 1 or C a s e (d) We have H p rn (mod q), K p rsb (mod q), q N, q H, q p, q B, q K, so that ( ) ( ) pr prs 1, 1, q q and thus (s/q) 1, contradicting (s/q) 1 (ii) By (i), v q (HA + KM) 0 or v q (HA KM) 0 If both are zero we are finished Otherwise, without loss of generality, we may assume v q (HA+KM) a > 0, so that v q (HA KM) 0 Thus q a H A K M As H A K M H (K p rsb ) K (H p rn ) r(k N sh B ), and q r, we deduce q a K N sh B Then, by Lemma 33, as (s/q) 1, we conclude that a is even Lemma 36 Let p be a prime and A, B, H, K, M, N, r, s nonzero integers such that Let q be an odd prime such that H p M + rn, K p A + sb q p, q HK, q rs Then q does not divide both of HA ± KM P r o o f Suppose q is an odd prime such that q p, q HK, q rs, q (HA + KM, HA KM) Clearly by interchanging the roles of H, M, N, r and K, A, B, s respectively, if necessary, we may suppose that q r Now q (HA + KM) (HA KM) KM so that as q, q K, we have q M Then from H p M + rn we see that q H p, which is impossible as q H and q p Lemma 37 Let D, D be positive integers with D D 3 (mod 4), D D, and D/D 5 (mod 8) Suppose that B, C, H, K, L, N are nonzero integers such that H and K are odd and (34) H C K L D (K N (D/D )H B ), (35) B C (mod ), (B, C) 1 or, (36) L N (mod ), (L, N) 1 or Define the nonnegative integers r and s by
Representation of primes by binary quadratic forms 31 (37) r HC + KL, s HC KL Then r and s are both even P r o o f From (34) and (37) we see that r+s K N (D/D )H B By Lemma 3(iii), as D/D 5 (mod 8), r + s is even We assume r and s are both odd and obtain a contradiction Replacing K by K, if necessary, we may suppose that r s We consider two cases: (i) r > s, (ii) r s C a s e (i): r > s From (37) we see that s (HC +KL)±(HC KL), so that s 1 HC, s 1 KL But H and K are both odd, so s 1 C, s 1 L If s 3 then C, L and so, by (35) and (36), we have B, N Thus 4 K N (D/D )H B and so, by (34), 4 H C K L, that is, r+s 4, contradicting r > s 3 If s 1 (so that r 3) C and L are odd so that, by (35) and (36), B and N are odd Thus K N (D/D )H B, and so, by (34), H C K L, that is, r + s, a contradiction C a s e (ii): r s From (37) we have r HC ± KL so that, as K, H are both odd, r C, r L If r 3, then by (35) and (36), we see that B, N Thus 4 K N (D/D )H B, 4 H C K L, r + s 4, r s, a contradiction If r 1 then either L, C or L, C If L, C holds then, by (36), we have N Then 3 D K N H C, 4 K L, 3 (D K N H C ) + K L DH B (by (34)), so that 3 B, which is impossible If L, C holds then, by (35), B Then 3 DH B K L, 4 H C, 3 (DH B K L )+H C D K N, so that 3 N, which is impossible Lemma 38 Let E, E be positive integers with E E 3 (mod 4), E E, and E/E 5 (mod 8) Suppose that A, B, H, K, M, N are nonzero integers such that H and K are odd and (38) H A K M E (K N (E/E )H B ), (39) A B (mod ), (A, B) 1, (310) M N (mod ), (M, N) 1 Define the nonnegative integers r and s by (311) r HA + KM, s HA KM Then r and s are both odd P r o o f From (38) and (311) we see that r+s K N (E/E )H B By Lemma 3(iii), r + s is even We assume r and s are both even and obtain a contradiction Replacing K by K, if necessary, we may suppose that r s We consider two cases: (i) r > s, (ii) r s C a s e (i): r > s If s 0 then HA KM is odd and thus HA + KM is odd, so that r 0, contradicting r > s Thus s From (311) we see that s (HA + KM) ± (HA KM), so that, as H and K are odd, s 1 A,
3 J B Muskat et al s 1 M Thus B and N are odd and so K N (E/E )H B, as E/E 5 (mod 8), that is, H A K M But s 1 A, s 1 M so (s 1)+3 H A K M, giving (s 1) + 3, s 1, a contradiction C a s e (ii): r s If r s 0 then HA ± KM are odd, so either A odd, M even or A even, M odd If A odd, M even, then B even, N odd, so H A E K N, K M EH B, contradicting (38) If A even, M odd, then B odd, N even, so H A E K N, K M EH B, contradicting (38) Thus r s (even) We have r HA+KM, r HA KM, so that r H A K M Furthermore, r A, r M By (39) and (310) both B and N are odd Thus K N (E/E )H B, so that H A K M, contradicting r We are now ready to prove Theorem P r o o f o f T h e o r e m We consider each of the 1 cases (A), (B),, (L) separately C a s e (A): D m, D m 1, m 7 Let p be an odd prime such that ( 1/p) (/p) 1, that is, p 1 (mod 8) From (3) and (31) we have (3A1) (3A) H p M + m 3 N, H > 0, (M, N) 1, (H, p) 1, K p A + m B, K > 0, (A, B) 1, (K, p) 1 By (33) we have ( D/K) ( m /K) 1 so that ( ) ( ) m 1 (3A3) K K Eliminating p from (3A1) and (3A) yields (3A4) (HA + KM)(HA KM) m 3 (K N H B ) As H, A, K, M are odd, exactly one of HA ± KM is divisible by but not by 4 We choose the sign of A so that HA +KM Then, from (3A4), we see that m 4 HA KM We factor (HA + KM)/ into primes as follows: (3A5) (HA + KM)/ ε (/q i )1 q e i i (/r j ) 1 r f j j, where ε ±1, e i, f j are positive integers, and q i, r j are distinct odd primes By Lemma 35(ii) with s, each f j in (3A5) is even We conclude that (3A6) (HA + KM)/ ±1 (mod 8) Next, by (3A4) and Lemma 31 (with m 5 ), we have 1 (HA + KM)(KM + 4KN) ( 1 (HA + KM) + KN) (mod 8) if m 7, 1 (HA + KM)(KM) ( 1 (HA + KM)) (mod 8) if m 8,
Representation of primes by binary quadratic forms 33 so that appealing to (3A6) we have { K(M + 4N) ±1 (mod 8) if m 7, KM ±1 (mod 8) if m 8, and thus (3A7) ( ) ( ) M + 4N ( ) K M if m 7, if m 8 Finally, recalling that K is odd and represented by the form class S 1 p of discriminant D, we have C p is a fourth power in H( D) S p is a square in H( D) (as H ( D) Z k, k ) Sp 1 is a square in H( D) ( ) ( ) 1 1 K K ( ) 1 (by (3A3)) K ( ) 1 if m 7, M + 4N ( ) 1 if m 8 (by (3A7)) M C a s e (B): D 4p m 1 1, D 4p m 1 1 1, m 1 1, 1 (mod 8) Let p be an odd prime satisfying ( 1/p) (p/ ) 1, so that p 1 (mod 4) and p From (3) and (31) we have (3B1) H p M + p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, (3B) K p A + p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1 By (33) we have ( D/K) ( 4p m 1 1 /K) 1, so that by the law of quadratic reciprocity ( ) ( ) m1 ( ) m1 1 p1 K (3B3) K K From (3B1) and (3B), we obtain (3B4) (HA + KM)(HA KM) p m 1 1 1 (K N H B ) By Lemma 36, does not divide both of HA ± KM Choose the sign of A so that HA + KM Define odd integers R and S by (3B5) R (HA + KM)/ r, S (HA KM)/ s p α 1,
34 J B Muskat et al where p α 1 HA KM Note that RS and, by (3B4), α m 1 1 We factor R into primes as follows: (3B6) R ε q e i i r f j j, ( /q i )1 ( /r j ) 1 where ε ±1, e i, f j are positive integers, and q i, r j are distinct odd primes By Lemma 35(ii) each f j is even Then from (3B6), by the law of quadratic reciprocity, we obtain ( ) ( ) R ε ( ) ei qi ( ) fj rj (3B7) ( /q i )1 ( /q i )1 ( p1 q i ) ei 1 ( /r j ) 1 By Lemma 31 and (3B4) we have (HA + KM)(KM + wkn) (3B8) (HA + KM + wkn) (mod ) if m 1 1, (HA + KM)KM (HA + KM) (mod ) if m 1, where w 1 (mod ) From (3B8) we see that ( )( )( )( ) HA + KM K M + wn 1 if m 1 1, p ( )( )( )( ) 1 HA + KM K M 1 if m 1 Thus, as (/ ) 1 and (HA + KM/ ) ( r R/ ) 1 (by (3B5) and (3B7)), we obtain ( ) M + wn ( ) K if m 1 1, (3B9) ( ) M if m 1 Finally, C p is a fourth power in H( D) ( ) ( ) 1 K 1 K ( ) K 1 (by (3B3))
Representation of primes by binary quadratic forms 35 ( ) M + wn 1 ( ) if m 1 1, where w 1 (mod ), M 1 if m 1 (by (3B9)) We remark that when m 1 1 it is possible for to divide M so that the symbol (M/ ) cannot be used for the criterion in this case We note that the value of the Legendre symbol (M + wn/ ) is independent of the choice of solution ±w of w 1 (mod ), as ( )( ) ( M + wn M wn M w N ) ( M + N ) ( H ) p 1 C a s e (C): D 4p m 1 1, D 4p m 1 1 1, m 1 (even), 7 (mod 8) Let p be an odd prime satisfying ( 1/p) (p/ ) 1, so that p 1 (mod 4) and p From (3) and (31) we have (3C1) (3C) H p M + p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, K p A + p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1 From (33) we have ( D/K) ( 4p m 1 1 /K) 1 so that ( ) 1 (3C3) 1, K and hence (3C4) K 1 (mod 4) Reducing (3C1) and (3C) modulo 8, we obtain A (p 1)/ (mod 4), B 1 (mod ) or (3C5) A 1 (mod ), B (p 1)/ (mod 4), M 1 (mod ), N (p 1)/ (mod 4) Replacing M by M if necessary, we may suppose that (3C6) M 1 (mod 4) From (3C1) and (3C) we obtain (3C7) (HA + KM)(HA KM) p m 1 1 1 (K N H B ) By Lemma 36, does not divide both of HA ± KM Choose the sign of A so that HA + KM, and define the odd integers R and S by (3C8) R (HA + KM)/ r, S (HA KM)/ s p α 1,
36 J B Muskat et al where p α 1 HA KM Clearly RS Note that { α m1 1, (3C9) r s 0 if A is even, min(r, s) 1 if A is odd We show that when A is odd { p 1 (mod 8) if r + s 4, (3C10) p 5 (mod 8) if r + s 3 This is clear from (3C5) as p 1 (mod 8) B N 0 (mod 4) 4 K N H B 4 (HA + KM)(HA KM) r + s 4, p 5 (mod 8) B N (mod 4) (N/) + (B/) Next we show that (3C11) K (N/) H (B/) 3 K N H B 3 (HA + KM)(HA KM) r + s 3 ( )( ) R S ( 1) α+1 From (3C7) and (3C8) we see that p α 1 H A K M p m 1 1 1 (K N H B ), so that p α m 1+1 1 K N H B Then, by Lemma 34, we observe that { p (α m 1+1)/ 1 N, p (α m 1+1)/ 1 B if α is odd, p (α m 1+)/ 1 N, p (α m 1)/ 1 B if α is even Define integers N 1 and B 1 by { N p (α m 1+1)/ 1 N 1, B p (α m 1+1)/ 1 B 1 if α is odd, N p (α m 1+)/ 1 N 1, B p (α m 1)/ 1 B 1 if α is even, where N 1 (α odd) and B (α even) Hence r+s RS (H A K M )/p α 1 (K N H B )/p α m 1+1 1 { K N1 H B1 if α is odd, K N1 H B1 if α is even, so that, as (/ ) 1, ( K ( ) RS N ) 1 1 if α is odd, p ( 1 H B1 ) 1 if α is even, completing the proof of (3C11)
Representation of primes by binary quadratic forms 37 Now factor R and S into primes: R ε (p (3C1) 1 /q i )1 S ε ( /q i )1 q e i i q g i i ( /r j ) 1 ( /r j ) 1 r f j j, r h j j, where ε, ε ±1, e i, f j, g i, h j are nonnegative integers, and q i, r j are distinct odd primes By Lemma 35(ii) for each j both f j and h j are even Hence, appealing to the law of quadratic reciprocity, we have ( ) ( ) R ε ( ) ei qi ( ) fj rj that is (3C13) ε ( /q i )1 ( /q i )1 (( 1 q i ( /r j ) 1 )( p1 q i )) ei, ( ) R ε( 1) E, where E ( /q i )1 q i 3 (mod 4) Next, taking the first equation in (3C1) modulo 4, we have R ε ( 1) e i ε( 1) E (mod 4), so that ( /q i )1 q i 3 (mod 4) (3C14) ε( 1) E ( 1) (R 1)/ Thus, from (3C13) and (3C14), we obtain ( ) R (3C15) ( 1) (R 1)/ Similarly we derive (3C16) ( ) S ( 1) (S 1)/ The next step is to show that ( ) R (3C17) ( 1) N/ We consider three cases e i
38 J B Muskat et al C a s e (i): A even We have R 1 HA + KM 1 A N (mod 4) (by (3C8) and (3C9)) (by (3C1), (3C4) and (3C6)) (by (3C5)), so that by (3C15) we obtain ( ) R ( 1) (R 1)/ ( 1) N/ C a s e (ii): A odd, r 1 We have R 1 1 (HA + KM) 1 KM + s 1 p α 1 S 1 (by (3C8)) s 1 p α 1 S { 0 N (mod 4) if p 1 (mod 8) N (mod 4) if p 5 (mod 8) so that by (3C15) we obtain ( ) R ( 1) (R 1)/ ( 1) N/ C a s e (iii): A odd, s 1 We have (by (3C4) and (3C6)) (by (3C10) and (3C5)), S (HA KM)/p α 1 ( r 1 R KM)/p α 1 (by (3C8)) ( 1) α ( r 1 R 1) (by (3C4) and (3C6)) { ( 1) α+1 if p 1 (mod 8) ( 1) α (by (3C10)) if p 5 (mod 8) that is α + (p + 5)/ (mod 4), (S 1)/ α + (p + 3)/4 (mod ), so that, by (3C5), (3C11) and (3C16), we obtain ( ) ( ) R ( 1) α+1 ( 1) Sp1 α+1+(s 1)/ ( 1) α+1+α+(p+3)/4 ( 1) (p 1)/4 ( 1) N/ This completes the proof of (3C17) Writing (3C7) in the form (HA + KM) (HA + KM)KM p m 1 1 1 (K N H B ), and appealing to (3C8), we see that r R r+1 RKM 0 (mod ),
so that Representation of primes by binary quadratic forms 39 ( ) ( RKM r+1 ) ( RKM r ) R 1, which implies by (3C17), ( ) ( )( K R M (3C18) Finally, ) ( 1) N/ ( Mp1 ) C p is a fourth power in H( D) ( ) ( ) 1 K 1 K ( ) K 1 (by (3C3)) ( 1) N/ ( M ) 1 (by (3C18)) We have shown that if M 1 (mod 4) then (3C19) C p is a fourth power in H( D) ( 1) N/ ( M ) 1 Clearly, if M 3 (mod 4), (3C19) becomes (as ( M/ ) (M/ )) (3C0) C p is a fourth power in H( D) ( 1) N/ ( M ) 1 Putting (3C19) and (3C0) together, we obtain C p is a fourth power in H( D) without any restriction on M ( ) M ( 1) (M 1+N)/ 1 ( ) ( 1) N/ p1 1 M C a s e (D): D 8p m 1 1, D 8p m 1 1 1, m 1 1, 1 (mod 8) Let p be an odd prime such that ( /p) (p/ ) 1, so that p 1, 3 (mod 8) and p From (3) and (31) we have (3D1) (3D) H p M + p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, K p A + p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1
40 J B Muskat et al By (33) we have ( D/K) ( 8p m 1 1 /K) 1, so that, by the law of quadratic reciprocity, we have ( ) ( ) m1 K (3D3) K Next, from (3D1) and (3D), we obtain (3D4) (HA + KM)(HA KM) p m 1 1 1 (K N H B ) The rest of the proof proceeds almost exactly as the proof of Case (B), immediately following (3B4), except that in the equivalent to (3B8) w is chosen so that w (mod ), and at the end we use ( /K) C a s e (E): D 8p m 1 1, D 4p m 1 1, m 1 (odd) 1, 7 (mod 8) Let p be an odd prime such that (/p) (p/ ) 1, so that p 1, 7 (mod 8) and p From (3) and (31) we have (3E1) H p M + p m 1 1 N, H > 0, (M, N) 1, (H, p ) 1, (3E) K p A + p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1 From (33) we have ( D/K) ( 8p m 1 1 /K) 1 so that (appealing to the law of quadratic reciprocity) ( ) ( ) ( ) p1 K (3E3) K K Reducing (3E1) and (3E) modulo 8, we see that A is odd and B 0 (mod ), M 1 (mod ), N 0 (mod 4) if p 1 (mod 8), (3E4) B 1 (mod ), M 0 (mod 4), N 1 (mod ) if p 7 (mod 8) From (3E1) and (3E) we obtain (3E5) (HA + KM)(HA KM) p m 1 1 (K N H B ) If p 1 (mod 8) exactly one of HA±KM is congruent to (mod 4); choose the sign of A so that HA + KM Thus α HA KM for some integer α If p 7 (mod 8) both of HA ± KM are odd We define the odd integers R and S by { R (HA + KM)/, S (HA KM)/ α if p 1 (mod 8), (3E6) R HA + KM, S HA KM if p 7 (mod 8) Factor R into primes: (3E7) R εp r 1 (/q i )1 q e i i (/r j ) 1 r f j j, where ε ±1, r 0, e i, f j are positive integers, and q i, r j are distinct odd primes all different from By Lemma 35(ii) each f j is even Hence, from
Representation of primes by binary quadratic forms 41 (3E7) modulo 8, we obtain R ±1 (mod 8) so that by (3E6), { HA + KM δ (mod 16) if p 1 (mod 8), (3E8) HA + KM δ (mod 8) if p 7 (mod 8), where δ ±1 We show next that ( ) ( (3E9) K M + N ) ( 1) N(+1)/8 by adapting the method used in the proof of [15, Theorem ] but without constraining the odd integers among A, B, H, K, M and N to be congruent to 1 (mod 4) We treat the case p 1 (mod 8) first By (3E1), (3E) and (3E4) we may express A 4a + ε A, B b, H 4h + ε H, K 4k + ε K, M 4m + ε M, N 4n, where ε A, ε H, ε K, ε M take only the values ±1 Note that ε A 1, ε A 1 (mod ), a a (mod ), A 8a + 1 (mod 16), and similarly for the others Then, from (3E8) we obtain (3E10) 4(ε H a + ε A h + ε M k + ε K m) + (ε A ε H + ε K ε M ) δ (mod 16) Taking (3E10) modulo 4, we see that (3E11) ε A ε H ε K ε M θ (say) Then, taking (3E10) modulo 8 and dividing by, we obtain (as 4ε H a 4a (mod 8), ) δ θ + (a + h + k + m) (mod 4), so that (3E1) δ θ( 1) a+h+k+m Thus, from (3E10), (3E11) and (3E1), we have ε H a + ε A h + ε M k + ε K m θ(( 1) a+h+k+m 1)/ (mod 4), so that, as ε H θε A, ε A θε H, ε M θε K, ε K θε M, we have (3E13) ε A a + ε H h + ε K k + ε M m (( 1) a+h+k+m 1)/ (mod 4) Next we take (3E5) modulo 3 and divide by 8 to obtain (3E14) (a + h + k + m) + (ε A a + ε H h ε K k ε M m) n + b (mod 4) Taking (3E14) modulo, we deduce (3E15) a + h + k + m b (mod ) Then, from (3E13) (3E15), we obtain b + (k + m) n + b (( 1) b 1)/ (mod 4)
4 J B Muskat et al But b b (( 1) b 1)/ (mod 4) for any integer b so that k + m n (mod ) Thus ( ) ( ) ( 1) (K 1)/8 ( 1) k ( 1) m+n, K M + N which proves (3E9) in this case as N is even Now we treat the case p 7 (mod 8) By (3E4) we may express A 4a ± 1, B 4b ± 1, H 4h ± 1, K 4k ± 1, M 4m, N 4n ± 1, so that A 8a + 1 (mod 16), From (3E8) modulo 8 we obtain (3E16) a + h + m 0 (mod ) Next, from (3E5) modulo 16, we deduce (3E17) k + n a + h + ( + 1)/8 (mod ) Eliminating a + h from (3E16) and (3E17) yields k m + n + ( + 1)/8 (mod ) Thus ( ) ( ( 1) k ( 1) m+n+(p1+1)/8 K which proves (3E9) in this case as N is odd Finally, M + N ) ( 1) (+1)/8, C p is a fourth power in H( D) ( ) ( ) K 1 K ( ) 1 (by (3E3)) K ( ) ( 1) N(p1+1)/8 1 (by (3E9)) M + N C a s e (F): D 8p m 1 1, D 4p m 1 1 1, m 1 (even), 7 (mod 8) Let p be an odd prime such that ( /p) (p/ ) 1, so that p 1, 3 (mod 8) and p From (3) and (31) we have (3F1) (3F) H p M + p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, K p A + p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1 From (33) we have ( D/K) ( 8p m 1 1 /K) 1, that is, ( ) (3F3) 1, K 1, 3 (mod 8) K
Representation of primes by binary quadratic forms 43 Reducing (3F1) and (3F) modulo 8, we obtain (3F4) A 1 (mod ), and (3F5) B 0 (mod ), M 1 (mod ), N 0 (mod 4) if p 1 (mod 8), B 1 (mod ), M (mod 4), N 1 (mod ) if p 3 (mod 8) Replacing (M, N) by ( M, N) if necessary, we may suppose that (3F6) M + N 1 (mod 4) (3F7) From (3F1) and (3F) we obtain (HA + KM)(HA KM) p m 1 1 1 (K N H B ) By Lemma 36, does not divide both of HA ± KM Choose the sign of A so that HA + KM Define odd integers R and S by (3F8) R (HA + KM)/ r, S (HA KM)/ s p α 1, where p α 1 HA KM Clearly, RS We note that { α m1 1 1, (3F9) min(r, s) 1, r + s 3 if p 1 (mod 8), r s 0 if p 3 (mod 8) We first show that (3F10) ( )( ) R S ( 1) α+1 From (3F7), (3F8) and (3F9), we see that p α m 1+1 1 K N H B Thus, by Lemma 34, we have { p (α m 1+1)/ 1 N, p (α m 1+1)/ 1 B if α is odd, p (α m 1+)/ 1 N, p (α m 1)/ 1 B if α is even Define integers N 1 and B 1 by { N p (α m 1+1)/ 1 N 1, B p (α m 1+1)/ 1 B 1 if α is odd, N p (α m 1+)/ 1 N 1, B p (α m 1)/ 1 B 1 if α is even, so that Hence N 1 (α odd), B 1 (α even) { K r+s N1 H B1 if α is odd, RS K N1 H B1 if α is even,
44 J B Muskat et al so that (recalling (/ ) 1) ( K ( )( ) R S N ) 1 1 if α is odd, ( H B1 ) 1 if α is even, proving (3F10) We next show that ( ) R (3F11) ( ), R We factor R into primes as follows: (3F1) R ε ( /q i )1 q e i i ( ) S ( /r j ) 1 ( ) S r f j j, where ε ±1, e i, f j are positive integers, and q i, r j are distinct odd primes By Lemma 35(ii) each f j is even It is convenient to define (3F13) E k e i, k 1, 3, 5, 7 From (3F1) we have ( ) ( ) R ε ε ε ε ( /q i )1 ( /q i )1 ( /q i )1 ( /q i )1 q i k (mod 8) ( qi ) ei (as the f j are even) ( ) ei p1 (by the law of quadratic reciprocity) q i ( ( /q i )1 q i 5,7 (mod 8) that is, by (3F13), (3F14) q i ) ei ( 1) e i, ( ) R ε( 1) E 5+E 7 Now taking (3F1) modulo 4, we obtain, as each f j is even, R ε ε ( 1) e i ε( 1) E 3+E 7 (mod 4), ( /q i )1 q e i i ( /q i )1 q i 3 (mod 4)
that is, so (3F15) ε Representation of primes by binary quadratic forms 45 ( ) 1 ε( 1) E 3+E 7, R ( ) 1 ( 1) E 3+E 7 R Substituting (3F15) in (3F14), we obtain ( ) ( ) R 1 (3F16) ( 1) E 3+E 5 R Further, taking (3F1) modulo 8, we deduce R ε 3 e i that is, by (3F15), (3F17) ( /q i )1 q i 3 (mod 8) ε3 E 3 5 E 5 7 E 7 ε( 5) E 3 5 E 5 ( 1) E 7 ( /q i )1 q i 5 (mod 8) ε( 1) E 3+E 7 5 E 3+E 5 (mod 8), R 5 e i ( ) 1 5 E 3+E 5 (mod 8) R ( /q i )1 q i 7 (mod 8) Hence, from (3F17), we have ( ) ( ) E3 +E 5 (3F18) ( 1) E 3+E 5 R 5 Then, from (3F16) and (3F18), we deduce ( ) ( )( ) ( ) R 1, R R R as asserted in (3F11) The proof of (S/ ) ( /S) is similar The next step is to show that ( ) ( ) (3F19) ( 1) N( 1+1)/8 R M + N We consider two cases according as p 1 (mod 8) or p 3 (mod 8) C a s e 1: p 1 (mod 8) We set (recalling (3F1) (3F6)) A 4a + ε A, B b, H 4h + ε H, K 4k + ( 1) k, M 4m + 1, N 4n, 7 e i
46 J B Muskat et al where ε A ±1, ε H ±1 Substituting these in (3F7), reducing modulo 3, and dividing by 8, we obtain (3F0) (a + h + k + m + n) + (ε A a + ε H h ( 1) k k m) b (mod 4) Taking (3F0) modulo we have (3F1) b a + h + k + m (mod ) Next, substituting the above values for A, B, into the first equation in (3F8) when r 1 and into the second equation in (3F8) when s 1, we obtain, modulo 16, (3F) 4(ε H a + ε A h) + 4( 1) k m + 4k + (ε A ε H + ( 1) k ) R (mod 16) if r 1, 4(ε H a + ε A h) 4( 1) k m 4k + (ε A ε H + ( 1) k+1 ) p α 1 S (mod 16) if s 1 Looking at (3F) modulo 4, we deduce { εa ε H + ( 1) k (mod 4) if r 1, ε A ε H ( 1) k (mod 4) if s 1, that is, (3F3) ε A { ( 1) k ε H if r 1, ( 1) k+1 ε H if s 1 Substituting (3F3) in (3F), and dividing by, we obtain R ε H (a + ( 1) k h) + ( 1) k m + k + ( 1) k (mod 8) if r 1, (3F4) ( 1) α S ε H (a + ( 1) k+1 h) + ( 1) k+1 m k + ( 1) k+1 (mod 8) if s 1 Taking (3F4) modulo 4, we have { R (a + h + m) + 1 (mod 4) if r 1, (3F5) ( 1) α S (a + h + m) 1 (mod 4) if s 1, as k + ( 1) k 1 (mod 4) for all integers k Then, using (3F1) and (3F3), we may rewrite (3F0) as ε H (a + ( 1) k h) ( 1) (3F6) k m + k + (b + n) ( 1) k b (mod 4) if r 1, ε H (a + ( 1) k+1 h) ( 1) k+1 m k + (b + n) + ( 1) k b (mod 4) if s 1 Substituting (3F6) into (3F4) gives { R 4(b + k + m + n) ( 1) k b + ( 1) k (mod 8) if r 1, (3F7) ( 1) α S 4(b + k + m + n) + ( 1) k b + ( 1) k+1 (mod 8) if s 1
Representation of primes by binary quadratic forms 47 Now, from (3F1) and (3F5), we have { R (b + k) + 1 (mod 4) if r 1, (3F8) ( 1) α S (b + k) 1 (mod 4) if s 1 Applying (3F8) to the right side of (3F7), we obtain { R (R 1) + 4(m + n) ( 1) k b + ( 1) k (mod 8) if r 1, ( 1) α S (( 1) α S + 1) + 4(m + n) + ( 1) k b + ( 1) k+1 (mod 8) if s 1 By rearranging and then squaring, we obtain { R 4R + 4 8(m + n) 1 (mod 16) if r 1, S + 4( 1) α S + 4 8(m + n) 1 (mod 16) if s 1, as (b 1) 4b (b 1) + 1 1 (mod 16) Hence { (R 1) 4(R 1) 8(m + n) (mod 16) if r 1, (S 1) 4(S 1) 8(m + n) + 8(α + 1) (mod 16) if s 1 Thus if r 1 we have ( ) ( 1) (R 1)/+(R 1)/8 ( 1) m+n R ( ) ( 1) ((M+N) 1)/8, M + N and if s 1 we have ( ) ( 1) (S 1)/+(S 1)/8 ( 1) m+n+α+1 S ( ( 1) ((M+N) 1)/8 ( 1) α+1 M + N so that in both cases ( ) ( R M + N )( R )( ) S (by (3F10) and (3F11)) This completes the proof of (3F19) when p 1 (mod 8) as N is even in this case C a s e : p 3 (mod 8) We set (recalling (3F1) (3F6)) ) A 4a + ε A, B 4b + ε B, H 4h + ε H, K 4k + ( 1) k, M 4m +, N 4n 1, where ε A ±1, ε B ±1, ε H ±1 Taking R HA + KM modulo 4, we obtain (3F9) ε A ε H ( 1) (R+1)/
48 J B Muskat et al Then, taking R HA + KM modulo 8, and appealing to (3F9), we deduce ( ) (3F30) ( 1) a+h+k+m R From (3F7) modulo 16 we have (3F31) a + h + k n + ( + 1)/8 (mod ) Then, from (3F30) and (3F31), we obtain ( ) ( ( 1) m+n+(p1+1)/8 R M + N ) ( 1) (+1)/8, which is (3F19) as N is odd in this case This completes the proof of (3F19) Now, by (3F7) and (3F8), we have so that (3F3) Finally, r R r+1 RKM (HA + KM) (HA + KM)KM (HA + KM)(HA KM) p m 1 1 1 (K N H B ) 0 (mod ) (as m 1 ), ( ) r+1 ( )( )( ) R K M 1 C p is a fourth power in H( D) ( ) ( ) K 1 K ( ) K 1 (by (3F3)) ( )( ) R M 1 (by (3F3)) ( )( ) M 1 (by (3F11)), R that is, under the restriction (3F6), (3F33) C p is a fourth power in H( D) ( )( ) ( 1) N( 1+1)/8 M 1 (by (3F19)) M + N
Representation of primes by binary quadratic forms 49 In order to remove this restriction, the necessary and sufficient condition (3F33) must include the additional factor ( ) { 1 1 if M + N 1 (mod 4), M + N 1 if M + N 3 (mod 4), so that the condition can now be written ( )( ) ( 1) N(+1)/8 M 1 M + N C a s e (G): D 16p m 1 1, D 16p m 1 1 1, m 1 1, 1 (mod 4) Let p be an odd prime satisfying ( 1/p) (p/ ) 1, so that p 1 (mod 4) and p From (3) and (31) we have (3G1) (3G) H p M + 4p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, K p A + 4p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1 By (33) we have ( D/K) ( 16p m 1 1 /K) 1, so that by the law of quadratic reciprocity (3G3) ( ) 1 K ( ) m1 p1 K ( ) m1 K Reducing (3G1) and (3G) modulo 8, we obtain (3G4) A M 1 (mod ), B N (p 1)/4 (mod ) From (3G1) and (3G) we obtain (3G5) (HA + KM)(HA KM) 4p m 1 1 1 (K N H B ) By Lemma 36, does not divide both of HA ± KM Choose the sign of A so that HA + KM Define odd integers R and S by (3G6) R (HA + KM)/ r, S (HA KM)/ s p α 1, where p α 1 HA KM Clearly RS, α m 1 1, and appealing to (3G4) we see that (3G7) min(r, s) 1, r + s 4 From (3G5) and (3G6) we see that r+s K N H B Thus, by Lemma 3(iii), we have (3G8) r + s 0 (mod ) if 5 (mod 8) We factor R into primes: (3G9) R ε ( /q i )1 q e i i ( /r j ) 1 r f j j, where ε ±1, e i, f j are positive integers, and q i, r j are distinct odd primes By Lemma 35(ii) each f j is even From (3G9) we have by the law of
50 J B Muskat et al quadratic reciprocity (3G10) ( ) R 1 By Lemma 31 we have (HA + KM)(KM + wkn) (3G11) (HA + KM + wkn) (mod ) if m 1 1, (HA + KM)KM (HA + KM) (mod ) if m 1, where w 4 (mod ) Thus by (3G6) and (3G11), we have ( ) r+1 ( )( )( ) R K M + wn 1 if m 1 1, (3G1) ( ) r+1 ( )( )( ) R K M 1 if m 1 If 1 (mod 8) then (/ ) 1 If 5 (mod 8) then by (3G7) and (3G8), r is odd so that (/ ) r+1 ( 1) r+1 1 Hence, by (3G10) and (3G1), we have ( ) M + wn ( ) K if m 1 1, (3G13) ( ) M if m 1 Finally, C p is a fourth power in H( D) ( ) ( ) 1 K 1 K ( ) K 1 (by (3G3)) ( ) M + wn 1 ( ) if m 1 1, where w 4 (mod ) M 1 if m 1 (by (3G13)) C a s e (H): D 16p m 1 1, D 16p m 1 1 1, m 1 (even), 3 (mod 4) Let p be an odd prime such that ( 1/p) (p/ ) 1, so that p 1 (mod 4) and p From (3) and (31) we have (3H1) (3H) H p M + 4p m 1 1 1 N, H > 0, (M, N) 1, (H, p ) 1, K p A + 4p m 1 1 B, K > 0, (A, B) 1, (K, p ) 1
Representation of primes by binary quadratic forms 51 From (33) we have ( D/K) ( 16p m 1 1 /K) 1, so that ( ) 1 (3H3) 1, K 1 (mod 4) K Reducing (3H1) and (3H) modulo 8, we obtain (3H4) A M 1 (mod ), B N (p 1)/4 (mod ) From (3H1) and (3H) we obtain (3H5) (HA + KM)(HA KM) 4p m 1 1 1 (K N H B ) By Lemma 36, does not divide both of HA ± KM We choose the sign of A so that HA + KM, and define odd integers R and S by (3H6) R (HA + KM)/ r, S (HA KM)/ s p α 1, where p α 1 HA KM Clearly RS We note that α m 1 1 1, min(r, s) 1, (3H7) p 1 (mod 8) if r + s 4, p 5 (mod 8) if r + s 3 Proceeding exactly as in the proof of (3C11), we obtain ( ) r+s ( )( ) R S ( 1) α+1, that is, (3H8) ( )( ) ( ) r+s R S ( 1) α+1 Next, factoring R and S into primes and proceeding as in case (C) ((3C1) (3C16)), we obtain ( ) ( ) R S (3H9) ( 1) (R 1)/, ( 1) (S 1)/ From (3H8) and (3H9) we deduce (3H10) (R 1)/ + (S 1)/ (r + s)( 1 1)/8 + (α + 1) (mod ) Our next task is to show that (3H11) (M 1)/ + N (R 1)/ + (r + 1)( 1 1)/8 (mod ) From (3H7) either r 1 or s 1 We treat the case r 1 first In this case s by (3H7) Working modulo 4 we have
5 J B Muskat et al so that R KM + p α 1 s 1 S M + p α 1 s 1 S { if s M + 0 if s 3 { M + if p 5 (mod 8) M if p 1 (mod 8) M + N (mod 4) (by (3H6)) (by (3H3)) (as, S are odd) (by (3H7)) (by (3H4)), (M 1)/ + N (R 1)/ (mod ), proving (3H11) in this case We now turn to the case s 1 In this case r by (3H7) Working modulo 4 we have α + S ( 1) α S (as S 1 (mod )) so that p α 1 S (as 3 (mod 4)) (HA KM)/ r 1 R KM r 1 R M r 1 + M + { M + if r 3 p 1 (mod 8) M if r p 5 (mod 8) M + N (mod 4) (by (3H6)) (by (3H6)) (by (3H3)) (as R and M are odd) (by (3H7)) (by (3H4)), (M 1)/ + N (S 1)/ + (α + 1) (mod ) Appealing to (3H10) we obtain (M 1)/ + N (R 1)/ + (r + 1)( 1 1)/8 (mod ) as required From (3H5) and (3H6) we have so that r R r+1 RKM (HA + KM) (HA + KM)KM (HA + KM)(HA KM) 4p m 1 1 1 (K N H B ) 0 (mod ) (as m 1 ), ( r+1 ) ( RKM r R ) 1,
Representation of primes by binary quadratic forms 53 giving ( ) ( ) r+1 ( )( ) K R M ( 1) (r+1)(p 1 1)/8+(R 1)/ ( M ) (by (3H9)), that is, (3H1) Finally, ( ) ( ) K ( 1) (M 1)/+N Mp1 (by (3H11)) C p is a fourth power in H( D) ( ) ( ) 1 K 1 K ( ) K 1 (by (3H3)) ( ) M ( 1) (M 1)/+N 1 (by (3H1)) ( )( ) 1 M 1 M + N C a s e (I): D p m 1 1 pm, D p m 1 1 pm 1, m 1 (odd) 1, m 1, 3 (mod 4), 1 (mod 4), ( / ) 1 Let p be an odd prime such that (p/ ) (p/ ) 1, so that p, From (3) and (31) we have (3I1) H p M + MN + 1 4 (1 + pm 1 1 pm 1 )N, H > 0, (M, N) 1, (H, p ) 1, (3I) K p A + AB + 1 4 (1 + pm 1 1 pm )B, K > 0, (A, B) 1, (K, p ) 1 From (33) we have ( D/K) ( p m 1 1 pm /K) 1, so that, by the law of quadratic reciprocity, ( ) ( ) m K K (3I3) We set (3I4) C A + B, L M + N Using (3I4) in (3I1) and (3I), we obtain (3I5) (3I6) 4H p L + p m 1 1 pm 1 N, 4K p C + p m 1 1 pm B
54 J B Muskat et al We note that { 1 if N is odd, (3I7) (L, N) (M + N, N) (M, N) (, N) if N is even, { 1 if B is odd, (3I8) (C, B) (A + B, B) (A, B) (, B) if B is even From (3I5) and (3I6) we obtain (3I9) (HC + KL)(HC KL) p m 1 1 pm 1 (K N H B ) Lemma 36 implies that neither nor divides both of HC ± KL By changing the sign of C if necessary, we may suppose that HC + KL We define odd integers R and S by (3I10) R (HC + KL)/ r p β, S (HC KL)/s p α 1 p γ, where p α 1 HC KL, p β HC + KL, pγ HC KL We note that RS, RS, (3I11) α m 1 1, β + γ m 1 0, min(β, γ) 0 Next we factor R into primes: (3I1) R ε ( /q i )1 q e i i ( /r j ) 1 r f j j, where ε ±1, e i, f j are positive integers, and q i, r j are distinct odd primes From (3I5) (3I8), (3I10) and Lemma 35(ii) we see that each f j is even Then, from (3I1), we have ( ) ( ) R ε ( ) ei qi (3I13) 1, ( /q i )1 by the law of quadratic reciprocity Similarly we have ( ) S (3I14) 1 Next, by Lemma 31 and (3I9), we have (3I15) (KL ± HC)(KL + θkn) (KL ± HC + θkn) (mod ), where θ is a solution of the congruence (3I16) θ p m 1 1 pm 1 (mod )
Representation of primes by binary quadratic forms 55 Note that we may take θ 0 when m The congruence (3I16) is solvable when m 1 as ( p m 1 ) ( ) m1 1 p1 1 From (3I15) we have, as ( 1/ ) 1, ( )( )( )( ) HC ± KL K L + θn (3I17) 1 Recall from (3I11) that either β 0 or γ 0 Taking the + sign in (3I17) when β 0 and the sign when γ 0, and appealing to (3I10), we obtain ( ) r+1 ( )( )( ) R K L + θn 1 if β 0, (3I18) ( ) s+1 ( ) α ( )( )( ) p1 S K L + θn 1 if γ 0 Thus, by (3I13), (3I14) and (3I18), we have ( ) r+1 ( ) L + θn ( ) K (3I19) p ( ) s+1 ( ) L + θn if β 0, if γ 0 If 1 (mod 8) then (/ ) r+1 (/ ) s+1 (/ ) If 5 (mod 8), by Lemma 37, r and s are both even so (/ ) r+1 (/ ) s+1 (/ ) Hence (3I19) reduces to ( ) ( )( ) K L + θn (3I0) Now set w (1 + θ)/ (mod ) so that (3I1) L + θn M + (1 + θ)n (M + wn) (mod ), and w is a solution of the congruence w w + 1 4 (1 + pm 1 1 pm 1 ) 0 (mod ) Note w 1 (mod ) if m Hence, by (3I0) and (3I1), we have ( ) ( K M + wn (3I) )