On the number of representations of n by ax 2 + by(y 1)/2, ax 2 + by(3y 1)/2 and ax(x 1)/2 + by(3y 1)/2

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1 ACTA ARITHMETICA On the number of representations of n by ax + byy 1/, ax + by3y 1/ and axx 1/ + by3y 1/ by Zhi-Hong Sun Huaian 1 Introduction For 3, 4,, the -gonal numbers are given by p n n + n Thus, p 3 n n + n, p 4 n n, p n 3n n Let Z and N be the set of integers and the set of positive integers, respectively Let Z Z Z { x, y : x, y Z} For n N let rn fx, y { x, y Z : n fx, y} For 143 values of a, b the formula for rn axx 1/ + byy 1/ is nown see [3, 4] In [] the author determined rn 3x x/ + b3y y/ for b 1,, In this paper, using some results for binary quadratic forms in [4, ] we determine rn x + byy 1/ in the cases b 1,, 3, 4,, 6, 8, 9, 10, 11, 14, 1, 16, 1, 9, 30, 3, 39, 1, 6, 9, rn x + b3y y/ in the cases b 1,, 3, 4,, 7, 8, 13, 17 and rn x x/ + b3y y/ in the cases b 1,, 3,, 7, 10, 11, 14, 1, 19, 6, 31, 34, 3,, 9, 91, 11, 119, 4 For example, we have r n x yy 1 + b b for b 3,, 11, 9, 8n+b r n x y3y 1 + b 6b for b, 7, 13, 17, 4n+b xx 1 y3y 1 3b r n + b 1n+b+3/ where a m is the Legendre Jacobi Kronecer symbol for b 7, 11, 19, 31, 9, 010 Mathematics Subject Classification: Primary 11E; Secondary 11E16 Key words and phrases: binary quadratic forms, number of representations DOI: /aa [81] c Instytut Matematyczny PAN, 011

2 8 Z H Sun A negative integer d with d 0, 1 mod 4 is called a discriminant Let d < 0 be a discriminant The conductor of d is the largest positive integer f fd such that d/f 0, 1 mod 4 As usual we set wd, 4, 6 according as d < 4, d 4 or d 3 For a, b, c Z we denote by [a, b, c] the equivalence class containing the form ax + bxy + cy It is nown [1] that 11 [a, b, c] [c, b, a] [a, a + b, a + b + c] for Z For n N and a, b, c Z with a, c > 0 and b 4ac < 0 let 1 R[a, b, c], n { x, y Z : n ax + bxy + cy } That is, R[a, b, c], n rn ax +bxy+cy It is nown that R[a, b, c], n R[a, b, c], n If R[a, b, c], n > 0, we say that n is represented by [a, b, c] or ax + bxy + cy Let Hd be the form class group consisting of classes of primitive, integral binary quadratic forms of discriminant d For n N, following [] we define 13 δn, d d, Nn, d RK, n n K Hd When n is odd and a Z, we also define δn, a n a, where a is the Jacobi symbol In the paper we mainly use the formula for Nn, d and reduction formulas for RK, n in [] to determine rn ax + byy 1/, rn ax + by3y 1/ and rn axx 1/ + by3y 1/ for some special values of a, b In addition to the above notation, throughout this paper [x] denotes the greatest integer not exceeding x and a, b denotes the greatest common divisor of integers a and b For a prime p and n N, ord p n denotes the unique nonnegative integer α such that p α n ie p α n but p α+1 n Throughout this paper, p denotes a prime and products over p run through all distinct primes p satisfying any restrictions given under the product symbol For example the condition p 1 mod 4 under a product restricts the product to those distinct primes p which are of the form 4 +1 Basic lemmas Let d < 0 be a discriminant and n N Recall that δn, d n d By [, Lemma 41], we have d 11 + ord p n p 1 δn, d if ord q n for every prime q with d q 1, 0 otherwise When n is odd and d is not a discriminant, 1 is also true By 1, d n 1 implies δn, d 0

3 Number of representations of n by ax + byy 1/ 83 Lemma 1 [, Theorem 41] Let d < 0 be a discriminant with conductor f Let n N and d 0 d/f Then 0 if n, f is not a square, Nn, d m 1 1 d/m n wdδ p p m, d 0 p m if n, f m for m N In particular, when n, f 1 we have Nn, d wdδn, d 0 Lemma [4, Lemma ] Let a, b, n N with n i If a and 4 a bb, then { R[a, 0, b], n if n a mod 4, R[a, 0, 4b], n 0 otherwise If a, b and 8 b, then { R[a, 0, b], n if n a mod 8, R[a, 0, 4b], n 0 otherwise ii If a + b and 8 ab, then { R[a, 0, b], n if n a + b mod 8, R[4a, 4a, a + b], n 0 otherwise Lemma 3 [4, Theorem 1] Let b {6, 10, 1,, 8, 8}, b r b 0 b 0, n N and n Then { δn, b if n 1 mod 8, R[1, 0, 4b], n 0 otherwise, { δn, b if n b + 1 mod 8, R[4, 4, b + 1], n 0 otherwise, { δn, b if n R[ r+ b0 mod 8,, 0, b 0 ], n 0 otherwise, { R[ r+, r+, r δn, b if n + b 0 ], n r + b 0 mod 8, 0 otherwise Lemma 4 [4, Theorem 6] Let n N and n α n 0 with n 0 Then δn 0 /9, 0 if 9 n and n 0 ±1 mod, R[1, 0, 4], n δn 0, 0 if 3 n 1 and n 0 ±1 mod, 0 otherwise, δn 0 /9, 0 if 9 n and n 0 ±1 mod, R[, 0, 9], n δn 0, 0 if 3 n and n 0 ±1 mod, 0 otherwise

4 84 Z H Sun 3 Formulas for rn ax + byy 1/ Lemma 31 Let a, b, n N Then rn ax + byy 1/ { R[8a, 0, b], 8n + b if 8 b, R[a, 0, b/8], n + b/8 R[a, 0, b/], n + b/8 if 8 b Moreover, if 4 a and b, then rn ax + byy 1/ R[a, 0, b], 8n + b; if a, b and 4 a b/, then rn ax + byy 1/ R[a, 0, b/], 4n + b/ Proof It is clear that rn ax + byy 1/ r8n 8ax + by 1 b { x, y Z : 8n + b 8ax + by, y} R[8a, 0, b], 8n + b R[8a, 0, 4b], 8n + b When 8 b, we have 8a, 4b 8n + b and so R[8a, 0, 4b], 8n + b 0 If 4 a and b, by the above and Lemma we have rn ax + byy 1/ R[8a, 0, b], 8n + b R[a, 0, b], 8n + b; if a, b and 4 a b/, by the above and Lemma we have rn ax + byy 1/ R[8a, 0, b], 8n + b R[4a, 0, b/], 4n + b/ This completes the proof R[a, 0, b/], 4n + b/ Theorem 31 Let n N Then n + δ, if n 0 mod 4, rn x + 16yy 1/ δn +, if n 1 mod 4, 0 if n, 3 mod 4, rn x + 4yy 1/ δn + 1,, rn x + yy 1/ δ8n + 1, Proof If n x + 16yy 1/ for some x, y Z, then n x mod 8 and so n, 3 mod 4 Now assume n 0, 1 mod 4 As x n+ mod 4 is insolvable, we see that R[1, 0, 8], n+ 0 Thus, by Lemmas 1 and 31 we have

5 Number of representations of n by ax + byy 1/ 8 rn x + 16yy 1/ R[1, 0, ], n + R[1, 0, 8], n + R[1, 0, ], n + Nn +, 8 δn +, 8 { δn + /, if n 0 mod 4, δn +, if n 1 mod 4 As r4n x + 16yy 1/ rn x + 4yy 1/ and r16n x + 16yy 1/ rn x + yy 1/, by the above we deduce the remaining result So the theorem is proved Theorem 3 Let n N Then δn + 1, 1 if 4 n, rn x + 8yy 1/ 4δn + 1/, 1 if 8 n 1, 0 otherwise, rn x + yy 1/ δ4n + 1, 1, rn x + yy 1/ δ8n + 1, 1 Proof If n x + 8yy 1/ for some x, y Z, then n x mod 8 and so 4 n or 8 n 1 Now assume 4 n or 8 n 1 By Lemmas 1 and 31 we have rn x + 8yy 1/ R[1, 0, 1], n + 1 R[1, 0, 4], n + 1 Nn + 1, 4 Nn + 1, 16 { 4δn + 1, 4 δn + 1, 4 if 4 n, 4δn + 1, 4 0 if 8 n 1 This yields the result on rn x + 8yy 1/ Since r4n x + 8yy 1/ rn x + yy 1/ and r8n x + 8yy 1/ rn x + yy 1/, from the above we deduce the remaining result So the theorem is proved Theorem 33 Let n N, m {3, 7} and a {1,, m, m} Then r n ax + m yy 1 a m 4an+m Proof It is nown that H 4m {[1, 0, m]} and f 4m Thus, by Lemmas 31 and 1 we have rn x + myy 1/ R[1, 0, m], 4n + m N4n + m, 4m m m 4n+m Now replacing n with an we obtain the result 4n+m

6 86 Z H Sun Theorem 34 Let a, n N, a 30 and 4an α n 0 3 n 0 Then r n ax + 30 yy α n0 a 3 1 Proof If a 1, then by Lemma 31, [, Theorem 93] and 1 we have rn x + 30yy 1/ R[1, 0, 1], 4n α n0 3 n 0 3 α n 0 1 Observe that 3 α n 0 1 n 0 1 We see that the result is true for a 1 Since ran x + 30yy 1/ r n ax + 30 a yy 1/, the result follows Theorem 3 Let n N and b {3,, 11, 9} Then rn x + byy 1/ δ8n + b, b, rn bx + yy 1/ δ8n + 1, b Proof By Lemma 31 we have rn x +byy 1/ R[8, 0, b], 8n+b and rn bx +yy 1/ R[1, 0, 8b], 8n+1 Thus applying Lemma 3 we deduce the result Theorem 36 Let n N and a {1, 9} Then r n ax + 9 δ8n + 9/a, if 3 n a, yy 1 8n + 9/a a δ, if 9 n 9/a, 9 0 otherwise Proof From Lemma 31 we have rn x + 9yy 1/ R[, 0, 9], 8n + 9 and rn 9x + yy 1/ R[1, 0, 18], 8n + 1 Since H 7 {[1, 0, 18], [, 0, 9]} and f 7 3, applying [, Theorem 93] and 1 we deduce the result Lemma 3 [4, Theorem 4] Let m {, 7, 13, 17}, i {1,, 3, 6}, n N and in α i 3 β i n 0 with 6, n 0 1 Then R[i, 0, 6m/i], n > 0 if and only if ord q n 0 for every prime q with 6m q 1 and 1, 6m + 1 mod 4 if α i and β i, m + 3, 8m + 3 mod 4 if α i and β i, n 0 3m +, 3m + 8 mod 4 if α i and β i, m, m + 6 mod 4 if α i and β i Moreover, if R[i, 0, 6m/i], n > 0, then R[i, 0, 6m/i], n 1 + ord p n 0 6m p 1

7 Number of representations of n by ax + byy 1/ 87 Theorem 37 Let m {, 7, 13, 17}, a {1, 3, m, 3m}, n N and 8an + 3m 3 β n 0 with 3 n 0 Then r n ax + 3m a yy 1/ > 0 if and only if ord q n 0 for every prime q with 6m q 1 and { 1 mod 3 if m {7, 13} and β, n 0 mod 3 otherwise Moreover, if the above conditions hold, then r n ax + 3m a yy 1/ 6m p 11 + ord p n 0 Proof Clearly n β m mod 8 By Lemma 31 we have rn x +3myy 1/ R[, 0, 3m], 8n+3m Now applying Lemma 3 we deduce the result for a 1 Noting that ran x + 3myy 1/ r n ax + 3m a yy 1/ we deduce the remaining result Lemma 33 [4, Theorem 43] Let m {7, 13, 19}, i {1,,, 10}, n N and in α i β i n 0 with 10, n 0 1 Then R[i, 0, 10m/i], n > 0 if and only if ord q n 0 for every prime q with 10m q 1 and 1, 9, m, m mod 40 if α i and β i, + m, + 8m, + 18m, + 3m mod 40 if α i and β i, n 0 m +, m + 8, m + 18, m + 3 mod 40 if α i and β i, m, 9m, 10 + m, m mod 40 if α i and β i Moreover, if R[i, 0, 10m/i], n > 0, then R[i, 0, 10m/i], n 10m p ord p n 0 Theorem 38 Let m {7, 13, 19}, a {1,, m, m}, n N and 8an + m β n 0 with n 0 Then r n ax + m a yy 1/ > 0 if and only if ord q n 0 for every prime q with 10m q 1 and { ±1 mod if m 19 and β, n 0 ± mod otherwise Moreover, if the above conditions hold, then r n ax + m a yy 1/ 10m p ord p n 0 Proof Clearly n 0 m or m mod 8 according as β or β By Lemma 31 we have rn x + myy 1/ R[, 0, m], 8n + m Now applying Lemma 33 we deduce the result for a 1 Noting that ran x + myy 1/ r n ax + m a yy 1/ we deduce the remaining result

8 88 Z H Sun Theorem 39 Let n N Then rn x + 10yy 1/ 1 1/ φ 4 4n +, 4n+ where φ 4 m is given by q 1 q 4 1 q 0 1 or by [6, Theorem 4ii] φ 4 mq m q < 1 m1 Proof It is nown that f 80 and H 80 {[1, 0, 0], [4, 0, ], [3,, 7], [3,, 7]} Thus R[1, 0, 0], 4n++R[4, 0, ], 4n+ N4n+, 80 R[3,, 7], 4n+ If 4n+ 3x +xy+7y for some x, y Z, then 4n+ x +xy y x y mod 4 This is impossible Thus R[3,, 7], 4n + 0 Hence, using the above and Lemma 1 we see that R[1, 0, 0], 4n + + R[4, 0, ], 4n + N4n +, / 4n+ On the other hand, by [6, Theorem ] we have Hence 4n+ R[1, 0, 0], 4n + R[4, 0, ], 4n + φ 4 4n + R[4, 0, ], 4n + By Lemma 31 we have 4n+ 1 1/ φ 4 4n + rn x + 10yy 1/ R[8, 0, 10], 8n + 10 R[4, 0, ], 4n + Thus the result follows 4 Formulas for rn ax + b3y y/ Lemma 41 Let a, b, n N Then rn ax + b3y y/ R[4a, 0, b], 4n + b R[4a, 0, 4b], 4n + b In particular, if 3 b and 8 b, we have R[4a, 0, 9b], 4n + b + R[4a, 0, 36b], 4n + b rn ax + b3y y/ R[4a, 0, b], 4n + b

9 Number of representations of n by ax + byy 1/ 89 Moreover, if 4 a, b and 3 b, then Proof Clearly rn ax + b3y y/ R[6a, 0, b], 4n + b rn ax + b3y y/ r4n + b 4ax + b6y 1 { x, y Z : 4n + b 4ax + by, y, 3 y} { x, y Z : 4n + b 4ax + by, y} { x, y Z : 4n + b 4ax + b3y, y} R[4a, 0, b], 4n + b R[4a, 0, 4b], 4n + b R[4a, 0, 9b], 4n + b R[4a, 0, 36b], 4n + b If 3 b and 8 b, we then have rn ax + b3y y/ R[4a, 0, b], 4n+b If 4 a, b and 3 b, by Lemma we have R[4a, 0, b], 4n+b R[6a, 0, b], 4n + b So the lemma is proved Theorem 41 Let n N Then rn x + 43y y/ δ6n + 1, 6, rn x + 3y y/ δ4n + 1, 6 Proof As H 4 {[1, 0, 6], [, 0, 3]} and m x + 3y implies m 1 mod 6, using Lemma 1 we see that R[4, 0, 4], 4n + 4 R[1, 0, 6], 6n + 1 N6n + 1, 4 δ6n + 1, 6 Now putting a 1 and b 4 in Lemma 41 and applying the above we obtain rn x + 43y y/ δ6n + 1, 6 Replacing n by 4n we deduce the remaining result Theorem 4 Let n N Then δ3n + 1, 3 if 4 n, rn x + 83y y/ δ3n + 1/4, 3 if 8 n 1, 0 otherwise, rn x + 3y y δ1n + 1, 3 Proof If n x + 43y y for some x, y Z, then clearly 4 n or 8 n 1 Now assume 4 n or 8 n 1 From Lemma 41 we have rn x + 83y y/ R[4, 0, 8], 4n + 8 R[4, 0, 3], 4n + 8 R[1, 0, 3], 3n + 1 R[3, 0, 4], 3n + 1 N3n + 1, 1 R[3, 0, 4], 3n + 1 If 4 n, then 3n + 1 3x + 4y is insolvable Thus, by Lemma 1 we have N3n + 1, 1 R[3, 0, 4], 3n + 1 N3n + 1, 1 δ3n + 1, 3

10 90 Z H Sun If 8 n 1, then 3n mod 8 By Lemma 1, [, Theorem 93] and 1 we have N3n + 1, 1 R[3, 0, 4], 3n + 1 6δ3n + 1/4, 3 δ3n + 1/4, 3 4δ3n + 1/4, 3 Now combining the above we obtain the result for rn x + 43y y As r4n x + 43y y rn x + 3y y, applying the above we deduce the remaining result Theorem 43 Let n N and b {, 7, 13, 17} Then rn x + b3y y/ δ4n + b, 6b, rn bx + 3y y/ δ4n + 1, 6b Proof From Lemma 41 we have rn x + b3y y/ R[b, 0, 6], 4n + b It is easily seen that H 4b {[1, 0, 6b], [, 0, 3b], [3, 0, b], [6, 0, b]} and 4n + b cannot be represented by x + 6by, x + 3by and 3x + by Thus, using the fact that f 4b 1 and Lemma 1 we deduce R[b, 0, 6], 4n + b N4n + b, 4b δ4n + b, 4b Now combining the above we obtain rn x + b3y y/ δ4n + b, 6b Replacing n with bn we deduce the remaining result Theorem 44 Let n N Then rn x + 33y y/ δn 0, if 9 n 1 and 8n α n 0 3 n 0, δ8n + 1, if 3 n or n 4, 7 mod 9, 0 if n mod 3 and rn 3x + 3y y/ δ4n + 1, Proof From Lemmas 41 and we see that rn x + 33y y/ R[4, 0, 3], 4n + 3 R[4, 0, 7], 4n + 3 R[8, 0, 1], 8n + 1 R[8, 0, 9], 8n + 1 R[, 0, 1], 8n + 1 R[, 0, 9], 8n + 1 Since H 8 {[1, 0, ]}, by Lemma 1 we have R[, 0, 1], 8n + 1 δ8n + 1, It is nown that H 7 {[1, 0, 18], [, 0, 9]} and f 7 3 Thus, by [, Theorem 93] and 1 we have δ8n + 1, if 3 n, R[, 0, 9], 8n + 1 δ8n + 1/9, if 9 n 1, 0 otherwise

11 8n+1 Number of representations of n by ax + byy 1/ 91 When 9 n 1 and 8n α n 0 3 n 0, we have 3 α + n 0 8n α 1 n 0 Now combining all the above we obtain the result for rn x +33y y/ Note that r3n x + 33y y/ rn 3x + 3y y/ We then obtain the remaining result Formulas for rn ax x/ + b3y y/ Lemma 1 Let a, b, n N Then rn ax x/ + b3y y/ R[1a, 1a, 3a + b], 4n + 3a + b R[1a, 0, 4b], 4n + 3a + b R[1a, 1a, 3a + 9b], 4n + 3a + b + R[1a, 0, 36b], 4n + 3a + b If 4 a, we also have rn ax x/ + b3y y/ R[3a, 0, b], 4n + 3a + b R[3a, 0, 4b], 4n + 3a + b Proof Clearly R[3a, 0, 9b], 4n + 3a + b + R[3a, 0, 36b], 4n + 3a + b rn ax x/ + b3y y/ r4n + 3a + b 3ax 1 + b6y 1 { x, y Z : 4n + 3a + b 3ax + by, xy, 3 y} { x, y Z : 4n + 3a + b 3ax + by, xy} { x, y Z : 4n + 3a + b 3ax + b3y, xy} { x, y Z : 4n + 3a + b 3ax + y + by, y} { x, y Z : 4n + 3a + b 3ax + y + 9by, y} { x, y Z : 4n + 3a + b 1ax + 1axy + 3a + by, y} { x, y Z : 4n + 3a + b 1ax + 1axy + 3a + 9by, y} R[1a, 1a, 3a + b], 4n + 3a + b R[1a, 4a, 43a + b], 4n + 3a + b R[1a, 1a, 3a + 9b], 4n + 3a + b R[1a, 4a, 43a + 9b], 4n + 3a + b

12 9 Z H Sun Using 11 we see that [1a, 4a, 43a + b] [1a, 0, 4b] and [1a, 4a, 43a + 9b] [1a, 0, 36b] Thus the first part follows Now assume 4 a If 4n + 3a + b 3ax + b6y 1 for some x, y Z, then clearly x Thus, rn ax x/ + b3y y/ r4n + 3a + b 3ax 1 + b6y 1 r4n + 3a + b 3ax + b6y 1 { x, y Z : 4n + 3a + b 3ax + by, y, 3 y} { x, y Z : 4n + 3a + b 3ax + by, y} { x, y Z : 4n + 3a + b 3ax + b3y, y} R[3a, 0, b], 4n + 3a + b R[3a, 0, 4b], 4n + 3a + b R[3a, 0, 9b], 4n + 3a + b R[3a, 0, 36b], 4n + 3a + b This completes the proof Theorem 1 Let n N Then rn x x/ + 3y y/ δ6n + 1, 3 Proof From Lemmas 1 and 1 we see that rn x x/ + 3y y/ This yields the result Theorem Let n N Then R[1, 1, 4], 4n + 4 R[1, 0, 4], 4n + 4 R[3, 3, 1], 6n + 1 R[3, 0, 1], 6n + 1 N6n + 1, 3 N6n + 1, 1 6δ6n + 1, 3 δ6n + 1, 3 rn x x + 3y y/ δ4n + 7, 6, rn x x/ + 3y y δ4n +, 6 Proof For x, y Z it is clear that x + 6y mod 6 and x + 3y 1 mod 6 Since H 4 {[1, 0, 6], [, 0, 3]} and f 4 1, using Lemmas 1 and 1 we have rn x x + 3y y/ R[6, 0, 1], 4n + 7 N4n + 7, 4 δ4n + 7, 4

13 Number of representations of n by ax + byy 1/ 93 and rn x x/ + 3y y R[3, 0, ], 4n + R[3, 0, 8], 4n + R[, 0, 3], 4n + N4n +, 4 δ4n +, 4 This yields the result Theorem 3 Let n N Then rn x x/ + 33y y/ δ4n + 1, 1 if 3 n 1, δ4n + 1, 1 if 3 n 1 and 9 n, 0 if 9 n Proof Putting a 1 and b 3 in Lemma 1 we see that rn x x/ + 33y y/ R[1, 1, 6], 4n + 6 R[1, 1, 30], 4n + 6 R[,, 1], 4n + 1 R[,, ], 4n + 1 By Lemma 1 we have R[,, 1], 4n+1 N4n+1, 4 4δ4n+1, 1 As H 36 {[1, 0, 9], [,, ]}, by [, Theorem 93] and 1 we have δ4n + 1, 1 if 3 n 1, 1 R[,, ], 4n + 1 4δ4n + 1/9, 1 if 9 n, 0 otherwise If 9 n and 4n α n 0 3 n 0, then δ4n + 1, 1 δ4n + 1/9, α α 0 n 0 4n+1, 4n+1/9 Now putting all the above together we obtain the result Theorem 4 Let n N Then rn 3x x/ + 3y y/ δ1n +, 1 Proof Putting a 3 and b 1 in Lemma 1 we have rn 3x x/ + 3y y/ R[36, 36, 10], 4n + 10 R[18, 18, ], 1n + By 11, we have [18, 18, ] [, 18, 18] [,, ] [,, ] Thus R[18, 18, ], 1n + R[,, ], 1n + Now applying 1 we deduce the result

14 94 Z H Sun Theorem Let n N, a {1, } and 3n+a+3/4 α n 0 n 0 Then r n a x x + a 3y y 1 n 0 Proof Set b /a It is nown that H 60 {[1, 0, 1], [3, 0, ]} and f 60 For x, y Z we see that x +1y mod 3 and 3x +y 1 mod 3 Thus, using Lemmas 1 and 1 we have rn ax x/ + b3y y/ R[3a, 0, b], 4n + 3a + b R[3a, 0, 4b], 4n + 3a + b R[3a, 0, b], 4n + 3a + b R[1a, 0, 4b], 4n + 3a + b R[3a, 0, b], 4n + 3a + b R[3a, 0, b], 6n + 3a + b/4 N4n + 3a + b, 60 N6n + 3a + b/4, n+a+3/4 6n+3a+b/4 1 1 α + α n 0 n 0 This yields the result 6n+3a+b/4/4 Theorem 6 Let n N and b {7, 11, 19, 31, 9} Then rn x x/ + b3y y/ δ1n + b + 3/, 3b, rn bx x/ + 3y y/ δ1n + 3b + 1/, 3b 1 Proof As b 3 mod 4 we see that 3 + b 3b + 1 mod 4 Thus, R[3, 0, 4b], 4n b 0 and R[3b, 0, 4], 4n + 3b Hence, using Lemma 1 we get and rn x x/ + b3y y/ R[3, 0, b], 4n b R[3, 0, 4b], 4n b R[3, 0, 9b], 4n b + R[3, 0, 36b], 4n b R[3, 0, b], 4n b rn bx x/ + 3y y/ R[3b, 0, 1], 4n + 3b + 1 R[3b, 0, 4], 4n + 3b + 1 R[3b, 0, 9], 4n + 3b R[3b, 0, 36], 4n + 3b + 1 R[1, 0, 3b], 4n + 3b + 1

15 Number of representations of n by ax + byy 1/ 9 It is nown that we have H 1b {[1, 0, 3b], [3, 0, b], [,, 3b+1/], [6, 6, b + 3/]} and f 1b 1 One can easily see that 4n b cannot be represented by x + 3by, x + xy + 3b+1 y and 6x + 6xy + b+3 y, and 4n + 3b + 1 cannot be represented by 3x + by, x + xy + 3b+1 y and 6x + 6xy + b+3 y Thus, applying Lemma 1 we have and R[3, 0, b], 4n b N4n b, 1b δ4n b, 1b δ1n + b + 3/, 3b R[1, 0, 3b], 4n + 3b + 1 N4n + 3b + 1, 1b δ4n + 3b + 1, 1b δ1n + 3b + 1/, 3b Now combining all the above we deduce the result Theorem 7 Let n N, m {, 7, 13, 17} and a {1,, m, m} Then r n a x x + m a 3y y 6m 4n+3a+m/a Proof If 4n + 3a + m/a 3ax + 8m a y for some x, y Z, then x and so m a 1 4y m a 8m a y 0 mod 8 This is impossible So R[3a, 0, 8m/a], 4n + 3a + m/a 0 Hence, using Lemma 1 we see that r n a x x + m a 3y y R[3a, 0, m/a], 4n + 3a + m/a It is nown that H 4m {[1, 0, 6m], [, 0, 3m], [3, 0, m], [6, 0, m]} and f 4m 1 Let n N with n, 6 1 It is easily seen that n x + 6my n 1, 1 + 6m mod 4, n x + 3my n 3m +, 3m + 8 mod 4, n 3x + my n m + 3, 8m + 3 mod 4, n 6x + my n m, m + 6 mod 4 Since 1, 1 + 6m, 3m +, 3m + 8, m + 3, 8m + 3, m, m + 6 are distinct modulo 4, we see that n is represented by at most one class in H 4m As 4n+3a+m/a, 6 1, we see that 4n+3a+m/a cannot be represented by any class K H 4m with K [3a, 0, m/a] From the above and Lemma 1 we derive

16 96 Z H Sun r n a x x + m a 3y y R[3a, 0, m/a], 4n + 3a + m/a So the theorem is proved Theorem 8 Let n N Then N4n + 3a + m/a, 4m δ4n + 3a + m/a, 4m rn 1x x/ + 3y y/ δ1n + 3,, rn 3x x/ + 3y y/ δ1n + 7, Proof Suppose 1n + 3 α n 1 with n 1 From Lemma 1 and the fact that 4n + 46 mod 4 we have R[4, 0, 4], 4n and so rn 1x x/ + 3y y/ R[4, 0, 1], 4n + 46 R[4, 0, 4], 4n + 46 R[4, 0, 9], 4n R[4, 0, 36], 4n + 46 R[4, 0, 1], 4n + 46 R[1, 0, 4], α n 1 Since 4n mod 3, applying the above and Lemma 4 we see that { r n 1 x x + 3y y δn1, 0 if n 1 ± mod, 0 if n 1 ±1 mod If n 1 ±1 mod, as n 1 α n 1 1n mod 4 we see that 0 n 1 n 1 n 1 n 1 1 and so δn 1, 0 0 by 1 Since δn 1, 0 δn 1, 0, we always have r n 1 x x + 3y y 0 n 1 1n+3 Now suppose 1n + 7 β n with n From Lemma 1 and the fact that 4n + 14 mod 4 we have R[9, 0, 0], 4n and so rn 3x x/ + 3y y/ R[9, 0, ], 4n + 14 R[9, 0, 0], 4n + 14 R[9, 0, 4], 4n R[9, 0, 180], 4n + 14 R[9, 0, ], 4n + 14 R[, 0, 9], β n Since 4n + 14 mod 3, applying the above and Lemma 4 we see that { r n 3 x x + 3y y δn, 0 if n ± mod, 0 if n ±1 mod If n ±1 mod, as n α n 1n mod 4 we see that n n n 1 and so n 0 0 by 1 0 n

17 Number of representations of n by ax + byy 1/ 97 Since δn, 0 δn, 0, we always have r n 3 x x + 3y y 0 n The proof is now complete 1n+7 Theorem 9 Let n N and 4n+3 α n 0 n 0 If n 0 ±1 mod, then r n x x + 1 3y y r n x x + 3 3y y 0 If n 0 ± mod, then r n x x + 1 3y y r n x x + 3 3y y where n 1 is given by n 0 3 β n 1 3 n 1 and δn 1, if 9 n 6, 0 if 3 n, δn 0, otherwise, δn 1, if 9 n 6, 0 if 3 n 1, δn 0, otherwise, Proof By Lemma 1 and the fact that 4n + 18 mod 4 we have rn x x/ + 13y y/ R[3, 0, 1], 4n + 18 R[3, 0, 60], 4n + 18 R[3, 0, 13], 4n R[3, 0, 40], 4n + 18 R[3, 0, 1], 4n + 18 R[3, 0, 13], 4n + 18 R[1, 0, ], 8n + 6 R[1, 0, 4], 8n + 6 rn x x/ + 33y y/ R[1, 0, 3], 4n + 18 R[1, 0, 1], 4n + 18 R[1, 0, 7], 4n R[1, 0, 108], 4n + 18 R[1, 0, 3], 4n + 18 R[1, 0, 7], 4n + 8 R[1, 0, ], 8n + 6 R[, 0, 9], 8n + 6 For m N, m x + y x, y Z implies x and m x/ + y Thus R[1, 0, ], m R[1, 0, ], m Hence R[1, 0, ], 8n + 6 R[1, 0, ], α n 0 R[1, 0, ], α 1 n 0 R[1, 0, ], n 0

18 98 Z H Sun If n 0 ±1 mod, then x n 0 mod is insolvable and so R[1, 0, ], 8n + 6 R[1, 0, ], n 0 0 Hence, by the above we have rn x x/ + 13y y/ 0 and rn x x/ + 33y y/ 0 Now we assume n 0 ± mod Then n 0 ±4 mod and so n 0 cannot be represented by x + xy + 3y Since H 0 {[1, 0, ], [,, 3]}, from the above and Lemma 1 we see that R[1, 0, ], 8n + 6 R[1, 0, ], n 0 Nn 0, 0 δn 0, 0 From Lemma 4 we now that δn 0 /9, 0 if 9 n 6, R[1, 0, 4], 8n + 6 δn 0, 0 if 3 n, 0 otherwise, δn 0 /9, 0 if 9 n 6, R[, 0, 9], 8n + 6 δn 0, 0 if 3 n 1, 0 otherwise Since δn 0, 0 δn 0,, δn 0 /9, 0 δn 0 /9, and n 0 3 β n 1 3 n 1, we see that for n 6 mod 9, δn 0, 0 δn 0 /9, 0 n 0, n 0 /9 n 1 3 β β 1 n 1 Now putting all the above together we obtain the result Lemma [4, Theorem 61] Let i, n N, i 30 and in α i 3 β i γ i n 0 with n 0, 30 1 Let m {7, 11, 3} and [i, 0, 1m/i] if i, [ A i i, i, 1 i + 1m ] if i i/ Then RA i, n > 0 if and only if ord q n 0 for every prime q with 1m q 1, 1 n 0 1 m+1/4α i+β i, n α i+[m+/3]β i +γ i and n 0 1 α i+β i +[m 3/]γ i Moreover, if RA i, n > 0, then RA i, n 1 + ord p n 0 1m p 1 Theorem 10 Let n N, m {7, 11, 3}, a {1,, m, m} and 1n+3a+m/a/ γ n 0 n 0 Then n is represented by ax x/+ m a 3y y/ if and only if ord q n 0 for every prime q with 1m q 1

19 and Number of representations of n by ax + byy 1/ 99 n0 1 [m 3/]γ if a 1, 1 [m 3/]γ+1 if a, 1 1+[m 3/]γ+1 if a m, 1 1+[m 3/]γ if a m Moreover, if the above conditions hold, then r n ax x/ + m a 3y y/ 1 + ord p n 0 1m p 1 Proof As 3a + m/a mod 4, we see that R[3a, 0, 0m/a], 4n + 3a + m/a R[1a, 0, 0m/a], 4n + 3a + m/a 0 Hence, using Lemma 1 we get r n a x x + m a 3y y R[3a, 0, m/a], 4n + 3a + m/a From 1n + 3a + m/a/ γ n 0 we deduce that n 0 3a + m/a/ a3 + m/ am 1/ mod 4 and 1 γ n 0 γ n 0 m/a am mod 3 Thus 1 1 m 1 a/4, n 0 n0 am 1 γ 3 3 a 1 [m+/3]+γ 3 For a 1 we have 34n m 3 γ n 0 For a we have 14n m 3 γ+1 n 0 For a m we have 4n + + 3m γ+1 n 0 For a m we have 4n m γ n 0 Thus applying the above and Lemma we deduce the result Using Lemma 1 and [4, Theorems 6 and 63] one can similarly prove the following result Theorem 11 Let n N, m {13, 17}, a {1, 7, m, 7m} and 1n + 3a + 7m/a/ 7 γ n 0 7 n 0 Then n is represented by ax x/ + 7m a 3y y/ if and only if ord q n 0 for every prime q with 1m q 1 and { 3,, 6 mod 7 if a 1, 7, n 0 1,, 4 mod 7 if a m, 7m Moreover, if the above conditions hold, then r n ax x/ + 7m a 3y y/ 1 + ord p n 0 1m p 1 Let µn denote the Möbius function Using Lemma 1 and [4, Theorem 81] one can similarly deduce the following result Theorem 1 Let a, n N, a 4 and 1n+3a+4/a/ γ 7 δ n 0 with n 0, 3 1 Then n is represented by ax x/ + 4 a 3y y/

20 100 Z H Sun if and only if ord q n 0 for every prime q with 136 q 1 and n 0 n γ+δ µa Moreover, if the above conditions hold, then r n ax x/ + 4 a 3y y/ 136 p ord p n 0 Remar 1 Let n N and 4n + 1 α n 0 3 β n 1 with n 0 and 3 n 1 By [, Theorem 44] we have r n 3x x + 3y y 1 + n 0 Suppose n 1 α n Then n and n 0 3 β n Since n 0 1 mod 4 and n 1 1 β mod 4, we have n 0 n 0 n 0 3 β n n and n 1 n By 1, n 0 implies n 0 n p n p ordp n 1 Thus, by we have 3 r n 3x x + 3y y n n 1 Acnowledgements The author is supported by the Natural Sciences Foundation of China grant n 1 References [1] D A Cox, Primes of the Form x + ny : Fermat, Class Field Theory, and Complex Multiplication, Wiley, New Yor, 1989 [] Z H Sun, The expansion of Q 1 1 qa 1 q b, Acta Arith , 11 9 [3], On the number of representations of n by axx 1/ + byy 1/, J Number Theory , [4], Binary quadratic forms and sums of triangular numbers, Acta Arith , 7 97 [] Z H Sun and K S Williams, On the number of representations of n by ax + bxy + cy, ibid 1 006, [6],, Ramanujan identities and Euler products for a type of Dirichlet series, ibid 1 006, Zhi-Hong Sun School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu 3001, PR China zhihongsun@yahoocom Received on and in revised form on