Colloq. Math. 145(2016), no. 1, ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS. 1. Introduction. x(x 1) (1.1) p m (x) = (m 2) + x.

Transcription

1 Colloq. Math. 145(016), no. 1, ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS FAN GE AND ZHI-WEI SUN Abstract. For m = 3, 4,... those p m (x) = (m )x(x 1)/ + x with x Z are called generalized m-gonal numbers. Sun [13] studied for what values of positive integers a, b, c the sum ap 5 + bp 5 + cp 5 is universal over Z (i.e., any n N = {0, 1,,...} has the form ap 5 (x) + bp 5 (y) + cp 5 (z) with x, y, z Z). We prove that p 5 + bp 5 + 3p 5 (b = 1,, 3, 4, 9) and p 5 + p 5 + 6p 5 are universal over Z, as conjectured by Sun. Sun also conjectured that any n N can be written as p 3 (x) + p 5 (y) + p 11 (z) and 3p 3 (x) + p 5 (y) + p 7 (z) with x, y, z N; in contrast, we show that p 3 + p 5 + p 11 and 3p 3 + p 5 + p 7 are universal over Z. Our proofs are essentially elementary and hence suitable for general readers. For m = 3, 4,... we set 1. Introduction x(x 1) (1.1) p m (x) = (m ) + x. Those p m (n) with n N = {0, 1,,...} are the well-known m-gonal numbers (or polygonal numbers of order m). We call those p m (x) with x Z generalized m-gonal numbers. Note that (generalized) 3-gonal numbers are triangular numbers and (generalized) 4-gonal numbers are squares of integers. In 1638, Fermat asserted that each n N can be written as the sum of m polygonal numbers of order m. This was proved by Lagrange, Gauss and Cauchy in the cases m = 4, m = 3 and m 5 respectively (see Moreno and Wagstaff [10, pp ] or Nathanson [11, Chapter 1, pp. 3-34]). The generalized pentagonal numbers play a crucial role in Euler s famous recurrence for the partition function. For a, b, c Z + = {1,, 3,...} and i, j, k {3, 4,...}, Sun [13] called the sum ap i + bp j + cp k universal over N (resp., over Z) if for any n N the equation n = ap i (x)+bp j (y)+cp k (z) has solutions over N (resp., over Z). In 186 Liouville (cf. [4, p. 3]) determined all those universal ap 3 + bp 3 + cp 3. The second author [1] initiated the determination of those universal sums 010 Mathematics Subject Classification. Primary 11E5; Secondary 11B75, 11D85, 11E0, 11P3. Key words and phrases. Polygonal numbers, representations of integers, ternary quadratic forms. 1

2 F. GE AND Z.-W. SUN ap i + bp j + cp k with {i, j, k} = {3, 4}, and this project was completed via [1, 5, 9]. For almost universal sums ap i + bp j + cp k with {i, j, k} {3, 4}, see [8, 1, ]. It is known that generalized hexagonal numbers are identical with triangular numbers (cf. [6] or [13, (1.3)]). The second author recently established the following result. Theorem 1.1. (Sun [13, Theorem 1.1]) Suppose that ap k + bp k + cp k is universal over Z, where k {4, 5, 7, 8, 9,...}, a, b, c Z + and a b c. Then k = 5, a = 1 and (b, c) is among the following 0 ordered pairs: (1, c) (c {1,, 3, 4, 5, 6, 8, 9, 10}), (, ), (, 3), (, 4), (, 6), (, 8), (3, 3), (3, 4), (3, 6), (3, 7), (3, 8), (3, 9). Guy [6] realized that p 5 +p 5 +p 5 is universal over Z, and Sun [13] proved that the sums p 5 + p 5 + p 5, p 5 + p 5 + 4p 5, p 5 + p 5 + p 5, p 5 + p 5 + 4p 5, p 5 + p 5 + 5p 5, p 5 + 3p 5 + 6p 5 are universal over Z. So the converse of Theorem 1.1 reduces to the following conjecture of Sun. Conjecture 1.. (Sun [13, Remark 1.]) The sum p 5 +bp 5 +cp 5 is universal over Z if the ordered pair (b, c) is among (1, 3), (1, 6), (1, 8), (1, 9), (1, 10), (, 3), (, 6), (, 8), (3, 3), (3, 4), (3, 7), (3, 8), (3, 9). Our following result confirms this conjecture for six ordered pairs (b, c) for the first time. Theorem 1.3. For (b, c) = (1, 3), (, 3), (, 6), (3, 3), (3, 4), (3, 9), the sum p 5 + bp 5 + cp 5 is universal over Z. Remark. This result appeared in the initial preprint version of this paper posted to arxiv in 009. Sun [13] investigated those universal sums ap i + bp j + cp k over N. By Sun [13, Conjectures 1.10 and 1.13], p 3 + p 5 + p 11 and 3p 3 + p 5 + p 7 should be universal over N. Though we cannot prove this, we are able to show the following result.

3 ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS 3 Theorem 1.4. The sums p 3 + p 5 + p 11 and 3p 3 + p 5 + p 7 are universal over Z. Theorems 1.3 and 1.4 will be shown in Sections and 3 respectively. Our proofs are essentially elementary and hence suitable for general readers.. Proof of Theorem 1.3 Lemma.1. (Sun [13, Lemma 3.]) Let w = x + 3y 4 (mod 8) with x, y Z. Then there are odd integers u and v such that w = u + 3v. Lemma.. Let w = x + 3y with x, y odd and 3 x. Then there are integers u and v relatively prime to 6 such that w = u + 3v. Proof. It suffices to consider the case 3 y. Without loss of generality, we may assume that x y (mod 4) (otherwise we may use y instead of y). Thus (x y)/ and (x + 3y)/ = (x y)/ + y are odd. Observe that ( ) ( ) x + 3y x y (.1) x + 3y = + 3. As 3 x and 3 y, neither (x y)/ nor (x+3y)/ is divisible by 3. Therefore u = (x + 3y)/ and v = (x y)/ are relatively prime to 6. This concludes the proof. Lemma.3. (Jacobi s identity) We have ( ) ( ) x + y z x y (.) 3(x + y + z ) = (x + y + z) We need to introduce some more notation. For a, b, c Z +, we set E(ax + by + cz ) = {n N : n ax + by + cz for any x, y, z Z}. Proof of Theorem 1.3. Let b, c Z +. For n N we have n = p 5 (x) + bp 5 (y) + cp 5 (z) = 3x x + b 3y y + c 3z z 4n + b + c + 1 = (6x 1) + b(6y 1) + c(6z 1). If w Z is relatively prime to 6, then w or w is congruent to 1 modulo 6. Thus, p 5 + bp 5 + cp 5 is universal over Z if and only if for any n N the equation 4n + b + c + 1 = x + by + cz has integral solutions with x, y, z relatively prime to 6. Below we fix a nonnegative integer n. (i) By Dickson [3, Theorem III], (.3) E(x + y + 3z ) = {9 k (9l + 6) : k, l N}.

4 4 F. GE AND Z.-W. SUN So 4n + 5 = u + v + 3w for some u, v, w Z. As 3w 5 (mod 4), u or v is odd. Without loss of generality we assume that u. Since v + 3w 5 u 4 (mod 8), by Lemma.1 we can rewrite v + 3w as s + 3t with s, t odd. Now we have 4n + 5 = u + s + 3t with u, s, t odd. By u + s 5 (mod 3), both u and s are relatively prime to 3. Applying Lemma. we can express s + 3t as y + 3z with y, z relatively prime to 6. Thus 4n + 5 = u + y + 3z with u, y, z relatively prime to 6. This proves the universality of p 5 + p 5 + 3p 5 over Z. (ii) By Dickson [3, Theorem X], (.4) E(x + y + 3z ) = {4 k (16l + 10) : k, l N}. So 4n + 6 = u + v + 3w for some u, v, w Z. Clearly v and w have the same parity. Thus 4 v + 3w and hence u 6 (mod 4). So u is odd and v + 3w 6 u 4 (mod 8). By Lemma.1 we can rewrite v + 3w as s + 3t with s, t odd. Now we have 4n + 6 = u + s + 3t with u, s, t odd. Note that s + u > 0 and s + u 0 (mod 3). By [7, p. 173] or [13, Lemma.1], we can rewrite s + u as x + y with x and y relatively prime to 3. As x + y = s + u 3 (mod 8), both x and y are odd. By Lemma., x + 3t = r + 3z for some integers r, z Z relatively prime to 6. Thus 4n + 6 = r + y + 3z with r, y, z relatively prime to 6. It follows that p 5 + p 5 + 3p 5 is universal over Z. (iii) By Dickson [3, Theorem IV], (.5) E(x + 3y + 3z ) = {9 k (3l + ) : k, l N}. So 4n + 7 = u + 3v + 3w for some u, v, w Z. Since u 7 (mod 4), without loss of generality we assume that w. As u + 3v 7 3w 4 (mod 8), by Lemma.1 there are odd integers s and t such that u +3v = s +3t. Thus 4n+7 = s +3t +3w with s, t, w odd. Clearly, s is relatively prime to 6. By Lemma., s + 3t = x 0 + 3y for some integers x 0 and y relatively prime to 6, and x 0 + 3w = x + 3z for some integers x and z relatively prime to 6. Therefore 4n + 7 = x + 3y + 3z with x, y, z relatively prime to 6. This proves the universality of p 5 + 3p 5 + 3p 5 over Z. (iv) By [13, Theorem 1.7(iii)], 4n+8 = u +v +3w for some u, v, w Z with w. Clearly u v (mod ). Without loss of generality, we assume that u = r with r Z. Since (r) + v 8 (mod 3), both r and v are relatively prime to 3. As v and w are odd, v + 3w 4 (mod 8) and hence r is odd. By Lemma., we can rewrite v + 3w as x + 3y with x and y relatively prime to 6. Note that 4n + 8 = 4r + v + 3w = x + 3y + 4r

5 ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS 5 with x, y, r relatively prime to 6. It follows that p 5 + 3p 5 + 4p 5 is universal over Z. (v) By (.3), 4n + 13 = u + v + 3w for some u, v, w Z. Since 3w 13 1 (mod 4), without loss of generality we may assume that u is odd. As v + 3w 13 u 4 (mod 8), by Lemma.1 we can rewrite v +3w as s +3t with s and t odd. Thus 4n+13 = u +s +3t with u, s, t odd. Since u + s 13 1 (mod 3), without loss of generality we may assume that 3 u and s = 3r with r Z. By Lemma., u +3t = x +3y0 for some integers x and y 0 relatively prime to 6, also y0 + 3r = y + 3z for some integers y and z relatively prime to 6. Thus 4n+13 = x +3y0 +9r = x + 3y + 9z with x, y, z relatively prime to 6. This proves the universality of p 5 + 3p 5 + 9p 5 over Z. (vi) By the Gauss-Legendre theorem (cf. [11, pp. 17-3]), 8n + 3 = x + y + z for some odd integers x, y, z. Without loss of generality we may assume that x y (mod 4). By Jacobi s identity (.), we have 3(8n+3) = u +v +6w, where u = x+y +z, v = (x+y)/ z and w = (x y)/ are odd integers. As u + v is a positive integer divisible by 3, by [7, p. 173] or [13, Lemma.1] we can write u + v = a + b with a and b relatively prime to 3. Since a +b = u +v 3 (mod 8), both a and b are odd. By Lemma., b + 3w = c + 3d for some integers c and d relatively prime to 6. Thus 4n + 9 = a + b + 6w = a + c + 6d with a, c, d relatively prime to 6. It follows that p 5 + p 5 + 6p 5 is universal over Z. In view of the above, we have completed the proof of Theorem Proof of Theorem 1.4 Proof of Theorem 1.4. (i) Let n N. By part (v) in the proof of Theorem 1.3, there are integers u, v, w Z relatively prime to 6 such that 7n + 61 = 4(3n + ) + 13 = 9u + 3v + w. Clearly w 61 3v 7 (mod 9) and hence w ±7 (mod 9). So there are x, y, z Z such that 7n + 61 = 9(x + 1) + 3(6y 1) + (18z 7) and hence n = p 3 (x) + p 5 (y) + p 11 (z). (Note that p 11 (x) = 9(x x)/ + x = (9x 7x)/.) (ii) Let n N. It is easy to see that n = 3p 3 (x) + p 5 (y) + p 7 (z) 10n + 77 = 5(3(x + 1)) + 5(6y 1) + 3(10z 3).

6 6 F. GE AND Z.-W. SUN Suppose 10n + 77 = 5x + 5y + 3z for some x, y, z Z with z odd. Then x + y 77 3z (mod 4) and hence x and y are odd. Note that 3z 77 1 (mod 5) and hence z ±3 (mod 10). As 5x + 5y 77 5 (mod 3), exactly one of x and y is divisible by 3. Thus there are u, v, w Z such that 10n + 77 = 5(3(u + 1)) + 5(6v 1) + 3(10w 3). By the above, to prove the universality of 3p 3 + p 5 + p 7 over Z, we only need to show that 10n + 77 = 5x + 5y + 3z for some x, y, z Z with z odd. By (.3), there are u, v, w Z such that 10n + 77 = u + v + 3w. As 3w 77 1 (mod 4), u or v is odd, say, u. As v + 3w 77 u 4 (mod 8), by Lemma.1 we may assume that v and w are odd without loss of generality. We claim that 10n + 77 = a + b + 3c for some odd integers a, b, c with c ± (mod 5). This holds if w ± (mod 5). Suppose that w ± (mod 5). If w ±1 (mod 5), then u + v 77 3w 1 (mod 5) and hence u or v is divisible by 5. If w 0 (mod 5), then u + v 77 (mod 5) and hence u v 1 (mod 5). Without loss of generality, we assume that one of v and w is divisible by 5 and the other one is congruent to 1 or 1 modulo 5, we may also suppose that v w (mod 4) (otherwise we may use w instead of w). By the identity (.1), ( ) ( ) v + 3w v w v + 3w = + 3. Note that both (v w)/ and (v + 3w)/ = (v w)/ + w are odd. Also, (v w)/ is congruent to or modulo 5. This confirms the claim. By the above, there are odd integers a, b, c Z with c ± (mod 5) such that 10n + 77 = a + b + 3c. Since 3c 77 (mod 5), we have 5 a + b and hence a (b) (mod 5). Without loss of generality we assume that a b (mod 5). Then x = (a + b)/5 and y = (a b)/5 are odd integers, and a + b = (x + y) + (x y) = 5(x + y ). Now we have 10n + 77 = 5(x + y) + 3c with x, y, c odd. This concludes our proof of Theorem 1.4. Acknowledgements. The authors would like to thank Dr. Hao Pan for his helpful comments. The second author was supported by the National Natural Science Foundation of China (grant ).

7 ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS 7 References [1] W. K. Chan and A. Haensch, Alomost universal ternary sums of squares and triangular numbers, in: Quadratic and Higher Degree Forms, Dev. Math., Vol. 31, Springer, New York, 013, pp [] W. K. Chan and B.-K. Oh, Almost universal ternary sums of triangular numbers, Proc. Amer. Math. Soc. 137 (009), [3] L. E. Dickson, Integers represented by positive ternary quadratic forms, Bull. Amer. Math. Soc. 33 (197), [4] L. E. Dickson, History of the Theory of Numbers, Vol. II, AMS Chelsea Publ., [5] S. Guo, H. Pan and Z.-W. Sun, Mixed sums of squares and triangular numbers (II), Integers 7 (007), #A56, 5pp (electronic). [6] R. K. Guy, Every number is expressible as the sum of how many polygonal numbers? Amer. Math. Monthly 101 (1994), [7] B. W. Jones and G. Pall, Regular and semi-regular positive ternary quadratic forms, Acta Math. 70 (1939), [8] B. Kane and Z.-W. Sun, On almost universal mixed sums of squares and triangular numbers, Trans. Amer. Math. Soc. 36 (010), [9] B.-K. Oh and Z.-W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 19 (009), [10] C. J. Moreno and S. S. Wagstaff, Sums of Squares of Integers, Chapman & Hall/CRC, Boca Raton, FL, 006. [11] M. B. Nathanson, Additive Number Theory: The Classical Bases, Grad. Texts in Math., vol. 164, Springer, New York, [1] Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 17 (007), [13] Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (015), (Fan Ge) Department of Mathematics, University of Rochester, Rochester, NY 1467, U.S.A. address: fange.math@gmail.com (Zhi-Wei Sun, corresponding author) Department of Mathematics, Nanjing University, Nanjing 10093, People s Republic of China address: zwsun@nju.edu.cn