aa r X i v : . [ m a t h . N T ] D ec Preprint, arXiv:1504.01608
NATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Zhi-Wei SUN
Department of Mathematics, Nanjing UniversityNanjing 210093, People’s Republic of [email protected] http://math.nju.edu.cn/ ∼ zwsun Abstract.
Let a, b, c be positive integers. It is known that there are infinitelymany positive integers not representated by ax + by + cz with x, y, z ∈ Z . Incontrast, we conjecture that any natural number is represented by ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ with x, y, z ∈ Z if ( a, b, c ) = (1 , , , (2 , , ⌊ T x /a ⌋ + ⌊ T y /b ⌋ + ⌊ T z /c ⌋ with x, y, z ∈ Z , where T x denotes thetriangular number x ( x + 1) /
2. We confirm this general conjecture in some specialcases; in particular, we prove that (cid:26) x + y + (cid:22) z (cid:23) : x, y, z ∈ Z and 2 ∤ y (cid:27) = { , , , . . . } and (cid:26)(cid:22) x m (cid:23) + (cid:22) y m (cid:23) + (cid:22) z m (cid:23) : x, y, z ∈ Z (cid:27) = { , , , . . . } for m = 5 , , . We also pose several conjectures for further research; for example, we conjecturethat any integer can be written as x − y + z , where x , y and z are positiveintegers.
1. Introduction
Let N = { , , , . . . } be the set of all natural numbers (nonnegative integers).A well-known theorem of Lagrange asserts that each n ∈ N can be written asthe sum of four squares. It is known that for any a, b, c ∈ Z + = { , , , . . . } there are infinitely many positive integers not represented by ax + by + cz with x, y, z ∈ Z .A classical theorem of Gauss and Legendre states that n ∈ N can be writtenas the sum of three squares if and only if it is not of the form 4 k (8 l + 7) with Mathematics Subject Classification . Primary 11E25; Secondary 11B75, 11D85, 11E20,11P32.
Keywords . Representations of integers, the floor function, squares, polygonal numbers.Supported by the National Natural Science Foundation (Grant No. 11171140) of China. ZHI-WEI SUN k, l ∈ N . Consequently, for each n ∈ N there are x, y, z ∈ Z such that8 n + 3 = (2 x + 1) + (2 y + 1) + (2 z + 1) , i.e. , n = x ( x + 1)2 + y ( y + 1)2 + z ( z + 1)2 . Those T x = x ( x +1) / x ∈ Z are called triangular numbers . For m = 3 , , . . . ,those m -gonal numbers (or polygonal numbers of order m ) are given by p m ( n ) := ( m − (cid:18) n (cid:19) + n = ( m − n − ( m − n n = 0 , , , . . . ) , and those p m ( x ) with x ∈ Z are called generalized m -gonal numbers . Cauchy’spolygonal number theorem states that for each m = 5 , , . . . any n ∈ N can bewritten as the sum of m polygonals of order m (see, e.g., [N96, pp. 3-35] and [MW,pp. 54-57].)For any k ∈ Z , we clearly have T k = (2 k + 1) −
18 = (cid:22) (2 k + 1) (cid:23) . As any natural number can be expressed as the sum of three triangular numbers,each n ∈ N can be written as ⌊ x / ⌋ + ⌊ y / ⌋ + ⌊ z / ⌋ with x, y, z ∈ Z . B. Farhi[F13] conjectured that any n ∈ N can be expressed the sum of three elements ofthe set {⌊ x / ⌋ : x ∈ Z } and showed this for n n as the sum of three squares.In [F] Farhi provided an elementary proof of the conjecture and made a furtherconjecture that for each a = 3 , , , . . . any n ∈ N can be written as the sum ofthree elements of the set {⌊ x /a ⌋ : x ∈ Z } . This general conjecture of Farhi hasbeen solved for a = 3 , , , , Conjecture 1.1.
Let a, b, c ∈ Z + with a b c . (i) If the triple ( a, b, c ) is neither (1 , , nor (2 , , , then for any n ∈ N thereare x, y, z ∈ Z such that n = (cid:22) x a (cid:23) + (cid:22) y b (cid:23) + (cid:22) z c (cid:23) = (cid:22) x a + y b (cid:23) + (cid:22) z c (cid:23) = (cid:22) x a (cid:23) + (cid:22) y b + z c (cid:23) . (1.1)(ii) For any n ∈ N , there are x, y, z ∈ Z such that n = (cid:22) T x a (cid:23) + (cid:22) T y b (cid:23) + (cid:22) T z c (cid:23) = (cid:22) T x a + T y b (cid:23) + (cid:22) T z c (cid:23) = (cid:22) T x a (cid:23) + (cid:22) T y b + T z c (cid:23) . (1.2) ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Moreover, if the triple ( a, b, c ) is not among (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (3 , , , then for any n ∈ N there are x, y, z ∈ Z such that n = (cid:22) x ( x + 1) a (cid:23) + (cid:22) y ( y + 1) b (cid:23) + (cid:22) z ( z + 1) c (cid:23) = (cid:22) x ( x + 1) a + y ( y + 1) b (cid:23) + (cid:22) z ( z + 1) c (cid:23) = (cid:22) x ( x + 1) a (cid:23) + (cid:22) y ( y + 1) b + z ( z + 1) c (cid:23) . (1.3)In this paper we establish some results in the direction of Conjecture 1.1. Theorem 1.1. (i)
For each m = 4 , , any n ∈ N can be written as x + (2 y ) + ⌊ z /m ⌋ with x, y, z ∈ Z . Also, any n ∈ Z + can be expressed as x + y + ⌊ z / ⌋ with x, y, z ∈ Z and ∤ y . (ii) For any δ ∈ { , } , any n ∈ Z + can be expressed as x + y + ⌊ z / ⌋ with x, y, z ∈ Z and y ≡ δ (mod 2) . (iii) For each m = 2 , , , , any n ∈ N can be written as x + y + ⌊ z /m ⌋ with x, y, z ∈ Z . Also, for each m = 3 , , we have (cid:26) x + y + (cid:22) z ( z + 1) m (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.4)(iv) For each m = 5 , , , we have (cid:26) x + (cid:22) y m (cid:23) + (cid:22) z m (cid:23) : x, y, z ∈ Z (cid:27) = (cid:26)(cid:22) x m (cid:23) + (cid:22) y m (cid:23) + (cid:22) z m (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.5)(v) We have (cid:26) T x + T y + (cid:22) T z (cid:23) : x, y, z ∈ Z (cid:27) = (cid:26)(cid:22) T x (cid:23) + (cid:22) T y (cid:23) + (cid:22) T z (cid:23) : x, y, z ∈ Z (cid:27) = N and (cid:26) x ( x + 1) + y ( y + 1) + (cid:22) z ( z + 1)4 (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.6) Remark x = (3 x ) /
9, Theorem 1.1(iii) with m = 9 implies that any n ∈ N can be written as ⌊ x / ⌋ + ⌊ y / ⌋ + ⌊ z / ⌋ with x, y, z ∈ Z . Theorem1.1(iv) confirms Farhi’s conjecture for a = 5 , ,
15. The author [S15a, Remark1.8] conjectured that for any n ∈ N we can write 20 n + 9 as 5 x + 5 y + (2 z + 1) with x, y, z ∈ Z ; it is easy to see that (1.4) for m = 5 follows from this conjecture.As { T x + 2 T y + T z : x, y, z ∈ Z } = N by Liouville’s result, any n ∈ N can bewritten as T x + T y + T z / x, y, z ∈ Z .As a supplement to parts (i)-(iii) of Theorem 1.1, we pose the following con-jecture. ZHI-WEI SUN
Conjecture 1.2. (i)
Let n ∈ Z + . Then, for any integer m > and δ ∈ { , } ,we have n = x + y + ⌊ z /m ⌋ for some x, y, z ∈ Z with y ≡ δ (mod 2) . (ii) For any integer m > , we have (cid:26) x + (2 y ) + (cid:22) z ( z + 1) m (cid:23) : x, y, z ∈ Z (cid:27) = N . For each m = 4 , , . . . , any positive integer n can be represented by x + y + ⌊ z ( z + 1) /m ⌋ with x, y, z ∈ Z and ∤ y .Remark { x + (2 y ) + T z : x, y, z ∈ Z } = { x + (2 y ) + 2 T z : x, y, z ∈ Z } = N (cf. [S07, Section 4]).For any a ∈ Z + , clearly (cid:26)(cid:22) x a (cid:23) : x ∈ Z (cid:27) ⊇ (cid:26)(cid:22) ( ax ) a (cid:23) = ax : x ∈ Z (cid:27) . Theorem 1.2. (i)
For each m = 2 , , , we have (cid:26) x + 2 y + (cid:22) z m (cid:23) : x, y, z ∈ Z (cid:27) = N . (ii) For each m = 3 , , , , we have (cid:26) x + 3 y + (cid:22) z m (cid:23) : x, y, z ∈ Z (cid:27) = N . (iii) We have (cid:26) x + 5 y + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = (cid:26) x + 6 y + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N . (iv) We have (cid:26) x + 2 y + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = (cid:26) x + 3 y + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N and (cid:26) x + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N . Our following conjecture involving the ceiling function is quite similar to Con-jecture 1.1.
ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Conjecture 1.3.
Let a, b, c ∈ Z + with a b c . (i) If the triple ( a, b, c ) is not among (1 , , , (1 , , , (1 , , , then for any n ∈ N there are x, y, z ∈ Z such that n = (cid:24) x a (cid:25) + (cid:24) y b (cid:25) + (cid:24) z c (cid:25) . (ii) We have (cid:26)(cid:24) T x a (cid:25) + (cid:24) T y b (cid:25) + (cid:24) T z c (cid:25) : x, y, z ∈ Z (cid:27) = N . Moreover, if the triple ( a, b, c ) is neither (1 , , nor (1 , , , then for any n ∈ N there are x, y, z ∈ Z such that n = (cid:24) x ( x + 1) a (cid:25) + (cid:24) y ( y + 1) b (cid:25) + (cid:24) z ( z + 1) c (cid:25) . We are also able to deduce some results similar to Theorems 1.1-1.2 in thedirection of Conjecture 1.3. Here we just collect few results of this type.
Theorem 1.3. (i)
For each m = 2 , , , , , , we have (cid:26)(cid:24) x m (cid:25) + (cid:24) y m (cid:25) + (cid:24) z m (cid:25) : x, y, z ∈ Z (cid:27) = N . (1.7)(ii) We have (cid:26) x + 3 y + (cid:24) z (cid:25) : x, y, z ∈ Z (cid:27) = (cid:26) x + 3 y + (cid:24) z (cid:25) : x, y, z ∈ Z (cid:27) = N . (1.8)(iii) For any n ∈ N , there are x, y, z ∈ Z such that n = x ( x + 1) + y ( y + 1)3 + (cid:24) z ( z + 1)3 (cid:25) . (1.9) Also, any n ∈ N can be written as x (3 x + 1) + y (3 y + 1) + ⌈ z ( z + 1) / ⌉ with x, y, z ∈ Z , and hence (cid:26)(cid:24) x ( x + 1)3 (cid:25) + (cid:24) y ( y + 1)3 (cid:25) + (cid:24) z ( z + 1)3 (cid:25) : x, y, z ∈ Z (cid:27) = N . (1.10) Remark x + 3 y + ⌊ z / ⌋ with x, y, z ∈ Z .Now we state another theorem. ZHI-WEI SUN
Theorem 1.4. (i)
For any integer a > , we have (cid:26)(cid:22) x + y + z a (cid:23) : x, y, z ∈ Z (cid:27) = N and (cid:26)(cid:22) x ( x + 1) + y ( y + 1) + z ( z + 1) a (cid:23) : x, y, z ∈ Z (cid:27) = N . (ii) Let a ∈ Z + . If a is odd, then any n ∈ N can be written as x + y + z + ⌊ a ( x + y + z ) ⌋ with x, y, z ∈ Z . If ∤ a , then any n ∈ N can be written as x + y + z + ⌊ a ( x + y + z ) ⌋ with x, y, z ∈ Z . (iii) For any n ∈ N , there are x, y, z ∈ Z such that n = p ( x )2 + (cid:24) p ( y )2 (cid:25) + (cid:24) p ( z )2 (cid:25) . Hence { s ( x ) + s ( y ) + s ( z ) : x, y, z ∈ Z } = N , (1.11) where s ( x ) := (cid:24) p ( − x )2 (cid:25) = x + (cid:6) . x (cid:7) . Remark m = 19 ,
20, we have 111 = x + y + z + ⌊ ( x + y + z ) /m ⌋ forany x, y, z ∈ Z .The generalized octagonal numbers p ( x ) = x (3 x −
2) ( x ∈ Z ) have someproperties similar to certain properties of squares. For example, recently theauthor [S16] showed that any n ∈ N can be written as the sum of four generalizedoctagonal numbers; this result is quite similar to Lagrange’s theorem on sums offour squares. Note that (cid:22) p ( x )2 m (cid:23) = (cid:22) p ( x ) + 14 m (cid:23) = (cid:22) p (1 − x )4 m (cid:23) and (cid:22) p ( x ) m (cid:23) = (cid:22) (3 x − m (cid:23) (1.12)for any m ∈ Z + and x ∈ Z . Theorem 1.5. (i) n ∈ N can be written as p ( x ) + p ( y ) + 2 p ( z ) with x, y, z ∈ Z if and only if n does not belong to the set (cid:26) k +2 q −
23 (4 k + 2) : k ∈ N and q ∈ Z + (cid:27) . Also, each nonnegative even number can be represented by p ( x ) + 2 p ( y ) + 4 p ( z ) with x, y, z ∈ Z . Consequently, (cid:26) p ( x ) + (cid:22) p ( y )2 (cid:23) + (cid:22) p ( z )2 (cid:23) : x, y, z ∈ Z (cid:27) = N (1.13) ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ and (cid:26)(cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.14)(ii) We have (cid:26) p ( x ) + p ( y ) + (cid:22) p ( z )2 (cid:23) : x, y, z ∈ Z (cid:27) = N , (1.15) hence (cid:26) p ( x ) + p ( y ) + (cid:22) p ( z )8 (cid:23) : x, y, z ∈ Z (cid:27) = N (1.16) and (cid:26)(cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.17)(iii) For n ∈ N there are x, y, z ∈ Z such that n = p ( x ) + p ( y ) + p ( z )4 . (1.18)(iv) We have (cid:26) p ( x ) + p ( y ) + (cid:22) p ( z )5 (cid:23) : x, y, z ∈ Z (cid:27) = N (1.19) and hence (cid:26)(cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ Z (cid:27) = N . (1.20)We are going to prove Theorems 1.1-1.2 in the next section, and show Theorems1.3-1.4 in Section 3. Section 4 is devoted to our proof of Theorem 1.5. We posesome further conjectures in Section 5.
2. Proofs of Theorems 1.1-1.2
Lemma 2.1.
Suppose that n ∈ Z + is not a power of two. Then there are x, y, z ∈ Z with | x | < n , | y | < n and | z | < n such that x + y + z = n .Proof . In 1907 Hurwitz (cf. [D99, p. 271]) showed that |{ ( x, y, z ) ∈ Z : x + y + z = n }| =6 Y p> (cid:18) p ord p ( n )+1 − p − − ( p +1) / p ord p ( m ) − p − (cid:19) , (2.1)where ord p ( n ) is the order of n at the prime p . Note that( ± n ) + 0 + 0 = 0 + ( ± n ) + 0 = 0 + 0 + ( ± n ) . As n has an odd prime p , by (2.1) we have |{ ( x, y, z ) ∈ Z : x + y + z = n }| > p ord p ( n )+1 − p ord p ( n ) p − > p > x, y, z ∈ Z with x , y , z = n such that x + y + z = n .This concludes the proof. (cid:3) ZHI-WEI SUN
Lemma 2.2. (i)
Let u and v be integers with u + v a positive multiple of .Then u + v = x + y for some x, y ∈ Z with ∤ xy . (ii) For any n ∈ N with n ≡ ± , we can write n as x + 5 y + z with x, y, z ∈ Z and ∤ z .Remark Lemma 2.3.
Let n > be an integer with n ≡ , or n ≡ , . Then we can write n as x + 5 y + z with x, y, z ∈ Z such that x y (mod 2) if n ≡ , , and ∤ y if n ≡ ,
19 (mod 40) .Proof . As n ≡ n ≡ n isthe sum of three squares. As n is not a power of two, in view of Lemma 2.1 wecan always write n as w + u + v with u, v, w ∈ Z and w , u , v = n . Withoutloss of generality, we assume that 2 ∤ w and u ≡ v (mod 2). Clearly, u ≡ v ≡ n ≡ w ≡ − n (mod 5), then u + v ≡ n (mod 5)and hence u ≡ v ≡ n (mod 5). If w ≡ n (mod 5), then u + v is a positivemultiple of 5 and hence by Lemma 2.2 we can write it as s + t , where s and t are integers with s ≡ − n (mod 5) and t ≡ n (mod 5). When n ≡ s + t = u + v ≡ s ≡ t ≡ | w , thenone of u and v is divisible by 5 and the other is congruent to n modulo 5.By the above, there always exist x, y, z ∈ Z with z ≡ n (mod 5) such that n = x + y + z and that 2 | z if n ≡ x ≡ − y ≡ ( ± y ) (mod 5). Without loss of generality, we assume that x ≡ y (mod 5) and hence2 x ≡ − y (mod 5). Set ¯ x = ( x − y ) / y = (2 x + y ) /
5. Then n = x + y + z = 5¯ x + 5¯ y + z . If n ≡ | z and hence ¯ x ¯ y (mod 2). If n ≡ z n (mod 4) and hence ¯ x or ¯ y is odd. This concludes the proof. (cid:3) Remark n > n ≡ , n = 5 x + 5 y + (2 z ) with x, y, z ∈ Z if n is nota square.For convenience, we define E ( f ( x, y, z )) := { n ∈ N : n = f ( x, y, z ) for any x, y, z ∈ Z } for any function f : Z → N . Proof of Theorem 1.1 . Let n be a fixed nonnegative integer.(i) By Dickson [D39, pp. 112-113], E (4 x + 16 y + z ) = [ k ∈ N { k + 2 , k + 3 , k + 12 } ∪ { k (8 l + 7) : k, l ∈ N } . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ So, there are x, y, z ∈ Z such that 4 n + 1 = 4 x + 16 y + z and hence n = x + (2 y ) + ⌊ z / ⌋ .For r ∈ { , } , if 6 n + r = 6 x + 24 y + z with x, y, z ∈ Z , then z ≡ r (mod 6)and n = x + (2 y ) + ⌊ z / ⌋ . By Dickson [D39, pp. 112-113], E (6 x + 24 y + z ) = [ k ∈ N { k + 3 , k + 5 , k + 12 } ∪ { k (3 l + 2) : k, l ∈ N } . If both 6 n +1 and 6 n +4 belong to this set, then one of them has the form 32 k +12and hence we get a contradiction since 32 k + 12 ± , n ≡ , n + 1 ≡ , n + 1 = 5 x + 5 y + z with x, y, z ∈ Z and x y (mod 2), thus x or y is odd and n = x + y + ⌊ z / ⌋ . By Dickson [D39, pp. 112-113], E ( x + y + 5 z ) = { k (8 l + 3) : k, l ∈ N } . If n ≡ n ≡ x, y, z ∈ Z such that n = x + y + 5 z = x + y + ⌊ (5 z ) / ⌋ and one of x and y is odd since 5 z ≡ z n (mod 4). If n ≡ n + 4 ≡
19 (mod 40) and hence by Lemma2.3 there are x, y, z ∈ Z with 2 ∤ y such that 5 n + 4 = 5( x + y ) + z and hence n = x + y + ⌊ z / ⌋ with y odd.(ii) By [D39, pp. 112-113], there are x, y, z ∈ Z such that 8 n +1 = 8 x +32 y + z and hence n = x + (2 y ) + ⌊ z / ⌋ .Suppose that n ∈ Z + . As conjectured by Sun [S07] and proved by Oh andSun [OS], there are x, y, z ∈ Z with y odd such that n = x + y + T z and hence n = x + y + ⌊ (2 z + 1) / ⌋ .(iii) If 2 n ≡ n
6∈ { k (8 l + 7) : k, l ∈ N } . If 2 n n + 1
6∈ { k (8 l + 7) : k, l ∈ N } . So, for some δ ∈ { , } , we have 2 n + δ k (8 l + 7) : k, l ∈ N } and hence (by the Gauss-Legendre theorem) 2 n + δ = x + y + z for some x, y, z ∈ Z with z ≡ δ (mod 2). Note that x ≡ y (mod 2)and 2 n + δ = 2 (cid:18) x + y (cid:19) + 2 (cid:18) x − y (cid:19) + z . Therefore, n = (cid:18) x + y (cid:19) + (cid:18) x − y (cid:19) + z − δ (cid:18) x + y (cid:19) + (cid:18) x − y (cid:19) + (cid:22) z (cid:23) . By Dickson [D39, pp. 112-113], E (3 x + 3 y + z ) = { k (3 l + 2) : k, l ∈ N } . So, there are x, y, z ∈ Z such that 3 n + 1 = 3( x + y ) + z and hence n = x + y + ⌊ z / ⌋ . Clearly 9 n + 1 ≡ n + 7 (mod 2) but 9 n + 1 n + 7 (mod 4). So, for some r ∈ { , } , we have 9 n + r
6∈ { k (8 l + 7) : k, l ∈ N } and hence (by the Gauss-Legendre theorem) there are x, y, z ∈ Z such that 9 n + r = (3 x ) + (3 y ) + z andtherefore n = x + y + ⌊ z / ⌋ .By Dickson [D39, pp. 112-113], E (21 x +21 y + z ) = [ k,l ∈ N { k (8 l +7) , k (3 l +2) , k (7 l +3) , k (7 l +5) , k (7 l +6) } . For each r = 1 , ,
16, if 21 n + r belongs to the above set then it has the form4 k (8 l + 7) with k, l ∈ N . If { n + 1 , n + 4 , n + 16 } ⊆ { k (8 l + 7) : k, l ∈ N } , then 21 n + 4 and 21 n + 16 are even since 21 n + 4 n + 16 (mod 8), hence21 n + 1 ≡ n + 4 ≡ r ∈ { , , } and x, y, z ∈ Z we have 21 n + r = 21( x + y ) + z andhence n = x + y + ⌊ z / ⌋ .By Dickson [D39, pp. 112-113], E (12 x + 12 y + z ) = [ k ∈ N { (4 k + 2 , k + 3 } ∪ { k (3 l + 2) : k, l ∈ N } . So, for some x, y, z ∈ Z we have 12 n + 1 = 12( x + y ) + (2 z + 1) and hence n = x + y + z ( z + 1)3 = x + y + (cid:22) z ( z + 1)3 (cid:23) . This proves (1.4) for m = 3.By Jones and Pall [JP], there are x, y, z ∈ Z such that 16 n + 1 = 16 x + 16 y +(2 z + 1) and hence n = x + y + (2 z + 1) −
116 = x + y + (cid:22) z ( z + 1)4 (cid:23) . This proves (1.4) for m = 4.By [S15a, Theorem 1.7(ii)], n can be written as x + y + p ( z ) with x, y, z ∈ Z .Note that p ( z ) = z (3 z − z (3 z − . So (1.4) holds for m = 6.(iv) Now we prove (1.5) for m = 5. By Dickson [D39, pp. 112-113], E (5 x + y + z ) = { k (8 l + 3) : k, l ∈ N } . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ As 5 n + 2 n + 4 (mod 4), for a suitable choice of r ∈ { , } we can write 5 n + r as 5 x + y + z with x, y, z ∈ Z . If r = 2, then y ≡ z ≡ n = x + y −
15 + z −
15 = x + (cid:22) y (cid:23) + (cid:22) z (cid:23) . If r = 4, then we may assume that y ≡ z ≡ n = x + y z −
45 = x + (cid:22) y (cid:23) + (cid:22) z (cid:23) . If { n + 5 , n + 6 , n + 9 } ⊆ E := { k (8 l + 7) : k, l ∈ N } , then we must have5 n + 6 ≡ n + 9 ≡ r ∈ { , , } the number 5 n +5+ r is the sum of three squares. If 5( n + 1) + r = m for some m ∈ Z + which is not apower of two, then by Lemma 2.1 we have 5( n + 1) + r = x + y + z for some x, y, z ∈ Z with x , y , z = 5( n + 1) + r . If 5( n + 1) + r = (2 k ) for some k ∈ Z + ,then r ∈ { , } , 5( n + 1) + (5 − r ) = 4 k + 5 − r ≡ − r ≡ ± n + 1) + (5 − r ) E . So, for a suitable choice of r ∈ { , , } , we canwrite 5( n + 1) + r = x + y + z with x, y, z ∈ Z and x , y , z = 5( n + 1) + r .Clearly, one of x , y , z , say z , is congruent to r modulo 5. Then x + y is apositive multiple of 5. By Lemma 2.2, x + y = ¯ x + ¯ y for some ¯ x, ¯ y ∈ Z with5 ∤ ¯ x ¯ y . Without loss of generality we may assume that ¯ x ≡ y ≡ n = ¯ x −
15 + ¯ y −
45 + z − r (cid:22) ¯ x (cid:23) + (cid:22) ¯ y (cid:23) + (cid:22) z (cid:23) . Now we show (1.5) for m = 6. By Dickson [D39, pp. 112-113], E (6 x + y + z ) = { k (9 l + 3) : k, l ∈ N } . So, there are x, y, z ∈ Z such that 6 n + 4 = 6 x + y + z . Clearly, exactly oneof y and z , say y , is divisible by 3. Note that y and z have the same parity. If y ≡ z ≡ y ≡ z ≡ y ≡ z ≡ y ≡ z ≡ n = x + y + z −
46 = x + (cid:22) y (cid:23) + (cid:22) z (cid:23) . Assume that n is even. Then 6 n + 9 ≡ n + 9 = x + y + z with x, y, z ∈ Z and 3 ∤ xyz . Clearly, exactly one of x, y, z , say x , is odd. Thus x ≡ y ≡ z ≡ n = x −
16 + y −
46 + z −
46 = (cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) . Now suppose that n is odd. Then 3 n + 4 ≡ n + 4 = x + y + 2 z with x, y, z ∈ Z and 3 ∤ xyz .Without loss of generality, we may assume that x ≡ y (mod 3) (otherwise we mayuse − y to replace y ). Clearly, x y (mod 2). Thus 6 n + 8 = ( x + y ) + ( x − y ) +(2 z ) with ( x + y ) ≡ x − y ) ≡ z ) ≡ n = ( x + y ) −
16 + ( x − y ) −
36 + (2 z ) −
46 = (cid:22) ( x + y ) (cid:23) + (cid:22) ( x − y ) (cid:23) + (cid:22) (2 z ) (cid:23) . Now we prove (1.5) for m = 15. By Dickson [D39, pp. 112-113], E (3 x + y + z ) = { k (9 l + 6) : k, l ∈ N } . So, there are x, y, z ∈ Z such that 3 n + 1 = 3 x + y + z and hence15 n + 5 = 15 x + (2 + 1 )( y + z ) = 15 x + (2 y − z ) + ( y + 2 z ) . As (2 y − z ) + ( y + 2 z ) = 5( y + z ) is a positive multiple of 5, by Lemma 2.2there are u, v ∈ Z with 5 ∤ uv such that (2 y − z ) + ( y + 2 z ) = u + v . Withoutloss of generality, we assume that u ≡ v ≡ n + 5 = 15 x + u + v with u ≡ v ≡ n = x + u −
115 + v −
115 = x + (cid:22) u (cid:23) + (cid:22) v (cid:23) . If { n + 6 , n + 9 , n + 15 } ⊆ E := { k (8 l + 7) : k, l ∈ N } , then wemust have 15 n + 6 ≡ n + 9 ≡ r ∈ { , , } thenumber 15 n + 5 + r is the sum of three squares. In view of [S16, Lemma 2.2], wecan write 15 n + 5 + r = x + y + z with x, y, z ∈ Z and 3 ∤ xyz . It is easy tosee that one of x , y , z , say z , is congruent to r modulo 5. Then x + y is apositive multiple of 5, and hence by Lemma 2.2 we can write x + y = ¯ x + ¯ y with ¯ x, ¯ y ∈ Z and 5 ∤ ¯ x ¯ y . Without loss of generality, we may assume that ¯ x ≡ y ≡ x ≡ y ≡ z ≡ r (mod 15). Therefore n = ¯ x −
115 + ¯ y −
415 + z − r
15 = (cid:22) ¯ x (cid:23) + (cid:22) ¯ y (cid:23) + (cid:22) z (cid:23) . (v) Clearly, (cid:26)(cid:22) T x (cid:23) : x ∈ Z (cid:27) ⊇ (cid:26) p ( x ) = T x − x ∈ Z (cid:27) . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ By [S15a, Theorem 1.14], { T x + T y + p ( z ) : x, y, z ∈ Z } = N . It is also knownthat { p ( x ) + p ( y ) + p ( z ) : x, y, z ∈ Z } = N (cf. Guy [Gu] and [S15a]).Now it remains to prove (1.6). Clearly, for some r ∈ { , } , 2 n + r is not atriangular number. Hence, by [S07, Theorem 1(iii)] there are x, y, z ∈ Z with x y (mod 2) such that 2 n + r = x + y + T z . Thus 4 n +2 r = ( x + y ) +( x − y ) + z ( z +1)with x ± y odd and z ( z + 1) ≡ r −
1) (mod 4). Write x + y = 2 u + 1 and x − y = 2 v + 1 with u, v ∈ Z . Then n = (2 u + 1) −
14 + (2 v + 1) −
14 + z ( z + 1) − r − u ( u + 1) + v ( v + 1) + (cid:22) z ( z + 1)4 (cid:23) . In view the above, we have completed the proof of Theorem 1.1. (cid:3)
Proof of Theorem 1.2 . Let n be a fixed natural number.(i) By a known result first observed by Euler (cf. [D99, p. 260] and also [P]),there are x, y, z ∈ Z such that 2 n + 1 = 2 x + 4 y + z and hence n = x + 2 y + ⌊ z / ⌋ .Suppose that n = x + 2 y + ⌊ (3 z ) / ⌋ = x + 2 y + 3 z for all x, y, z ∈ Z . Then n is even by a known result (cf. [D39, p. 112-113] or [P]). By [D39, p. 112-113], E (3 x + 6 y + z ) = { k + 2 : k ∈ N } ∪ { k (16 l + 14) : k, l ∈ N } . Since 3 n + 1 is odd, for some x, y, z ∈ Z we have 3 n + 1 = 3 x + 6 y + z andhence n = x + 2 y + ⌊ z / ⌋ .By [D39, p. 112-113], E (4 x + 8 y + z ) = [ k ∈ N { k + 2 , k + 3 } ∪ { k (16 l + 14) : k, l ∈ N } . So there are x, y, z ∈ Z such that 4 n + 1 = 4 x + 8 y + z and hence n = x + 2 y + ⌊ z / ⌋ .By [D39, p. 112-113], E (5 x + 10 y + z ) = [ k,l ∈ N { k (5 l + 2) , k (5 l + 3) } . Thus, for some x, y, z ∈ Z we have 5 n + 1 = 5 x + 10 y + z and hence n = x + 2 y + ⌊ z / ⌋ .(ii) By [D39, p. 112-113], E (3 x + y + z ) = { k (9 l + 6) : k, l ∈ N } . So, there are x, y, z ∈ Z such that 3 n + 1 = 3 x + (3 y ) + z and hence n = x + 3 y + ⌊ z / ⌋ . By [D39, p. 112-113], E (4 x + 12 y + z ) = [ k ∈ N { k + 2 , k + 3 } ∪ { k (9 l + 6) : k, l ∈ N } . Choose δ ∈ { , } such that 4 n + δ x, y, z ∈ Z wehave 4 n + δ = 4 x + 12 y + z and hence n = x + 3 y + ⌊ z / ⌋ .If 6 n + r = 6 x + 18 y + z for some r ∈ { , , , } and x, y, z ∈ Z , then n = x + 3 y + ⌊ z / ⌋ . Now suppose that 6 n + r = 6 x + 18 y + z for any r ∈ { , , , } and x, y, z ∈ Z . By [D39, p. 112-113], S := E (6 x + 18 y + z ) = [ k ∈ N { k + 2 , k + 3 } ∪ { k (8 l + 5) : k, l ∈ N } . So 6 n + 1 or 6 n + 4 is congruent to 5 modulo 8. If 6 n + 4 ≡ n + 1 ≡ n + 1 ∈ S . So, 6 n + 1 ≡ n + 3 ≡ n + 3 ∈ S , we must have 3 | n . As 6 n ≡ n ≡ n ∈ S we have 6 n = 4(8 q + 5) for some q ∈ Z .As 6 n + 4 = 4(8 q + 6) S , we get a contradiction.As conjectured by Sun [S07] and confirmed in [GPS], there are x, y, z ∈ Z suchthat n = x + 3 y + T z and hence n = x + 3 y + ⌊ (2 z + 1) / ⌋ .(iii) By [D39, p. 112-113], E (8 x + 40 y + z ) coincides with [ k ∈ N { k + 2 , k + 3 , k + 5 , k + 28 } ∪ [ k,l ∈ N { k (25 l + 5) , k (25 l + 20) } . Choose δ ∈ { , } such that 8 n + δ n + δ E (8 x +40 y + z ).So, for some x, y, z ∈ Z we have 8 n + δ = 8 x + 40 y + z and hence n = x + 5 y + ⌊ z / ⌋ .By [D39, p. 112-113], E (4 x + 24 y + z ) = [ k ∈ N { k + 2 , k + 3 } ∪ { k (9 l + 3) : k, l ∈ N } . Choose δ ∈ { , } such that 4 n + δ n + δ E (4 x +24 y + z ).Hence there are x, y, z ∈ Z such that 4 n + δ = 4 x + 24 y + z and thus n = x + 6 y + ⌊ z / ⌋ .(iv) By [JP] or [D39, p. 112-113], for some x, y, z ∈ Z we have 8 n + 1 = 16 x +16 y + z and hence n = 2 x + 2 y + ⌊ z / ⌋ .In view of [D39, p. 112-113], E (6 x + 9 y + z ) = { k + 2 : k ∈ N } ∪ { k (9 l + 3) : k, l ∈ N } . So, there are x, y, z ∈ Z such that 3 n + 1 = 6 x + 9 y + z and hence n =2 x + 3 y + ⌊ z / ⌋ . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ By [D39, p. 112-113], E (3 x + 3 y + 2 z ) = { k (3 l + 1) : k, l ∈ N } . So there are x, y, z ∈ Z such that 6 n + 5 = 3 x + 3 y + 2 z . Since x y (mod 2),without loss of generality we may assume that 2 | x and 2 ∤ y . Thus n = 2 (cid:16) x (cid:17) + y −
12 + z −
13 = 2 (cid:16) x (cid:17) + (cid:22) y (cid:23) + (cid:22) z (cid:23) . So far we have completed the proof of Theorem 1.2. (cid:3)
3. Proofs of Theorems 1.3-1.4
Proof of Theorem 1.3 . (i) Clearly, 0 = ⌈ /m ⌉ + ⌈ /m ⌉ + ⌈ /m ⌉ , 1 = ⌈ / ⌉ + ⌈ / ⌉ + ⌈ / ⌉ and 2 = ⌈ / ⌉ + ⌈ / ⌉ + ⌈ / ⌉ . for any m ∈ { , , , } . So wejust consider required representations for n ∈ { , , , . . . } .If n is even, then 2 n − ≡ x, y, z with 2 ∤ yz such that 2 n − x ) + y + z and thus n = 2 x + y + 12 + z + 12 = (2 x ) (cid:24) y (cid:25) + (cid:24) z (cid:25) . When n ≡ n − ≡ x, y, z ∈ Z with 2 ∤ z such that 2 n − x ) +(2 y ) + z and thus n = 2 x + 2 y + z + 12 = (2 x ) y ) (cid:24) z (cid:25) . If n ≡ n − ≡ x, y, z such that 2 n − x + y + z and thus n = x + 12 + y + 12 + z + 12 = (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . This proves (1.7) for m = 2.Now we show (1.7) for m = 3. Clearly we cannot have { n − , n − } ⊆{ k (8 l + 7) : k, l ∈ N } and hence either 3 n − n − n − x + y + z for some x, y, z ∈ Z , then exactlyone of x, y, z (say, x ) is divisible by 3, hence n = 3 (cid:16) x (cid:17) + y + 23 + z + 23 = (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . When 3 n − x + y + z with x, y, z ∈ Z not all zero, by [S16, Lemma 2.2]there are u, v, w ∈ Z with 3 ∤ uvw such that 3 n − u + v + w and hence n = u + 23 + v + 23 + w + 23 = (cid:24) u (cid:25) + (cid:24) v (cid:25) + (cid:24) w (cid:25) . As 4 n −
6∈ { k (8 l + 7) : k, l ∈ N } , by the Gauss-Legendre theorem thereare x, y, z ∈ Z such that 4 n − x ) + (2 y ) + (2 z + 1) and hence n = x + y + ⌈ (2 z + 1) / ⌉ . This proves (1.7) for m = 4.Now we prove (1.7) for m = 5 by modifying our proof of the last equalityin (1.5). If { n − , n − , n − } ⊆ E := { k (8 l + 7) : k, l ∈ N } , then5 n − ≡ n − ≡ r ∈ { , , } we can write 5 n − − r > n − − r = m for some integer m > n − − r = x + y + z for some x, y, z ∈ Z with x , y , z = 5 n − − r . If 5( n − − r = (2 k ) for some k ∈ Z + , then r ∈ { , } , and 5( n − − (5 − r ) = 4 k + 2 r − ≡ r − ≡ ± n − − (5 − r ) E . So, for a suitable choice of r ∈ { , , } , we canwrite 5( n − − r = x + y + z with x, y, z ∈ Z and x , y , z = 5( n − − r .Clearly, one of x , y , z , say z , is congruent to − r modulo 5. Then x + y isa positive multiple of 5. By Lemma 2.2, x + y = ¯ x + ¯ y for some ¯ x, ¯ y ∈ Z with 5 ∤ ¯ x ¯ y . Without loss of generality, we may assume that ¯ x ≡ y ≡ n = ¯ x + 45 + ¯ y + 15 + z + r (cid:24) ¯ x (cid:25) + (cid:24) ¯ y (cid:25) + (cid:24) z (cid:25) . Now we show (1.7) for m = 6. If n is odd, then 6 n − ≡ n − x + y + z with x, y, z ∈ Z , 2 ∤ x , 2 | y , 2 | z and 3 ∤ xyz , therefore n = x + 56 + y + 26 + z + 26 = (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . Now assume that n is even. Then 6 n − ≡ n −
10 = x + y + z with x, y, z ∈ Z , 2 ∤ xy and 2 | z . Notethat exactly one of x, y, z is divisible by 3. If 3 ∤ xy and 3 | z , then x ≡ y ≡ z ≡ n = x + 56 + y + 56 + z (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . If 3 ∤ z , then exactly one of x and y , say x , is divisible by 3, hence x ≡ y ≡ z ≡ n = x + 36 + y + 56 + z + 26 = (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Now we prove (1.7) for m = 15. By the proof of the last equality in (1.5) for m = 15, for a suitable choice of r ∈ { , , } we have 15( n − r = x + y + z for some x, y, z ∈ Z with x ≡ y ≡ z ≡ r (mod 15).It follows that n = x + 1415 + y + 1115 + z + 15 − r
15 = (cid:24) x (cid:25) + (cid:24) y (cid:25) + (cid:24) z (cid:25) . (ii) Now we turn to prove (1.8). Apparently, 0 = 0 + 3 × + ⌈ / ⌉ . Let n ∈ Z + . If 2 n − ≡ ∤ n . So, we may choose δ ∈ { , } suchthat 2 n − δ
6∈ { k (8 l + 5) : k, l ∈ N } . By [D39, p. 112-113], E (2 x + 6 y + z ) = { k (8 l + 5) : k, l ∈ N } . So there are x, y, z ∈ Z such that 2 n − δ = 2 x + 6 y + z and hence n = x + 3 y + ⌈ z / ⌉ .Obviously, 0 = 0 + 3 × + ⌈ / ⌉ . Let n ∈ Z + . By [D39, p. 112-113], T := E (10 x + 30 y + z ) = [ k,l ∈ N { k (8 l + 5) , k (9 l + 6) , k (5 l + 2) , k (5 l + 3) } . If 10 n − r T for some r ∈ { , , , , , } , then there are x, y, z ∈ Z such that10 n − r = 10 x + 30 y + z and hence n = x + 3 y + ⌈ z / ⌉ . Now we supposethat 10 n − r ∈ T for all r = 0 , , , , , | n ( n + 1), then by 10 n − ∈ T we have 10 n − ≡ n − ≡ n − ∈ T . When n ≡ n − ∈ T we must have 10 n − ≡ n ≡ n ≡ n ∈ T , and thus by 10 n − ∈ T we have 10 n − ≡ n ≡ δ ∈ { , } with n ≡ δ (mod 2). Then 12 n + 5 − δ E (3 x + y + z ) = { k (9 l + 6) : k, l ∈ N } . So, there are u, v, w ∈ Z such that 12 n + 5 − δ = 3 u + v + w . If v and w areboth even, then 5 ≡ u (mod 4) which is impossible. Without loss of generality,we assume that w = 2 z + 1 with z ∈ Z . Then3 u + v ≡ n + 5 − δ − ≡ . Hence, by [S15a, Lemma 3.2] we can write 3 u + v as 3(2 x + 1) + (2 y + 1) with x, y ∈ Z . Therefore,12 n +5 − δ = 3(2 x +1) +(2 y +1) +(2 z +1) = 12 x ( x +1)+4 y ( y +1)+4 z ( z +1)+5 and hence 3 n − δ = 3 x ( x + 1) + y ( y + 1) + z ( z + 1) . Note that m ( m + 1) m ∈ Z . If y ( y + 1) , z ( z + 1) − δ ≡ | y ( y + 1). Then n = x ( x + 1) + y ( y + 1)3 + z ( z + 1) + δ x ( x + 1) + y ( y + 1)3 + (cid:24) z ( z + 1)3 (cid:25) . Let δ ∈ { , } with n ≡ δ (mod 2). Then 12 n + 3 − δ is congruent to 0 or 2modulo 3. As 12 n + 3 − δ ≡ u, v, w such that12 n + 3 − δ = u + v + w . If δ = 0, then by [S16, Lemma 2.2] we can write u + v + w as r + s + t with r, s, t ∈ Z and gcd( rst,
6) = 1. So, there are x, y, z ∈ Z such that12 n +3 − δ = (2 x +1) +(2 y +1) +(2 z +1) = 4 x ( x +1) +4 y ( y +1) +4 z ( z +1) +3and 2 x + 1 , y + 1 x, y x ( x + 1) and y ( y + 1)are divisible by 3. Thus n = x ( x + 1)3 + y ( y + 1)3 + z ( z + 1) + δ (cid:24) x ( x + 1)3 (cid:25) + (cid:24) y ( y + 1)3 (cid:25) + (cid:24) z ( z + 1)3 (cid:25) . Note that { m ( m + 1) / m ∈ Z & 3 | m ( m + 1) } = { q (3 q + 1) : q ∈ Z } .The proof of Theorem 1.3 is now complete. (cid:3) Proof of Theorem 1.4 . (i) Let n ∈ N . If 2 n + 1 ∈ { k (8 l + 7) : k, l ∈ N } ,then 2 n ≡ n
6∈ { k (8 l + 7) : k, l ∈ N } . So, for some δ ∈ { , } we have 2 n + δ k (8 l + 7) : k, l ∈ N } , and hence by the Gauss-Legendre theorem there are x, y, z ∈ Z such that 2 n + δ = x + y + z and hence n = ⌊ ( x + y + z ) / ⌋ . Note also that n = T x + T y + T z for some x, y, z ∈ Z .This proves the desired result for a = 2.Now we handle the case a >
2. Clearly, for some r ∈ { , } we have an + r k (8 l + 7) : k, l ∈ N } , hence for some x, y, z ∈ Z we have an + r = x + y + z and thus n = ⌊ ( x + y + z ) /a ⌋ . Take δ ∈ { , } with an ≡ δ (mod 2). Then,there exist x, y, z ∈ Z such that ( an + da ) / T x + T y + T z and hence n = ⌊ ( x ( x + 1) + y ( y + 1) + z ( z + 1)) /a ⌋ .(ii) Suppose that a is odd. As 16 n + 3 a ≡ n + 3 a can be expressed as the sum of three odd squares. For any oddinteger w , either w or − w is congruent to a modulo 4. Thus, there are x, y, z ∈ Z such that16 n +3 a = (4 x + a ) +(4 y + a ) +(4 z + a ) , i.e. , n = 2( x + y + z )+ a ( x + y + z ) . Hence n = x + y + z + ⌊ a ( x + y + z ) ⌋ as desired. ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Now assume that gcd( a,
6) = 1. Choose δ ∈ { , } such that n ≡ δ (mod 2).As 12(3 n + δ ) + 3 a ≡ u, v, w such that 12(3 n + δ ) + 3 a = u + v + w . Applying [S16, Lemma 2.2], we can write u + v + w as r + s + t , where r, s, t are integers with r ≡ u ≡ u ≡ , s ≡ v ≡ , t ≡ w ≡ , and 3 ∤ rst. Thus r or − r has the form 6 x + a , s or − s has the form 6 y + a , and t or − t hasthe form 6 z + a , where x, y, z ∈ Z . Therefore,12(3 n + δ ) + 3 a =(6 x + a ) + (6 y + a ) + (6 z + a ) =12(3 x + ax + 3 y + ay + 3 z + 3 z ) + 3 a and hence n = x + y + z + a ( x + y + z ) − δ x + y + z + j a x + y + z ) k . Now we suppose that 2 | a and 3 ∤ a . If 9 n +3( a/ +3 r ∈ { k (8 l +7) : k, l ∈ N } for all r = 1 , ,
3, then 9 n +3( a/ +6 ≡ n +3( a/ +9 ≡ r ∈ { , , } and u, v, w ∈ Z we have 9 n + 3( a/ + 3 r = u + v + w . By[S16, Lemma 2.2] we can write 9 n + 3( a/ + 3 r = ¯ u + ¯ v + ¯ w , where ¯ u, ¯ v, ¯ w ∈ Z and 3 ∤ ¯ u ¯ v ¯ w . So there are x, y, z ∈ Z such that9 n + 3 r + 3 (cid:16) a (cid:17) = (cid:16) x + a (cid:17) + (cid:16) y + a (cid:17) + (cid:16) z + a (cid:17) , i.e., 3 n + r − x (3 x + a ) + y (3 y + a ) + z (3 z + a ) . It follows that n = x + y + z + a ( x + y + z ) − ( r − x + y + z + j a x + y + z ) k . (iii) Obviously, 0 = p (0) / ⌈ p (0) / ⌉ + ⌈ p (0) / ⌉ . Now we let n > δ ∈ { , } with n δ (mod 2). As 6 n − δ is congruent to 1 or 2 modulo 4,by the Gauss-Legendre theorem we can write 6 n − δ as the sum of three squaresand hence by [S16, Lemma 2.2] there are x, y, z ∈ Z such that6 n − δ = (3 x − + (3 y − + (3 z −
1) = 3 p ( x ) + 1 + (3 p ( y ) + 1) + (3 p ( z ) + 1) . Clearly, 3 x − , y − , z − x − ≡ , y − ≡ z − ≡ − δ ≡ n (mod 2) . Then p ( x ) = ((3 x − − / p ( y ) is odd, and p ( z ) ≡ − δ (mod 2).Therefore n = p ( x )2 + p ( y ) + 12 + p ( z ) + δ p ( x )2 + (cid:24) p ( y )2 (cid:25) + (cid:24) p ( z )2 (cid:25) . This concludes our proof. (cid:3)
4. Proof of Theorem 1.5
For a, b, c, n ∈ Z + , define r ( a,b,c ) ( n ) = |{ ( x, y, z ) ∈ Z : ax + by + cz = n }| (4.1)and H ( a,b,c ) ( n ) := Y p ∤ abc (cid:18) p ord p ( n )+1 − p − − (cid:18) − abcp (cid:19) p ord p ( n ) − p − (cid:19) , (4.2)where ( · p ) is the Legendre symbol. Clearly, H ( a,b,c ) ( n ) > Y p ∤ abc p ord p ( n )+1 − − ( p ord p ( n ) − p − Y p ∤ abc p ord p ( n ) . (4.3)In 1907 Hurwitz (cf. [D99, p. 271]) showed that r (1 , , ( n ) = 6 H (1 , , ( n ) which isjust (2.1). In 2013 S. Cooper and H. Y. Lam [CL] deduced some similar formulasfor r (1 , , ( n ) , r (1 , , ( n ) , r (1 , , ( n ) , r (1 , , ( n ) . Lemma 4.1.
For any integer n > , there are x, y, z ∈ Z with | x | < n and | y | < n such that x + y + 2 z = n .Proof . By Cooper and Lam [CL, Theorem 1.2], r (1 , , ( n ) = (cid:26) H (1 , , ( n ) if 2 ∤ n, H (1 , , ( n ) if 2 | n. (4.4)If n is odd, then there is an odd prime p dividing n , hence r (1 , , ( n ) = 4 H (1 , , ( n ) > n is even, then r (1 , , ( n ) = 12 H (1 , , ( n ) > > x + y + 2 z = n for ( x, y, z ) = ( ± n, , , (0 , ± n, x, y, z ∈ Z with x , y = n such that x + y + 2 z = n . This concludes theproof. (cid:3) Lemma 4.2.
Suppose that n ∈ Z + is not a power of two. Then there are x, y, z ∈ Z with | x | < n and | y | < n such that x + y + 5 z = n .Proof . As conjectured by Cooper and Lam [CL, Conjecture 8.1] and proved byGuo et al. [GPQ], r (1 , , ( n ) = 2(5 ord ( n )+1 − H (1 , , ( n ) . (4.5)If 5 | n , then 2(5 ord ( n )+1 − >
4. If n has a prime divisor p = 2 ,
5, then H (1 , , ( n ) > n > r (1 , , ( n ) > x + y + 5 z = n for ( x, y, z ) = ( ± n, , , (0 , ± n, x, y, z ∈ Z with x , y = n such that x + y + 5 z = n . This ends the proof. (cid:3) ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Remark
Proof of Theorem 1.5 . (i) Let n ∈ N . Clearly, n = p ( x ) + p ( y ) + 2 p ( z ) if andonly if 3 n + 4 = (3 x − + (3 y − + 2(3 z − . In view of [D39, pp. 112-113], E ( x + y + 2 z ) = { k (16 l + 14) : k, l ∈ N } . If 3 n + 4 = 4 k (16 l + 14) for some k, l ∈ N , then for some q ∈ Z + we have l = 3 q − n = 4 k (16(3 q −
1) + 14) −
43 = 4 k +2 q −
23 (4 k + 2) . If n has the form 4 k +2 q − (4 k + 2) with k ∈ N and q ∈ Z + , then n = p ( x ) + p ( y ) + 2 p ( z ) for all x, y, z ∈ Z .Now assume that n is not of the form 4 k +2 q − (4 k + 2) with k ∈ N and q ∈ Z + .Then there are r, s, t ∈ Z such that 3 n + 4 = r + s + 2 t . In view of Lemma 4.1,we may assume that r , s = 3 n + 4. Clearly r and s cannot be both divisible by3. Without loss of generality, we assume that 3 ∤ r . As s + 2 t = 3 n + 4 − r isa positive multiple of 3, by [S15a, Lemma 2.1] we can rewrite it as u + 2 v with u, v ∈ Z and 3 ∤ uv . Thus there are x, y, z ∈ Z such that3 n + 4 = r + u + 2 v = (3 x − + (3 y − + 2(3 z − =3 p ( x ) + 1 + (3 p ( y ) + 1) + 2(3 p ( z ) + 1)and hence n = p ( x ) + p ( y ) + 2 p ( z ).By the above, there are x, y, z ∈ Z with 2 n +1 = p ( x )+ p ( y )+2 p ( z ). Withoutloss of generality, we may assume that p ( x ) is even and p ( y ) = y (3 y −
2) is odd.Clearly, w = (1 − y ) / ∈ Z and p ( y ) − p ( w ). So, 2 n = p ( x ) + 2 p ( z ) +4 p ( w ). Note also that n = p ( x )2 + p ( y ) −
12 + p ( z ) = (cid:22) p ( x )2 (cid:23) + (cid:22) p ( y )2 (cid:23) + p ( z ) . Therefore (1.13) and (1.14) hold.(ii) Fix a nonnegative integer n . If 6 n + 5 ≡ n + 8 ≡ δ ∈ { , } we have 6 n + 5 + 3 δ E ( x + y + z ) = { k (8 l + 7) : k, l ∈ N } and hence 6 n + 5 + 3 δ = u + v + w for some u, v, w ∈ Z . Two of u, v, w have the same parity. Without loss ofgenerality, we assume that u + v = 2 s and u − v = 2 t for some s, t ∈ Z . Hence6 n + 5 + 3 δ = w + 2 s + 2 t . If (6 n + 5 + 3 δ ) = 2 m for some m ∈ Z + , then byLemma 4.1 there are r, s , t ∈ Z with s , t = m such that m = s + t + 2 r and hence 6 n + 5 + 3 δ = (2 r ) + 2 s + 2 t with 2 s , t = 6 n + 5 + 3 δ . So, we maysimply suppose that 6 n + 5 + 3 δ = w + 2 s + 2 t with 2 s , t = 6 n + 5 + 3 δ .Clearly, one of s and t is not divisible by 3. Without loss of generality we assumethat t = (3 x − with x ∈ Z As w + 2 s = 6 n + 5 + 3 δ − t is a positive multiple of 3, by [S15a, Lemma 2.1] we can write w + 2 s as (3 z − + 2(3 y − with y, z ∈ Z . Thus6 n +5+3 δ = (3 z − +2(3 y − +2(3 x − = 3 p ( z )+1+2(3 p ( x )+3 p ( y )+2)and hence n = p ( x ) + p ( y ) + p ( z ) − δ p ( x ) + p ( y ) + (cid:22) p ( z )2 (cid:23) . This proves (1.15). In view of (1.12), both (1.16) and (1.17) follow from (1.15).(iii) Let n ∈ N . As 12 n + 9 ≡ n + 9 as the sum of three squares. In view of [S16, Lemma 2.2], thereare u, v, w ∈ Z with 3 ∤ uvw such that 12 n + 9 = u + v + w . Clearly, exactly oneof u, v, w is odd. Without loss of generality we may assume that u = 2(3 x − v = 2(3 y −
1) and w = 3 z − x, y, z ∈ Z . Thus12 n + 9 = 4(3 x − + 4(3 y − + (3 z − = 12 p ( x ) + 12 p ( y ) + 3 p ( z ) + 9and hence (1.18) follows.(iv) By Dickson [D39, pp. 112-113], E (5 x + 5 y + z ) = [ k ∈ N { k + 2 , k + 3 } ∪ { k (8 l + 7) : k, l ∈ N } . If 15 n + 11 + 3 r belongs to this set for all r = 0 , ,
3, then 15 n + 11 is odd, hence15 n + 11 ≡ n + 11 + 3 ≡ r ∈ { , , } such that 15 n + 11 + 3 r E (5 x + 5 y + z ).Hence, for some u, v, w ∈ Z we have 15 n + 11 + 3 r = 5 u + 5 v + w . If 15 n +11 + 3 r = 5 m for some positive integer m which is not a power of two, then byLemma 4.2 there are u , v , w ∈ Z with u , v = m such that m = u + v +5 w and hence 15 n + 11 + 3 r = 5 u + 5 v + (5 w ) with 5 u , v = 15 n + 11 + 3 r . If15 n + 11 + 3 r = 5 × a for some a ∈ N , then a > r = 3, 15 n + 11 + 3 × × a − ≡ n + 11 + 3 E (5 x + 5 y + z ). So, we maysimply assume that 15 n + 11 + 3 r = 5 u + 5 v + w with 5 u , v < n + 11 + 3 r .Clearly, u or v is not divisible by 3. Without loss of generality we suppose that u = (3 x − for some x ∈ Z . As 5 v + w = 15 n + 11 + 3 r − u > v + w as 5(3 y − + (3 z − with y, z ∈ Z . Thus15 n + 11 + 3 r =5(3 x − + 5(3 y − + (3 z − =5(3 p ( x ) + 1) + 5(3 p ( y ) + 1) + 3 p ( z ) + 1and hence n = p ( x ) + p ( y ) + p ( z ) − r p ( x ) + p ( y ) + (cid:22) p ( z )5 (cid:23) . This proves (1.19). In view of (1.12), (1.20) follows from (1.19).The proof of Theorem 1.5 is now complete. (cid:3)
ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋
5. Some further conjectures
Conjecture 5.1.
For any n ∈ N , there are x, y, z ∈ N such that n + 3 = x + y + z and x ≡ , . Also, for any n ∈ N with n = 20 , there are x, y, z ∈ Z with x ≡ ± such that x + y + z = 8 n + 3 .Remark n ∈ N can be writtenas the sum of two triangular numbers and a hexagonal number, equivalently,8 n + 3 = (4 x − + y + z for some x, y, z ∈ N . Conjecture 5.2.
Let a > be an integer with a = 4 , . Then any positive integercan be written as the sum of three elements of the set {⌊ x /a ⌋ : x ∈ Z } one ofwhich is odd.Remark a = 4 , Conjecture 5.3.
Let T := n x + j x k : x ∈ Z o = (cid:26)(cid:22) k ( k + 1)4 (cid:23) : k ∈ N (cid:27) . Then each n = 2 , , , . . . can be expressed as r + s + t , where r, s, t are elementsof T with r s t and ∤ s . Also, for any ordered pair ( b, c ) among (1 , , (1 , , (1 , , (1 , , (1 , , (1 , , (1 , , (2 , , (2 , , each n ∈ N can be written as x + by + cz with x, y, z ∈ T .Remark { T x : x ∈ Z } = { p ( − x ) = x (2 x + 1) : x ∈ Z } . Conjecture 5.4. (i)
Let α be a positive real number with α = 1 and α . .Define S ( α ) := { x + ⌊ αx ⌋ : x ∈ Z } . Then any positive integer can be written as the sum of three elements of S ( α ) oneof which is odd. (ii) Let < α β γ . such that two of α, β, γ are different from or { α, β, γ } = { , /m } for some m = 2 , , , . . . . Then any n ∈ N can be written as x + y + z + ⌊ αx ⌋ + ⌊ βy ⌋ + ⌊ γz ⌋ with x, y, z ∈ Z . In particular, if a, b, c ∈ Z + are not all equal to one, then n x + y + z + j xa k + j yb k + j zc k : x, y, z ∈ Z o = N . Remark S (11 / S (8 /
5) one ofwhich is odd.
Conjecture 5.5.
Any integer n > can be written as p + ⌊ k ( k + 1) / ⌋ , where p is a prime and k is a positive integer.Remark p + T x , where p is prime or zero, and x is an integer.Motivated by Theorem 1.5, we pose the following conjecture. Conjecture 5.6.
Let a, b, c ∈ Z + . Then (cid:26)(cid:22) p ( x ) a (cid:23) + (cid:22) p ( y ) b (cid:23) + (cid:22) p ( z ) c (cid:23) : x, y, z ∈ Z (cid:27) = N . When ( a, b, c ) = (1 , , , (1 , , , (2 , , , we have (cid:26)(cid:22) p ( x ) a (cid:23) + (cid:22) p ( y ) b (cid:23) + (cid:22) p ( z ) c (cid:23) : x, y, z ∈ Z (cid:27) = N . If ( a, b, c ) = (1 , , , (2 , , , then (cid:26)(cid:22) p ( x ) a (cid:23) + (cid:22) p ( y ) b (cid:23) + (cid:22) p ( z ) c (cid:23) : x, y, z ∈ Z (cid:27) = N . Now we present a general conjecture related to Theorems 1.1-1.3.
Conjecture 5.7. (i)
Let a and b be positive integers. If c ∈ Z + is large enough,then (cid:26) ax + by + (cid:22) z c (cid:23) : x, y, z ∈ Z (cid:27) = (cid:26) ax + by + (cid:24) z c (cid:25) : x, y, z ∈ Z (cid:27) = N . Also, for any sufficiently large c ∈ Z + we have (cid:26) ax + by + (cid:22) z ( z + 1) c (cid:23) : x, y, z ∈ Z (cid:27) = N and (cid:26) ax + by + (cid:24) z ( z + 1) c (cid:25) : x, y, z ∈ Z (cid:27) = N . (ii) For a, b, c ∈ Z + with a b + c , if ( a, b, c ) = (1 , , , (3 , , , (4 , , then (cid:26) ax + (cid:22) y b (cid:23) + (cid:22) z c (cid:23) : x, y, z ∈ Z (cid:27) = N . ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ For a, b ∈ Z + , we define S ∗ ( a, b ) := (cid:26) c ∈ Z + : (cid:26) ax + by + (cid:24) z c (cid:25) : x, y, z ∈ Z (cid:27) = N (cid:27) ,S ∗ ( a, b ) := (cid:26) c ∈ Z + : (cid:26) ax + by + (cid:22) z c (cid:23) : x, y, z ∈ Z (cid:27) = N (cid:27) ,T ∗ ( a, b ) := (cid:26) c ∈ Z + : (cid:26) ax + by + (cid:24) z ( z + 1) c (cid:25) : x, y, z ∈ Z (cid:27) = N (cid:27) ,T ∗ ( a, b ) := (cid:26) c ∈ Z + : (cid:26) ax + by + (cid:22) z ( z + 1) c (cid:23) : x, y, z ∈ Z (cid:27) = N (cid:27) . Based on our computation we conjecture that S ∗ (1 ,
1) = { , , } , S ∗ (1 ,
2) = { , } , S ∗ (1 ,
3) = { , } , S ∗ (1 ,
4) = { , , , } ,S ∗ (1 ,
5) = { , , , } , S ∗ (1 ,
6) = { , , , } , S ∗ (1 ,
7) = { , , , } ,S ∗ (1 ,
8) = { , . . . , , } , S ∗ (1 ,
9) = { , . . . , } , S ∗ (1 ,
10) = { , . . . , , , } ,S ∗ (2 ,
2) = { , . . . , , , } , S ∗ (2 ,
3) = { , , } ; S ∗ (1 ,
2) = { } , S ∗ (1 ,
3) = { , , } , S ∗ (1 ,
4) = { , , , } , S ∗ (1 ,
5) = { , , , , } ,S ∗ (1 ,
6) = { , } , S ∗ (1 ,
7) = { , , , , } , S ∗ (1 ,
8) = { , , , , } ,S ∗ (1 ,
9) = { , , , , , } , S ∗ (1 ,
10) = { , , , , } ,S ∗ (1 ,
11) = { , , , , , , } , S ∗ (1 ,
12) = { , , , , , , } S ∗ (2 ,
2) = { , , , , , , } , S ∗ (2 ,
3) = { , , } ,S ∗ (2 ,
4) = { , , , } , S ∗ (2 ,
5) = { , , , } ; T ∗ (1 ,
1) = T ∗ (1 ,
2) = ∅ , T ∗ (1 ,
3) = 1 , T ∗ (1 ,
4) = { } , T ∗ (1 ,
5) = T ∗ (1 ,
6) = { , } ,T ∗ (1 ,
7) = { , , } , T ∗ (1 ,
8) = { } , T ∗ (1 ,
9) = T ∗ (1 ,
10) = T ∗ (1 ,
11) = { , , } ,T ∗ (2 ,
2) = { , } , T ∗ (2 ,
3) = { , } , T ∗ (2 ,
4) = { , , } , T ∗ (3 ,
4) = { , , } ; T ∗ (1 ,
2) = ∅ , T ∗ (1 ,
3) = { } , T ∗ (1 ,
5) = { , , } , T ∗ (1 ,
6) = { , } ,T ∗ (1 ,
7) = { , , } , T ∗ (1 ,
8) = { } , T ∗ (1 ,
10) = T ∗ (2 ,
3) = { , , } . Also, our computation suggests that (cid:26) x + 4 y + (cid:22) z c (cid:23) : x, y, z ∈ Z (cid:27) = N for any integer c >
42, and that (cid:26) x + 4 y + (cid:22) z ( z + 1) c (cid:23) : x, y, z ∈ Z (cid:27) = N for any integer c >
27. Note that 179 = 4 x + 4 y + ⌊ z / ⌋ for any x, y, z ∈ Z and that 29 = 4 x + 4 y + ⌊ z ( z + 1) / ⌋ for all x, y, z ∈ Z .Motivated by Theorem 1.4(i), we pose the following conjecture similar to Con-jecture 1.1. Conjecture 5.8.
Let a, b, c be positive integers with a b c . If c > , then (cid:26)(cid:22) x a + y b + z c (cid:23) : x, y, z ∈ Z (cid:27) = N If ( a, b, c ) = (1 , , , (1 , , , (1 , , , (1 , , , then (cid:26)(cid:22) x ( x + 1) a + y ( y + 1) b + z ( z + 1) c (cid:23) : x, y, z ∈ Z (cid:27) = N . Conjecture 5.9.
We have (cid:26) w + (cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ N (cid:27) = N and (cid:26) w + (cid:22) x (cid:23) + (cid:22) y (cid:23) + (cid:22) z (cid:23) : x, y, z ∈ N (cid:27) = N . Our following conjecture is a natural extension of Goldbach’s Conjecture.
Conjecture 5.10.
For any positive integers a and b with a + b > , any integer n > can be written as ⌊ p/a ⌋ + ⌊ q/b ⌋ with p and q both prime.Remark { a, b } = { , } , Conjecture 5.10 reduces to Lemoine’sConjecture which states that any odd number greater than 5 can be written as p + 2 q with p and q both prime. In the case a = b = 2, Conjecture 5.10 reducesto the Goldbach Conjecture.Let us conclude this paper with one more conjecture. Conjecture 5.11.
Let S = nj x k : x − and x + 1 are twin prime o = nj x k : 3 x − and x + 1 are twin prime o . Then, any positive integer can be written as the sum of two distinct elements of S one of which is even. Also, any positive integer can be expressed as the sum ofan element of S and a positive generalized pentagonal number.Remark Conjecture 5.12.
Any integer n > can be written as x + y + ϕ ( z ) with x, y ∈ N , z ∈ Z + , and max { x, y } or z prime. Also, any n ∈ Z + can be written as x + y + T z with x, y ∈ N and z ∈ Z + .Remark n = 1 , . . . , . See [S15c, A262311 andA262813] for related data. ATURAL NUMBERS REPRESENTED BY ⌊ x /a ⌋ + ⌊ y /b ⌋ + ⌊ z /c ⌋ Conjecture 5.13.
Any integer m can be written as x − y + z with x, y, z ∈ Z + .Remark m ∈ Z with | m | , see [S15c] forrelated data. For example,0 = 4 − +16 , − +1003 , and 11019 = 4325 − +3719409 . Conjecture 5.14.
Any n ∈ N can be written as w + x + y +2 z with w, x, y, z ∈ N . Also, any n ∈ N can be written as w + 2 x + y + 2 z with w, x, y, z ∈ N . Remark n = 1 , . . . , × , see [S15c, A262827and A262857] for related data. References [CP] S. Cooper and H. Y. Lam,
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