Fermat's polygonal number theorem for repeated generalized polygonal numbers
Soumyarup Banerjee, Manav Batavia, Ben Kane, Muratzhan Kyranbay, Dayoon Park, Sagnik Saha, Hiu Chun So, Piyush Varyani
aa r X i v : . [ m a t h . N T ] M a y FERMAT’S POLYGONAL NUMBER THEOREM FOR REPEATEDGENERALIZED POLYGONAL NUMBERS
SOUMYARUP BANERJEE, MANAV BATAVIA, BEN KANE, MURATZHAN KYRANBAY,DAYOON PARK, SAGNIK SAHA, HIU CHUN SO, AND PIYUSH VARYANI
Abstract.
In this paper, we consider sums of generalized polygonal numbers with repeats,generalizing Fermat’s polygonal number theorem which was proven by Cauchy. In particular,we obtain the minimal number of generalized m -gonal numbers required to represent everypositive integer and we furthermore generalize this result to obtain optimal bounds whenmany of the generalized m -gonal numbers are repeated r times, where r ∈ N is fixed. Introduction
Fermat famously conjectured in 1638 that every positive integer may be written as thesum of at most m m -gonal numbers; that is, for P m ( x ) := ( m − x − ( m − x (the x -th m -gonalnumber , where x ∈ N with N := N ∪ { } ) there exists an x = ( x , x , . . . , x m ) ∈ N m suchthat m X j =1 P m ( x j ) = n for every n ∈ N ; we call a Diophantine equation which represents every positive integer uni-versal . The m = 4 case of Fermat’s claim was Lagrange’s celebrated four squares theorem,proven in 1770, Gauss famously proved the m = 3 case, sometimes known as the EurekaTheorem, in 1796, and Cauchy finally resolved the general case in 1813 [5]. Guy [7] investi-gated the question of the optimality of Fermat’s polygonal number theorem. That is to say,for which ℓ ∈ N is the sum(1.1) ℓ X j =1 P m ( x j ) = n universal? More generally, Guy [7] considered sums of the type (1.1) with more generalinputs x j ∈ Z ( P m ( x ) with x ∈ Z is known as a generalized m -gonal number ) and used asimple argument based on the fact that the smallest generalized m -gonal number other than0 and 1 is m − ℓ ≥ m − m ≥
8, while Cauchy’s theorem implies that the
Date : May 11, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Fermat’s polygonal number theorem, polygonal numbers, Diophantine equations,universal quadratic polynomials.The research presented here was conducted while the second, fourth, sixth, seventh, and eighth authorswere undergraduate researcher assistants at the University of Hong Kong and they thank the university forits hospitality. The internship of the second author was additionally supported by the Hong Kong IndianChamber of Commerce, who he thanks for their generous support. The research of the third author wassupported by grants from the Research Grants Council of the Hong Kong SAR, China (project numbersHKU 17316416, 17301317, and 17303618). inimal choice satisfies ℓ ≤ m . Comparison of Guy’s and Cauchy’s theorems hence leaves asmall gap between the upper and lower bounds. In this paper, we ask where the true answerlies within this gap in the case of generalized m -gonal numbers. For a ∈ N ℓ and m ≥ x ∈ Z ℓ )(1.2) P m, a ( x ) := ℓ X j =1 a j P m ( x j ) . One may think of this as a weighted sum of generalized polygonal numbers or as a sum ofgeneralized polygonal numbers where the first generalized m -gonal number is repeated a times, the second is repeated a times, and so on. Using this second interpretation, we see byGuy’s work [7] that if P m, a is universal, then P ℓj =1 a j ≥ m −
4; an upper bound for P ℓj =1 a j is not clear, however. We consider the specific case when a r,ℓ ,ℓ := ( , r ) = (1 , , . . . , , r, r, . . . , r ) , where 1 is repeated ℓ times and r is repeated ℓ := ℓ − ℓ times. Let ℓ m denote the minimal ℓ for which (1.1) is universal when we more generally allow x ∈ Z ℓ and similarly for r ≥ ℓ ∈ N denote the optimal minimal choice ℓ = ℓ m,r,ℓ for which the sum of generalized m -gonal numbers P m, a r,ℓ ,ℓ defined in (1.2) is universal. Our main result is the following. Theorem 1.1. (1)
For m / ∈ { , } we have ℓ m = m − if m ≥ , if m ∈ { , , } , if m ∈ { , } . (2) For ≤ r < m − we have ℓ m,r,r − = (cid:24) m − r (cid:25) + ( r − . (3) We have ℓ m, , = j m k for m ≥ ,ℓ m, , = ( m − for m ≥ with m , m − for m ≥ with m ≡ ,ℓ m, , = (cid:24) m − (cid:25) + 2 for m ≥ ,ℓ m, , = (cid:24) m − (cid:25) + 3 for m ≥ ,ℓ m, , = (cid:24) m − (cid:25) + 4 for m ≥ . emarks. (1) Using Guy’s argument, for m sufficiently large (depending on r ), if P m, a r,ℓ ,ℓ is universal,then one must have ℓ ≥ r − r − P m, a r,ℓ ,ℓ . Hence Theorem 1.1 (2) is optimal in the ℓ aspect.The restriction on r is chosen so that we have at least 6 variables which are not repeated.The cases 2 ≤ r ≤ m . Indeed, a more careful case-by-case checking shows that onemay take m ≥
27 for r = 4, m ≥
34 for r = 5, and m ≥
40 for r = 6, but we havechosen the weaker restrictions on m appearing in Theorem 1.1 (3) in order to present theproof in a more systematic way. These improved lower bounds for m form a theoreticallimit on the extent to which the method in this paper may be applied; that is to say,reducing the bound on m beyond the stated bounds m ≥ m ≥ m ≥ m ≥ m ≥
40 for r = 2, r = 3, r = 4, r = 5, and r = 6, respectively, would require adifferent method than the one presented in this paper (or at least a serious modificationthat likely depends on the choice of m ) because we would not have enough variables toapply a crucial lemma that applies to the generic case. Motivated by this, the second,fourth, sixth, seventh, and eighth authors [1] have relaxed the conditions to ℓ = r + 4in order to guarantee at least 6 such variables for r ≥
2, thereby extending the methodin this paper to compute ℓ m,r,r +4 without any restriction on r or m .(2) The second restriction in Theorem 1.1 (2) is somewhat artificial. Namely, if r ≥ m − r − ≥ m − m -gonal numbers preceding the r -times repeatedgeneralized m -gonal numbers, and the original r − m ∈ { , } in Theorem 1.1 (1).A certain modification of Lemma 2.2 might work for m = 9, but the m = 7 case seemsto require a different method because the dimension is too small to use a modificationof Lemma 2.2. Together with K.-L. Kong, the first and sixth authors are investigatingthe usage of modular forms techniques to resolve these remaining cases.The case r = 3 in Theorem 1.1 (3) is exceptional both because ℓ m, , > ℓ m and becausethe dependence on r for r = 3 is vastly different than the generic dependence on large r inTheorem 1.1 (2). The primary reason for this is the fact that P m (2) = m ≡ m − P m ( −
1) (mod 3) . Because of this, it turns out that either 3 m −
12 or 2 m − P m, a , ,ℓ for ℓ < m −
2. Guy exploited a similar property for m − ℓ m ≥ m − m −
12 or 2 m − P m, a with arbitrary a ∈ N ℓ . Generalizing the diagonal case of theConway–Schneeberger fifteen theorem, Liu and the third author [10] proved that there existsa unique minimal γ m ∈ N such that P m, a is universal if and only if it represents every n ≤ γ m . It was shown in [10] that m − ≤ γ m ≪ m ε , and this was improved by the fifthauthor and Kim [11], who showed that there exists an absolute constant c ≥ m − ≤ γ m ≤ cm . It is natural to wonder about the optimal choice of c (perhaps onlyholding for m sufficiently large). The case r = 3 leads to the conclusion that c ≥ m ≡ c ≥ orollary 1.2. If m ≥ , then we have γ m ≥ ( m − if m , m − if m ≡ . Remark.
Using techniques from the arithmetic theory of quadratic forms, the constant γ m has been explicitly computed for some small m . In particular, we have γ = γ = 8 byBosma and the third author [4], γ = 15 by the Conway–Schneeberger fifteen theorem [6, 2], γ = 109 by Ju [8], and γ = 60 by Ju and Oh [9]. In light of the work in [11] and the lowerbound in Corollary 1.2, it may be interesting to systematically investigate other choices of a in order to obtain an improvement on the lower bound for c .The paper is organized as follows. In Section 2 we give some helpful preliminary informa-tion about quadratic forms and quadratic polynomials. In Section 3 we prove Theorem 1.1(1) and Theorem 1.1 (2), giving the stronger version of Fermat’s polygonal number theoremin the r = 1 case and its generalization for large r . Finally, in Section 4, we consider smallchoices of r >
1, for which a different technique is necessary, and the resulting bound for γ m given in Corollary 1.2. Acknowledgements
The authors thank Min-Joo Jang and Sudhir Pujahari for helpful conversations and theanonymous referee for a careful reading of the paper.2.
Preliminaries
The sums of polygonal numbers appearing in (1.2) are a special case of a natural class offunctions known as quadratic polynomials. In order to define these, recall that a homoge-neous polynomial Q of degree 2 is known as a quadratic form . If Q ( x ) ∈ Z whenever x ∈ Z ℓ ,then we call Q integer-valued , and it is moreover known as integral if the associated Grammatrix (i.e., the matrix A for which Q ( x ) = x T A x ) has integer coefficients ( warning : indifferent contexts, authors write Q ( x ) = x T A x , so one needs to be careful about a factorof 2 whenever comparing in the literature). We call such a quadratic form positive-definite if it only attains non-negative values and vanishes if and only if x = . A totally-positivequadratic polynomial is a function of the form P ( x ) = Q ( x ) + L ( x ) + c, where Q is a positive-definite quadratic form, L is a linear function defined over Z , and c is a constant, such that P ( x ) ≥ x ∈ Z ℓ and P ( x ) = 0 if and only if x = . Wefurthermore assume that P attains integer values for x ∈ Z ℓ .For a totally-positive quadratic polynomial P , we set r P ( n ) := { x ∈ Z ℓ : P ( x ) = n } . Note that if P = Q is a quadratic form with associated Gram matrix A , then for each matrix B ∈ GL ℓ ( Z ) satisfying B T AB = A and each x such that Q ( x ) = n , we have Q ( B x ) = x T B T AB x = x T A x = Q ( x ) = n. e call B an automorph of Q and set ω Q to be the number of automorphs of Q . The matrix B is a special case of an isometry between two quadratic forms Q and Q ′ ; we say that Q ′ isisometric to Q over a ring R if there exists B ∈ GL ℓ ( R ) such that B T AB = A ′ , where A and A ′ are the Gram matrices of Q and Q ′ , respectively. The set of isometry classes of a givendiscriminant is finiteThe first check for representations of n by a quadratic polynomial is to test local conditions.Namely, if P ( x ) = n is not solvable with x ∈ Z ℓp for some prime p (or, equivalently, modulo p j for some j ), then clearly P ( x ) = n is not solvable with x ∈ Z ℓ . An integer is saidto be locally represented if it is represented over Z p for all primes p . Minkowski beganthe study of the local-global principle; this asks for which locally-represented integers n doglobal representations (representations over the integers in this setting) exist. Siegel defineda natural weighted average r gen( Q ) ( n ) := 1 P rj =1 1 ω Qj r X j =1 r Q j ( n ) ω Q j , where the sum runs over all of the isometry classes of positive-definite quadratic forms Q j which are isometric to Q over Z p for all p (the set of such forms is known as the genus of Q and r is known as the class number of Q ). Siegel [12, 13] and Weil [15] then computedso-called local densities (roughly speaking, these “count” the number of representations over Z p and vanish precisely when no such representations exist) to give an explicit formula for r gen( Q ) ( n ). We need only the following well-known special form of their results. Theorem 2.1 (Siegel, Weil) . We have that r gen( Q ) ( n ) > if and only if n is locally repre-sented. Moreover, if the class number of Q is one, then r Q ( n ) > if and only if n is locallyrepresented. The following lemma plays a crucial role in the proof of Theorem 1.1.
Lemma 2.2.
The sum P j =1 P m ( x j ) represents every integer in the set ( m − N .Proof. Consider x ∈ Z in the hyperplane P j =1 x j = 0. For x in this hyperplane, we have X j =1 P m ( x j ) = m − X j =1 x j − m − X j =1 x j = m − X j =1 x j + ( − x − x − x − x ) ! = ( m − X ≤ i ≤ j ≤ x i x j . The quadratic form P ≤ i ≤ j ≤ x i x j has class number one and represents every integer locally,and is hence universal by Theorem 2.1 (alternatively, one may simply use the 290-theoremof Bhargava and Hanke [3] and verify that every integer up to 290 is represented by thisquadratic form, and thus the form is universal). (cid:3) . The extension of Fermat’s polygonal number theorem for r = 1 andlarge r In this section, we prove parts (1) and (2) of Theorem 1.1, giving the generalization ofFermat’s polygonal number theorem answering Guy’s question and covering the generic casefor r ≥ The case r = 1 . We next make use of Lemma 2.2 in order to prove Theorem 1.1 (1).
Proof of Theorem 1.1 (1).
The case m = 3 was proven by Gauss, the case m = 4 was provenby Lagrange, and Guy [7] uses Legendre’s classification of the integers which are sums ofthree squares to resolve the m = 5 case. Guy also points out that the set of generalizedhexagonal numbers is precisely the set of triangular numbers and hence the m = 6 casefollows from the m = 3 case. The m = 8 case is proven by Sun in [14, Theorem 1.1].Now assume that m ≥
10. Since we know that ℓ m ≥ m − m − m -gonalnumbers. Let n ∈ N be given and write it as n = ( m − k + k with 0 ≤ k ≤ m −
3. By Lemma 2.2, every multiple of m − m -gonal numbers. Hence if k may be written as a sum of m − m -gonal numbers, then we may choose x ∈ Z m − for which P j =1 P m ( x j ) = ( m − k and P m − j =6 P m ( x j ) = k , yielding the claim. This is possible for 0 ≤ k ≤ m − k = m − P m ( −
1) = m − m − ≤ k ≤ m −
4. We thus write k = m − − k with2 ≤ k ≤
6. For k ≥ k − n = ( m − k + k = ( m − k − k ) + k ( m −
2) + ( m − − k )= ( m − k − k + 1) + k ( m − , from which we conclude that n may be written as the sum of 5 + k generalized m -gonalnumbers (again using Lemma 2.2). If 5 + k ≤ m − m ≥ k + 9 which is automaticallytrue for m ≥ n is represented as long as k ≥ k −
1. On the otherhand, if m < k + 9, then we note that P m ( − k ) = ( m − k + k − k = ⇒ k = m − − k = P m ( − k ) − ( m − (cid:18) k + k − (cid:19) and write n = ( m − k + k = ( m − (cid:18) k + 1 − k + k (cid:19) + P m ( − k ) . Using Lemma 2.2, n may hence be written as the sum of 6 ≤ m − m -gonalnumbers as long as k ≥ k + k − n may be represented in the finitely many cases 0 ≤ k < k − ≤ k < k + k −
1) when m ≥ k + 9 (resp. 10 ≤ m < k + 9), with 2 ≤ k ≤
6. Firstsuppose that m ≥ k + 9. For 0 ≤ k < k − m − k + k = mk + m − − k − k. or m ≥ k + k (in particular, since k ≤ k − k ≤
6, this holds for m ≥
16) wesee that n may be represented by using k choices of m and m − − k − k choices of 1, forwhich we need (using that 0 ≤ k < k − k ≥ k + m − − k − k = m − − k − k ≤ m − − k ≤ m − m = 15, k = 4, and k = 6, for which onemay check by hand that 59 = 12 + 3 ·
15 + 2 · ≤ m < k + 9 ≤
15 and 0 ≤ k < k + k − ≤ n which need to be checked, and this may be doneby hand. (cid:3) Remark.
After reducing the proof to a check of finitely-many cases, we simply check the re-maining cases by hand for 10 ≤ m < k + 9 ≤
15 and 0 ≤ k <
20. One may instead drop therestriction 10 ≤ m <
15 (leaving m arbitrary as a variable) and use (the following list is com-plete for P m ( x ) ≤ m −
35 because the sequence ( P m (0) , P m (1) , P m ( − , P m (2) , P m ( − , . . . )is increasing for m > { P m ( x ) : x ∈ Z } = { , , m − , m, m − , m − , m − , m − , m − , m − , m − , m − , m − , m − , . . . } to systematically write ( m − k + ( m − − k ) (thinking of this as a polynomial in m ) asa linear combination of the polynomials occurring in (3.1) for each choice of 0 ≤ k < Inequalities for large r . For n ∈ N and r < m −
3, we write n ∈ N in the form(3.2) n = ( m − k + rk + k , where 0 ≤ k ≤ (cid:4) m − r (cid:5) and − ≤ k ≤ r −
6. In order to obtain an upper bound, we needthe following extension of Lemma 2.2.
Lemma 3.1.
Suppose that ≤ r < m − . For k ∈ N and − ≤ k ≤ r − , the integer k ( m −
2) + k ∈ N is represented by the sum of at most r − generalized m -gonal numbersunless − ≤ k ≤ − and k ≤ | k | − .Proof. Using Lemma 2.2, we may represent ( m − k with the first 5 variables. If 0 ≤ k ≤ r −
6, then we may represent k by taking P m ( x j ) ∈ { , } for the remaining r − m -gonal numbers.Now suppose that − ≤ k ≤ −
1. We note that P m ( x ) = ( m − P ( − x ) + x. Hence in particular we have P m ( k ) = ( m − P ( − k ) + k . We may therefore rewrite (3.2) as n = ( m −
2) ( k − P ( − k )) + rk + P m ( k ) , and for k ≥ P ( − k ) we conclude that n − rk may be represented with the first 6 ≤ r − t remains to show that n is represented for the cases k < P ( − k ) and − ≤ k ≤ − m − k + k ∈ { m − , m − , m − , m − , m − , m − , m − , m − , m − , m − } we may use (3.1) (thinking of ( m − k + k ∈ Z [ m ] as a polynomial in m ) to find arepresentation ( m − k + k = X j =1 P m ( x j )in 5 < r − A, B ) position corresponds to A ( m −
2) + B . If we have a representation A ( m −
2) + B = d X j =1 P m ( x j )in d variables and A ( m −
2) + B + P m ( x ) = C ( m −
2) + D , then in the ( C, D ) location ofthe graph we write (
C, D ) xd +1 to indicate that we have a representation of C ( m −
2) + D in d + 1 variables where we take x d +1 = x . One may then reconstruct the representation of C ( m −
2) + D by recursively working backwards through the graph; for example, if we have( C, D ) − d +1 , then we obtain the representation by looking at ( C − , D + 1) ∗ d and continuingrecursively until we have d = 1. To summarize, one traverses backwards through the graphas follows: ( C, D ) − d +1 → ( C − , D + 1) ∗ d , ( C, D ) d +1 → ( C, D − ∗ d , ( C, D ) d +1 → ( C − , D − ∗ d , ( C, D ) − d +1 → ( C − , D + 2) ∗ d , ( C, D ) − d +1 → ( C − , D + 4) ∗ d . his yields the following graph encoding the representations (we add the unnecessary entries( C, −
1) in order to include the representations of some integers in the ( − , −
5) (1 , −
4) (1 , −
3) (1 , −
2) (1 , − − (2 , −
5) (2 , −
4) (2 , −
3) (2 , − − (2 , − (3 , −
5) (3 , −
4) (3 , − − (3 , − − (3 , − (4 , −
5) (4 , − − (4 , − − (4 , − (5 , − − (5 , − − (5 , − − (6 , − − (6 , − − (6 , − − (7 , − − (7 , − − (8 , − − (8 , − (9 , − − (9 , − (10 , − − (10 , − − (11 , − − (12 , − − (13 , − − (14 , − − (15 , − − (cid:3) For the exceptional cases − ≤ k ≤ − k ≤ | k | −
1, we use the following lemma.
Lemma 3.2. If ≤ r < m − , k ≥ , and k ( m −
2) + k satisfies − ≤ k ≤ − and ≤ k ≤ | k | − , then n = k ( m −
2) + rk + k may be represented by r − X j =1 P m ( x j ) + r r + k − X j = r P m ( x j ) . In particular, we may take ℓ ≥ k − .Proof. For some 0 ≤ j ≤ k we have n = ( m − k + ( k − r + ( r + k ) = ( k − j ) m + j ( m −
3) + ( r + k − k + 3 j ) + ( k − r = ( k − j ) P m (2) + jP m ( −
1) + ( r + k − k + 3 j ) P m (1) + ( k − r. If r + k − k + 3 j ≥
0, then( k − j ) P m (2) + jP m ( −
1) + ( r + k − k + 3 j ) P m (1)is the sum of r + k − k + 3 j generalized m -gonal numbers. Hence if the system of equations r + k − k + 3 j ≤ r − ,r + k − k + 3 j ≥ | k | + 2 k ≤
6, then since r − ≥ j = 0. For7 ≤ | k | + 2 k ≤
10 the inequality k ≤ | k | − | k | + k >
3, and hence we maytake j = 1 in that case. Finally, if 11 ≤ | k | + 2 k ≤
13, then | k | + k >
6, so we may take j = 2 in this case. We are now ready to obtain an upper bound for ℓ m,r,r − for large r . Proposition 3.3. If ≤ r < m − , then we have ℓ m,r,r − ≤ (cid:4) m − r (cid:5) + ( r − .Proof. The claim is equivalent to proving that P m, a r,r − ,ℓ is universal for ℓ = (cid:4) m − r (cid:5) .Since k ≤ ℓ , we may represent rk with the r -times repeated variables all having x j ∈{ , } (i.e., P m ( x j ) ∈ { , } ), and Lemma 3.1 implies that n − rk = ( m − k + k may berepresented by the initial r − − ≤ k ≤ − k ≤ | k | − n = k ( m −
2) + k r + k with 0 ≤ k ≤ | k | − − ≤ k ≤ −
1. If k ≥
1, then Lemma 3.2 implies that n is represented.It remains to resolve the k = 0 case for − ≤ k ≤ − ≤ k ≤ | k | −
1. Inother words, we need to check the representations of the 10 integers k ( m −
2) + k ∈{ m − , m − , m − , m − , m − , m − , m − , m − , m − , m − } . We write m = rs + t for some 0 ≤ t ≤ r −
1. This gives n = k ( m −
2) + k = ( k − P m (2) + rs + t + k − k . If t + k − k ≥
0, then we are done because k − t + k − k ) < t ≤ r −
1. If t ≤ | k | + k ,then we may write (note that s ≥ m = t ≤ r −
1, which contradicts theassumption that m − > r ≥ n = ( k − j − P m (2) + jP m ( −
1) + ( s − r + t + k − k + 3 j + r. Noting that s − ≤ (cid:4) m − r (cid:5) , we are done as long as t + k − k + r + 3 j ≥ ,k − j − ≥ ,t + k − k + 3 j ≤ , with the last inequality coming from the fact that n − ( s − r must be represented by k − j − j + t + k − k + 3 j + r ≤ r − m -gonal numbers. As in the proofof Lemma 3.2 this holds for some j ∈ { , } .We finally deal with the case | k | + k < t < | k | + 2 k . Since t > | k | + k ≥ m − rs + t − ≥ rs and hence s = (cid:4) m − r (cid:5) . In this case, we rewrite n = k P m ( −
1) + k + k = ( k − − j ) P m ( −
1) + jP m (2) + rs + t − − j + k + k . Writing t = k + | k | + t ′ with 1 ≤ t ′ < k , we are done as long as t − k + k − j ≥ , k − − j + t ′ ≥ ,k − − j ≥ , ⇔ k − − j ≥ ,t + 2 k + k − j − ≤ r − , k + t ′ − j − ≤ r − , with the last inequality coming from the fact that we must write n − rs as the sum of atmost k − t − − j + k + k generalized m -gonal numbers. Setting δ := 1 if t ′ = k − δ = 0 otherwise, we claim that j = t ′ − δ satisfies the above system of inequalities. Since j ≤ t ′ ≤ k −
1, the second inequality automatically holds. The first inequality2 k − − j + t ′ = 2 ( k − t ′ ) − δ ≥ olds because k − t ′ ≥
1, with k − t ′ = 1 if and only if δ = 1. The third inequality becomes3 k − t ′ + 3 δ − ≤ r − . Note that since t ′ ≥
1, we have − t ′ + 3 δ ≤ − t ′ = k − ≤
2. In the exceptionalcase k ≤ t ′ = k −
1, we have3 k − t ′ + 3 δ − k + 1 ≤ < r − − t ′ + 3 δ ≤ − k ≤
4, and r ≥
7, so we find that3 k − t ′ + 3 δ − ≤ k − ≤ ≤ r − , and the claim follows. (cid:3) We next use Guy’s argument to obtain a lower bound for ℓ m,r,r − . Proposition 3.4.
We have ℓ m,r,r − ≥ (cid:6) m − r (cid:7) + ( r − .Proof. Following Guy [7], if P m, a is universal, then it must necessarily represent m −
4. Wewrite m − r − X j =1 P m ( x j ) + r ℓ X j = r P m ( x j )(3.4)By (3.1), we have P m (0) = 0, P m (1) = 1, and P m ( x ) > m − x / ∈ { , } , so anyrepresentation of m − x j ∈ { , } . Thus (3.4) yields the inequality m − ≤ r − r ( ℓ − r + 1) , from which we conclude that ℓ m,r,r − ≥ ℓ ≥ m − r + r − . (3.5)This yields the claim. (cid:3) We are now ready to Prove Theorem 1.1 (2).
Proof of Theorem 1.1 (2).
The upper bound in Proposition 3.3 and the lower bound inProposition 3.4 match unless r | m −
3. In the remaining case, we write m − rs and note that the claim is equivalent to proving that P m, a r,r − ,ℓ is universal for ℓ = s − n . By Lemma 3.1, if 0 ≤ k ≤ s −
1, then n − rk may berepresented by the initial r − m -gonal numbers and we only require k ≤ ℓ variables to represent rk , unless k < k ≤ | k | −
1. For k < k ≤ | k | −
1, weuse Lemma 3.2 to see that n is represented with ℓ = s − k = 0, while for k = 0we use the splitting (3.3) (with t = 3) with j ∈ { , } to obtain a representation.For k = s , rewrite n = k ( m −
2) + m − k = ( k + 1)( m −
2) + k + 1 . Again using Lemma 3.1, we see that n is represented by r − m -gonal numbersunless ( − ≤ k ≤ − k + 1 ≤ | k | −
2) or k = r −
6. We use (3.3) in the case − ≤ k ≤ −
2. In the case of k = r −
6, we then rewrite n = ( k + 1)( m −
2) + r − . In this case, Lemma 3.1 implies that n is represented unless k ≤
3, while Lemma 3.2 with k = 1 yields the claim for k ≤ Small choices of r In this section, we consider cases for small r . Proof of Theorem 1.1 (3).
We first assume that r = 2 and m ≥
14. Note that since m − m − m − P m, a , ,ℓ must have ℓ ≥ (cid:4) m (cid:5) −
1, or in otherwords ℓ ≥ (cid:4) m (cid:5) , yielding the lower bound ℓ m, , ≥ (cid:4) m (cid:5) .It remains to show that the form P m, a , ,ℓ with ℓ := (cid:4) m (cid:5) is indeed universal. We present n in the form(4.1) n = 2( m − k + 2 k + k . with − ≤ k ≤ m − k ∈ { , } (so 2 k + k precisely attains every residue mod-ulo 2( m −
2) once). By Lemma 2.2, we may represent 2( m − k as a sum of the type2 P j =2 P m ( x j ). If 0 ≤ k ≤ ℓ −
6, then we conclude that n may be represented by P m, a , ,ℓ .We next consider ℓ − ≤ k ≤ m −
8. Choosing j ∈ {− , } such that P m ( j ) ≡ k (mod 2),we may rewrite (4.1) as n = 2( m − k + 2 (cid:18) k + k − P m ( j )2 (cid:19) + P m ( j ) . Setting k := k + k − P m ( j )2 , the inequalities for k imply that ℓ − − m ≤ k ≤ m − − m −
42 = m − . Since k is an integer, we conclude that − ≤ k ≤ ℓ −
6. For 0 ≤ k ≤ ℓ − − ≤ k ≤ − r = 7) to conclude that( m − k + k may be written as the sum of at most 6 ≤ ℓ − m ≥
14 we have ℓ ≥
7) generalized m -gonal numbers unless k ≤ | k | − − ≤ k ≤ −
1. In this case we write n = 2 ( k ( m −
2) + k ) + k and, since − ≤ k ≤ −
1, Lemma 3.1 (choosing r = 7) implies that k ( m −
2) + k may bewritten as the sum of at most 6 ≤ ℓ − m ≥
14 we have ℓ ≥
7) generalized m -gonalnumbers unless k ≤ | k | − k ∈ { P m (0) , P m (1) } . It remains to show that for each − ≤ k ′ ≤ − ≤ k ≤ | k ′ | − − ≤ j ′ ≤ P m ( x ) + 2 ℓ X j =2 P m ( x j ) = n = 2 (( m − k + k ′ ) + P m ( j ′ ) . For each n of the form (4.2), we claim that we may choose x , . . . , x d +1 (with d ∈ N ) sothat P m ( x ) ≡ n (mod 2) and(4.3) 0 ≤ n − P m ( x ) − d +1 X j =2 P m ( x j ) ≤ ℓ − d. ote that if we may choose x j in this way, then since n ′ := n − P m ( x ) − d +1 X j =2 P m ( x j )is even and less than 2( ℓ − d ), we may write n ′ as a sum of at most ℓ − d twos, giving arepresentation of n with ℓ variables. It remains to choose the first d of the x j s appropriately.We collect the choices of the set X n such that x ∈ X n , x , . . . , x d +1 and the correspondingbounds on n ′ in Table 4.1. The bounds on n ′ are proven, for example, for 2 m − ≤ n ≤ m − x ∈ X n = {− , } by writing (using m ≥ ≤ m −
14 = 2 m − − m ≤ n − P m ( x ) ≤ m − − ( m −
3) = m − . Recalling that ℓ = (cid:4) m (cid:5) −
1, we have m − < m − ≤ ℓ (in general, it suffices to showthat m − ≤ n ′ ≤ ℓ − d ), and we see that (4.3) holds. Table 4.1.
Individual case checking for r = 2Interval with n x ∈ X n d x , . . . , x d +1 Bounds on n ′ m − ≤ n ≤ m − { , } m − ≤ n ′ ≤ m − m − ≤ n ≤ m − {− , } m − ≤ n ′ ≤ m − m − ≤ n ≤ m − { , } − m − ≤ n ′ ≤ m − m − ≤ n ≤ m − {− , } − m − ≤ n ′ ≤ m − m − ≤ n ≤ m − { , } , m − ≤ n ′ ≤ m − m − ≤ n ≤ m − { , } − , m − ≤ n ′ ≤ m − m − ≤ n ≤ m − {− , } − m − ≤ n ′ ≤ m − m − ≤ n ≤ m − { , } − , − , m − ≤ n ′ ≤ m − n = 8 m − {− , } − , − m − ≤ n ′ ≤ m − n = 8 m − m odd {− } − , n ′ = m − n = 8 m − m even { } − , − n ′ = m − m − ≤ n ≤ m − { , } − , − , − , m − ≤ n ′ ≤ m − n = 9 m − {− } − , − , − , − n ′ = 0 n = 9 m − {− } − , − , − n ′ = 0We next consider the r = 3 case. In this case, first note that Theorem 1.1 (1) implies that3 P m − j =3 P m ( x j ) represents every element of 3 N . Taking P m ( x ) , P m ( x ) ∈ { , } , we get arepresentation of every positive integer.We note that any representation of 3 m −
10 must have P m ( x ) ≡ P m ( x ) ≡ m ≡ m ≡ m − ≡ x = x = 1so that 3 m −
12 = 3 ℓ X j =3 P m ( x j ) . Dividing by 3 and using Guy’s argument [7] again, this implies that ℓ − ≥ m −
4, or inother words ℓ ≥ m − or m ≡ n = 3 m −
12 and similarly note that any representation of3 m −
12 = P m ( x ) + P m ( x ) + 3 ℓ X j =3 P m ( x j )must have x = x = 0, again implying that ℓ ≥ m − m ≡ n = 2 m − m − < m − m − P m ( x ) + P m ( x ) + 3 ℓ X j =3 P m ( x j )must have x = 0, x = 1, and P m ( x j ) ≤
1, from which we conclude that ℓ ≥ m − .We note that any representation of n < m −
2) must satisfy P m ( x ) + P m ( x ) ∈ { , m − , m + 1 } if n ≡ , { , m − , m − , m, m − } if n ≡ , { , m − , m, m − } if n ≡ . From this we can conclude that P ( x ) + P ( x ) + 3 P m − − j =3 P m ( x j ) represents every positiveinteger up to 3( m −
2) except integers in K := { m − , m − , m − , m − , m − , m − , m − , m − , m − , m − , m − , m − , m − , m − } . By Lemma 2.2, n = 3( m − k + k with k ∈ N , 0 ≤ k < m −
2) and k K isrepresented by P ( x ) + P ( x ) + 3 P m − j =3 P m ( x j ).On the other hand, for each k ∈ K we write k = 3( j ( m − − k ) + α with j ∈ { , } and α ∈ { , , m − } and we see that k ≤ k = 3 m −
22, in whichcase k = 6. Thus for every 3 m − = k ∈ K , Lemma 3.1 implies that k − α + 3 j ( m −
2) isrepresented as 3 times the sum of at most 6 generalized m -gonal numbers for j ≥ k − j . Using(3.1) we may check the smaller choices of j directly. For the remaining case k = 3 m − k + 3 j ( m −
2) = 3(( j + 1)( m − −
6) + 2= 3( j ( m − − m −
3) + 2 = 3( j ( m − − P m ( − . Thus, using Lemma 3.1 to represent j ( m − −
5, for every j ≥ k +3 j ( m −
2) as long as ℓ ≥
9. There remain finitely many choices of j for each k ∈ K and wecheck these as in the r = 2 case. Now suppose that 4 ≤ r ≤
6. We first obtain lower boundsfor ℓ m,r,r − by using Guy’s argument [7] for the exceptional choices of n in Table 4.2.We define sets S r by S := { , , , } ∪ { m + j : − ≤ j ≤ } ∪ { m − , m − , m − , m − , m, m + 1 }∪ { m + j : − ≤ j ≤ − ≤ j ≤ } ∪ { m − , m − } , able 4.2. Exceptional n for 4 ≤ r ≤ r m (mod r ) n lower boundfor ℓ m,r,r − m m − (cid:6) m − (cid:7) + 24 m ≡ m − (cid:6) m − (cid:7) + 35 all m − (cid:6) m − (cid:7) + 36 all m − (cid:6) m − (cid:7) + 4 S := { , , , , } ∪ { m + j : − ≤ j ≤ } ∪ { m + j : − ≤ j ≤ }∪ { m + j : − ≤ j ≤ − − ≤ j ≤ } ∪ { m + j : − ≤ j ≤ } ,S := { , , , , , } ∪ { m + j : − ≤ j ≤ } ∪ { m + j : − ≤ j ≤ }∪ { m + j : − ≤ j ≤ } ∪ { m + j : − ≤ j ≤ }∪ { m + j : − ≤ j ≤ } ∪ { m + j : − ≤ j ≤ − } . The sets S r are precisely the integers less than r ( m −
2) which are represented by P r − j =1 P m ( x j ).For each n ∈ N , we choose s ∈ S r and k , k ∈ N with k minimal such that n = s + r ( m − k + rk . By Lemma 2.2, we obtain a representation of n with r − k variables, taking x j = 1for the last k variables. If k ≤ ℓ − ℓ = ℓ m,r,r − as given in the statement of thetheorem, then n may be represented. We check in Tables 4.3, 4.4, and 4.5 that k ≤ ℓ . Forthose ℓ − ≤ k ≤ ℓ , we rewrite(4.4) n = s + r ( m − k − k )+ rkm + r ( k − k ) = s + r ( m − k − k )+ rkP m (2)+ r ( k − k ) P m (1) . Having chosen m large enough so that ℓ ≥
14, we see that for k ≤ k − k ≥ ℓ − − k ≥ . We then choose k := min(5 , k ). If k = k , then (4.4) gives a representation of n with r − k + ( k − k ) = r − k − k ≤ ℓ variables. On the other hand, if k = 5, thenLemma 2.2 may be employed to represent r ( m − k −
5) and we obtain a representationof n in (4.4) with r − k − ≤ ℓ variables. Table 4.3.
Bounds for k ≤ ℓ in the r = 4 caseInterval of n < m − s ∈ S k bound0 ≤ n ≤ m − { , , , } k ≤ m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − ( m − = m − { } { } k = 02 m − { m + j : − ≤ j ≤ } k ≤ m − − ( m − = m − m − ≤ n ≤ m − { m − , m − , k ≤ m − − (2 m − = m − m − , m, m + 1 } m − ≤ n ≤ m − { m + j : − ≤ j ≤ − } k ≤ m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (3 m − = m − able 4.4. Bounds for k ≤ ℓ in the r = 5 caseInterval of n < m − s ∈ S k bound0 ≤ n ≤ m − { , , , , } k ≤ m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − ( m − = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (2 m − = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ − k ≤ m − − (3 m − = m − or − ≤ j ≤ m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (4 m − = m − Table 4.5.
Bounds for k ≤ ℓ in the r = 6 caseInterval of n < m − s ∈ S k bound0 ≤ n ≤ m − { , , , , , } k ≤ m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − m = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (2 m − = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (3 m − = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ } k ≤ m − − (4 m − = m − m − ≤ n ≤ m − { m + j : − ≤ j ≤ − k ≤ m − − (5 m − = m − or − ≤ j ≤ − − ≤ j ≤ } m − ≤ n ≤ m − { m + j : − ≤ j ≤ − } k = 0 (cid:3) We are now ready to conclude the corollary.
Proof of Corollary 1.2.
The proof of Theorem 1.1 (3) immediately implies Corollary 1.2 be-cause either 3 m −
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Department of Mathematics, University of Hong Kong, Pokfulam, Hong Kong
E-mail address : [email protected] Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai,Maharashtra 400076, India
E-mail address : [email protected] Department of Mathematics, University of Hong Kong, Pokfulam, Hong Kong
E-mail address : [email protected] Department of Mathematics, Hong Kong Baptist University, Kowloon Tong, Kowloon,Hong Kong
E-mail address : [email protected] Department of Mathematical Sciences, Seoul National University, Seoul 151-747, Repub-lic of Korea
E-mail address : [email protected] Department of Mathematics, Indian Institute of Science Education and Research, Thiru-vananthapuram, Vithura, Kerala 695551, India
E-mail address : [email protected] Department of Mathematics, University of Hong Kong, Pokfulam, Hong Kong
E-mail address : [email protected] Department of Mathematics, Indian Institute of Technology, Roorkee, Roorkee, Ut-tarakhand 247667, India
E-mail address : [email protected]@gmail.com