This paper proves that primitive regular m-gonal forms do not exist for large m by constructing prime sequences with specific properties related to quadratic fields.
Contribution
It introduces a novel method of constructing prime sequences inert in quadratic fields to establish non-existence results for regular m-gonal forms.
Findings
01
No primitive regular m-gonal forms for sufficiently large m
02
Constructed prime sequences satisfy key inequalities
03
Linked prime inertness to form regularity
Abstract
Inspired by Dickson's classification of regular ternary quadratic forms, we prove that there are no primitive regular m-gonal forms when m is sufficiently large. In order to do so, we construct sequences of primes that are inert in a certain quadratic field and show that they satisfy a certain inequality bounding the next prime by the product of the previous primes, a question of independent interest.
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Taxonomy
TopicsAlgebraic Geometry and Number Theory · Finite Group Theory Research · Analytic Number Theory Research
Full text
Regular ternary polygonal forms
Zilong He
Department of Mathematics, University of Hong Kong, Pokfulam, Hong Kong
Inspired by Dickson’s classification of regular diagonal ternary quadratic forms, we prove that there are no primitive regular ternary m-gonal forms when m is sufficiently large. In order to do so, we construct sequences of primes that are inert in a certain quadratic field and show that they satisfy a certain inequality bounding the next such prime by a product of the previous primes, a question of independent interest.
Key words and phrases:
Polygonal numbers, regular quadratic polynomials, Diophantine equations, inequalities for primes
2010 Mathematics Subject Classification:
11D09,11E12,11E20
The research of the second author is supported by grant project numbers 17316416, 17301317, and 17303618 of the Research Grants Council of Hong Kong SAR. Part of the research was also conducted while the second author was supported by grant project number 17302515 of the Research Grants Council of Hong Kong SAR
1. Introduction
Representations of integers as sums of polygonal numbers have a long history going back to Fermat. Fermat famously conjectured that every integer may be written as the sum of 3 triangular numbers, 4 squares, 5 pentagonal numbers and in general mm-gonal numbers; Lagrange proved the four squares theorem, Gauss and Legendre independently showed the triangular number theorem, and Cauchy finally proved the general case. For m≥3 and x∈Z, we denote by pm(x):=((m−2)x2−(m−4)x)/2
the x-th generalized m-gonal number and for a sequence a1,…,an∈N we define the m-gonal form (or polygonal form )
[TABLE]
Fermat’s polygonal number conjecture may then be restated by saying that for a=(1,…,1) of length m, the m-gonal form △m,a is universal, i.e., for every positive integer ℓ, the Diophantine equation △m,a(x)=ℓ
is solvable. More generally, let F be a field and R⊂F a ring. For an n-ary quadratic polynomial f(x1,⋯,xn)∈F[x1,⋯,xn] and ℓ∈F, we say that ℓ is represented by f if the equation f(x)=ℓ is solvable with x∈Rn, which we denote by ℓR→f, and not represented otherwise, which we denote by ℓR→f. It is well known that a natural number can be represented by the sum of three squares if and only if it is not of the form 4t(8ℓ+7), with the restriction coming from the fact that one cannot even solve the congruence equation x12+x22+x32≡7(mod8).
In investigating representations of integers by quadratic polynomials it is therefore natural to first exclude integers which cannot possibly be represented modulo a fixed integer and then separately investigate the “sporadic” integers for which congruence equations are always solvable but the equation over the integers is not solvable. In order to better encode this information, we let Zp be the p-adic integers, with Z∞:=R by convention. We say that ℓ is locally (resp. globally) represented by an n-ary rational quadratic polynomial f, if ℓZp→f for each prime p including p=∞ (resp. if ℓZ→f).
A general principle, known as the Minkowski local-global principle, states that one should “usually” find a global solution whenever one finds a local solution. The aforementioned example states that the form given by the sums of three squares always obeys the Minkowksi local-global principle. This led to L. E. Dickson [7] asking which other quadratic forms always obey the local to global principle. He dubbed such forms regular, starting an investigation and classification of such forms (his definition being equivalent to Jones’s definition [11, Corollary, p. 124]). To more formally define regularity, we adopt the following definition of Chan and Ricci [5].
Definition 1.1*.*
A quadratic polynomial f is said to be regular if it globally represents all rational numbers that are locally represented by f. We also call f irregular if f is not regular.
It was shown by Jagy, Kaplansky, and Schiemann [10] that there are at most 913 regular ternary (i.e., n=3) quadratic forms (some of these are still only conjectural, although the list has been shown to be correct by Lemke Oliver [16] under the assumption of GRH), up to obvious repeats coming from multiplying a regular form by a fixed constant or by an invertible change of variables (more precisely, an isometry over Z). It is hence natural to wonder how abundant regular m-gonal forms are. In order to exclude the obvious repeats mentioned above, we call an m-gonal form primitive if gcd(a1,⋯,an)=1 and its discriminant is defined by ∏i=1nai. Chan and B. K. Oh [4] showed that there are only finitely many primitive regular ternary triangular forms (m=3), a result which was later extended by Chan and Ricci [5] to finiteness results for ternary quadratic polynomials. In this paper, we improve their results by obtaining a quantitative bound in terms of m on the possible choices of (a,b,c) for which △m,(a,b,c) may be regular, leading to the following theorem.
Theorem 1.2**.**
There exists an absolute constant C such that for m>C, there are no primitive regular ternary m-gonal forms △m,(a,b,c) with (a,b,c)∈N3.
Remark 1.3*.*
Due to the bound in Theorem 1.2 and Chan and Ricci’s results in [5], there are only finitely many tuples (a,b,c,m)∈N4 with gcd(a,b,c)=1 (m≥3) for which △m,(a,b,c) is regular. It would be interesting to try to determine this finite set explicitly. There has been recent progress in this direction, as M. Kim and B.-K. Oh [15] have just completely determined all of the regular ternary triangular forms △3,(a,b,c), determining that there are precisely 49 of them (see [15, Theorem 4.10 and Table 4] for a full list).
In the classification of primitive regular ternary quadratic forms ax2+by2+cz2 (namely, △4,(a,b,c)) [7, 13], to rule out most of the irregular ones, Dickson made use of an inequality involving primes of a certain type [7, Theorem 5]. To be more explicit, for a given positive integer b, assume that pi’s are all the odd prime numbers not represented by x2+by2 in ascending order and choose i0 such that
[TABLE]
He proved the inequality pi+1<p1p2⋯pi holds for i≥i0 [7, footnote, p. 336]. To give a rough illustration how such an inequality applies to the regularity of such forms, suppose that △4,(1,b,c) is regular and pi0+1 is locally represented. Then it must be the case that c≤pi0+1 (since otherwise x2+by2+cz2=pi0+1 cannot be solvable), and the inequality yields an inequality on c depending on b (as the p1,…,pi0 are all primes smaller than b). Inspired by this, we deal with primitive ternary m-gonal forms by virtue of analogous technical inequalities involving inert primes (see (4.1)), thereby showing Theorem 1.2.
The paper is organized as follows. In Section 2, we establish Lemma 2.6 by Earnest’s trick, which will be used to deduce the inequality (4.1) involving inert primes with additional restrictions analogous to Dickson’s one. In Section 3, we introduce the Watson’s transformation and give the conditions on local representation by a (ternary) polygonal form. In Section 4, we prove Theorem 1.2 by bounding the discriminant abc.
2. Earnest’s trick
Let k1,k2,…,kr be pairwise relatively prime positive integers. Let χi be a Dirichlet character modulo ki and ηi∈{±1}. Define
[TABLE]
For an integer M relatively prime to Γ:=\mboxlcm(k1,k2,⋯,kr) and a nonnegative number x we furthermore set
[TABLE]
Following Earnest’s trick [9, p. 855–856], we give an explicit bound on Sx(H). In order to state the bound, we require some notation. Let U={1,2} and α=(α1,⋯,αr) be an element of the product set Ur. Define
χα=∏i=1r(ηiχi)αi. Then χα is clearly a Dirichlet character modulo Γ. Characters χ1,…,χr are said to be independent if χα is a nonprincipal character for any α=β0, where β0=(2,⋯,2). We also let ω(n) denote the number of distinct prime divisors of n and ϕ denote the Euler totient function.
Lemma 2.1**.**
Suppose that χ1,⋯,χr are independent. Then
[TABLE]
We need explicit estimates for character sums before showing Lemma 2.1 and use a version of Polya–Vinogradov inequality proved by Bachman and Rachakonda [1, Corollary, p. 66].
Proposition 2.2** (Bachman - Rachakonda).**
Let k∈N. If χ is a nonprincipal character of modulus k and x,y are real numbers with x<y, then
[TABLE]
independent of x and y.
We modify Proposition 2.2 slightly so that it is applicable to our situation.
Lemma 2.3**.**
Let k,M be integers with gcd(k,M)=g. Then for any nonprincipal character χ of modulus k, we have
[TABLE]
Proof.
Let g1 be the least positive integer for which M/g1 is an integer relatively prime to g and M/g1=p1γ1p2γ2⋯prγr, where p1,…,pr are distinct primes. Then, by inclusion-exclusion, we have (letting μ denote the Möbius μ-function)
As g and g1 have the same prime factors, we have ω(M/g1)=ω(M/g)=ω(M)−ω(g). Plugging this into (2.2) and noting that χ(n)=0 if gcd(n,g1)>1, it follows that
[TABLE]
where in the last line we have again used Proposition 2.2.
∎
First note that if n∈Sχ,η (defined in (2.1)), then
[TABLE]
for any α∈Ur. On the other hand, if n∈/Sχ,η, then there exists some j for which either χj(n)=0 or ηjχj(n)=−1. In the former case, χα(n)=0 for any α∈Ur, while in the latter case we split the cases αj=1 and αj=2 to obtain that (assuming without loss of generality that j=r for ease of notation)
[TABLE]
Hence we see that
[TABLE]
and so
[TABLE]
We use the inclusion-exclusion principle to bound the first term from below by
[TABLE]
where {y}:=y−⌊y⌋ denotes the fractional part of y∈R. Since the χj are independent, all of the characters in the second term are nonprincipal, and hence Lemma 2.3 may be used to obtain the lower bound
[TABLE]
Combining these, we obtain
[TABLE]
∎
Remark 2.4*.*
Given a discriminant D and n∈N, let p1,p2,…,ps be the distinct odd prime divisors of d:=∣D∣, ν0(n):=(n−4), ν1(n):=(n8), and χi(n):=(n/pi), where (⋅/pi) is the Legendre symbol, i=1,⋯,s. Then the value of the Kronecker symbol (D/n) is determined by the value at n of these characters (for D<0, see [6, Chap. 1, §3, p. 50]).
[TABLE]
It is not difficult to verify that Γ/ϕ(Γ)≤d/ϕ(d), ω(Γ)≤ω(d), and Γ≤d. Also, note that r≤ω(Γ)+1, where r denotes the number of characters. By Lemma 2.1, we see that S0(H)>0 if
[TABLE]
Besides an explicit bound for Sx(H), we also need explicit upper bounds for n/ϕ(n) and ω(n), which are given by Rosser and Schoenfeld [20, Theorem 15] and Robin [19, Théorème 12], respectively.
For a given non-square discriminant D, let M be a positive integer satisfying M≥2 and gcd(D,M)=1. Set d=∣D∣. Then there exists some prime q∈(0,C0d2/3M1/6) such that (D/q)=−1 and gcd(q,M)=1, where C0:=20664 is a constant.
Proof.
By assumption D≡0,1(mod4) and D is not a perfect square, so d≥3 and hence dM≥6. Consider the function f in terms of d and M given by
[TABLE]
By (2.3), we have S0(H)>0 when H≥f(d,M). To find an appropriate H, we estimate f(d,M) explicitly term by term by virtue of Proposition 2.5 and prove that certain simple functions are nonnegative via a simple application of calculus. Precisely, 2ω(dM)≤4 for 6≤dM<11,
[TABLE]
[TABLE]
[TABLE]
It follows that f(d,M)≤C0d2/3M1/6.
Now, apply Lemma 2.1 with H=C0d2/3M1/6 and ηi′s chosen so that ∏i=1rηi=−1. Then S0(H)≥1. Hence there exists an integer N0∈(0,H) such that (D/N0)=−1 and gcd(N0,M)=1. Accordingly, there exists some prime q dividing N0 such that (D/q)=−1 and gcd(q,M)=1, from which we conclude that q≤N0≤H=C0d2/3M1/6.
∎
3. Local representation over Zp
3.1. Notation and setup
First, we introduce and collect some notation and definitions for the remaining sections. For a given discriminant D, we let
[TABLE]
where (D/⋅) is the Kronecker symbol. Fix an integer m>3 and let H denote a hyperbolic plane. For ℓ,ℓ1,ℓ2∈N, we define the sets
[TABLE]
For given positive integers a,b and c, write
[TABLE]
It is not difficult to see that Pm(a,b,c)∩Gm(a,b,c)=∅. Also, set
[TABLE]
and the corresponding product to be 1 if the specified set is empty. Put ρ(ℓ):=2ω(ℓ)ℓ/ϕ(ℓ) and K(a,b,c):=24Pabρ(Pabc) for short. For convenience, we also let δ=1 if \mboxord2(m)≥2 and [math] otherwise, and introduce the notation {2}δ to mean the set {2} if δ=1 and ∅ otherwise.
The regularity of an m-gonal form △m,(a1,⋯,as) is closely related to the quadratic form with congruence conditions given by
[TABLE]
that arises from completing the square. In this paper we are particularly interested in the case s=3. Now we introduce the regularity of such ternary quadratic polynomials, following the definition of B.-K. Oh [17].
Definition 3.1*.*
Let h be a positive integer and n and k nonnegative integers. If a quadratic polynomial f globally represents all nonnegative integers of the form hn+k that are locally represented by f, then it is said to be (h,k)-regular.
Remark 3.2*.*
For a,b,c∈N, △m,(a,b,c) is regular if and only if φm,(a,b,c) is (h,k)-regular, where (h,k)=(8(m−2),(m−4)2(a+b+c)). Note that if n is locally represented by φm,(a,b,c), then n≡(m−4)2(a+b+c)(mod8(m−2)). Hence φm,(a,b,c) is regular if and only if φm,(a,b,c) is (h,k)-regular. Thus we also call φm,(a,b,c) regular instead of (8(m−2),(m−4)2(a+b+c))-regular.
3.2. Watson Transformations
Following the definitions in [5] and [18], let L and K be Z-lattices on nondegenerate quadratic spaces (V,Q) and (U,Q) over Q, respectively and v,u∈V. A set L+v is called a Z-coset (or a lattice translation); it is called integral if Q(L+v)⊆Z. Given a Z-coset, denote by n(L+v) the Z-ideal generated by Q(x+v) for all x∈L, and call it primitive if n(L+v)⊆Z. Clearly, a Z-coset L+v that is primitive must be integral. Two Z-cosets L+v and K+u are isometric if there exists an isometry σ:V→U such that σ(L)=K and σ(v)−u∈K. The conductor of a Z-coset is defined by the smallest positive integer c such that cv∈L. For n∈Q, n is said to be represented by a Z-coset L+v if there exists x∈L such that Q(x+v)=n. Let Lp be the localization of L at p. The representation of n∈Qp by a Zp-coset Lp+v is defined in the same manner. A Z-coset L+v is said to be regular if it represents all rational numbers that are represented by Lp+v for each prime p, including ∞.
Similar to the case of quadratic forms and lattices, there exists a one-to-one correspondence between the set of equivalence classes of primitive regular complete quadratic polynomials in n variables over Q and the set of isometry classes of primitive regular Z-cosets on quadratic spaces of dimension n over Q ([4, p. 12] or [5, p. 84]). Hence we have the corresponding concepts for quadratic polynomials (e.g. conductor, integrity, primitivity and completeness, see [5, p. 77]). We only introduce the equivalence here.
Definition 3.3*.*
Two quadratic polynomials f(x) and g(x) over Q in n variables are said to be equivalent if there exist T∈GLn(Z) and v∈Zn such that g(x)=f(xT+v).
Suppose that L is a ternary Z-lattice on a quadratic space (V,Q). As usual, we denote by d(L) the discriminant and n(L) the norm of L. For any positive integer m, define
[TABLE]
and
[TABLE]
for each prime p.
Let p be a odd prime. If p∤n(L), define the maps
[TABLE]
then λp sends L to another lattice on the scaled space V1/p or V1/p2. Such maps λp are called Watson’s transformations. We require several properties of Λm(L) and Λm(Lp) (see [5, Lemma 4.2]) and a basic fact ([18, Lemma 2.5]).
Lemma 3.4**.**
Let L be a Z-lattice, m an integer and p a prime. Then
(i)* Λm(L) is a sublattice of L and Λm(Lp) is a sublattice of Lp.*
(ii)* Λm(L)p=Λm(Lp).*
(iii)* Λm(Lp)=Lp for p∤m.*
(iv)* n(Λm(L))⊆mZ and n(Λm(Lp))⊆pZp.*
Lemma 3.5**.**
Let L+v and K+u be Z-cosets. If Lp+v⊆Kp+u for all primes p, then L+v⊆K+u. In particular, L+v=K+u if and only if Lp+v=Kp+u for all primes p.
The following lemma allows us to reduce the power of some prime factors of d(L) by such transformation ([5, Lemma 4.4] or [3, Lemma 2.5]).
Lemma 3.6**.**
Let L be a ternary Z-lattice and p an odd prime. If p2∣d(L), then d(λp(L))=d(L)/pt for some t∈{1,2,4}.
Remark 3.7*.*
For a Z-coset L+v of conductor c, if Lp is split by a hyperbolic plane H for an odd prime p not dividing c, then Lp+v=Lp represents all of the integers in Zp.
By Lemma 3.4 (i), (ii) and (iv), Λp(L)p=Λp(Lp)⊆{x∈Lp:Q(x)∈pZp} and when p is odd, the converse containment follows from [3, Lemma 3.1] under the assumptions that p2∣d(L) and Lp is not split by H. Hence we have the following.
Lemma 3.8**.**
Let L be a ternary Z-lattice and p an odd prime. If p2∣d(L) and Lp is not split by H, then Λp(L)p=Λp(Lp)={x∈Lp:Q(x)∈pZp}.
For Z-cosets, we have a result analogous to Lemma 3.6, which is proved by Chan and Ricci [5, Proposition 4.6] (or [18, Lemma 2.6]). From its proof and Lemma 3.8, we see that the condition “Lp+v does not behave well at p” in [5, Proposition 4.6] can be replaced by “p2∣d(L) and Lp is not split by H”. Hence we are able to reformulate their proposition and prove it by following their arguments.
Lemma 3.9**.**
Let L+v be a primitive regular ternary Z-coset with conductor c and p an odd prime with p∤c. Suppose that p2∣d(L) and Lp is not split by H. Then λp(L)+pjv is a primitive regular Z-coset of conductor c, where j is the order of p modulo c.
Proof.
Let L be on the quadratic space (V,Q) and j the order of p modulo c. We assert that
[TABLE]
For q∣c, since p∤c and pjv−v∈Lq, Λp(L)q+pjv=Lq+pjv=Lq+v by Lemma 3.4 (iii). For q∤pc, Λp(L)q+pjv=Λp(L)q=Lq by Lemma 3.4 (iii) again. For q=p, since p2∣d(L) and Lp is not split by H, Λp(Lp)={x∈Lp:Q(x)∈pZp} by Lemma 3.8. Clearly, Q(pjv)∈pZp and so pjv∈Λp(Lp)=Λp(L)p. Hence Λp(L)p+pjv=Λp(L)p. Therefore, (3.1) is proved.
Suppose that n is represented by the genus of Λp(L)+pjv. By (3.1), Λp(L)q+pjv=Lq+v for q∣c or q∤pc. By Lemma 3.4 (i) and (ii), Λp(L)p=Λp(Lp)⊆Lp=Lp+v and hence Λp(L)q+pjv⊆Lq+v for each prime q. So n is represented by the genus of L+v. Since L+v is regular, n is represented by L+v. Therefore, n=Q(x+v) for some x∈L. Since x+v∈Lq for q∤pc and x+v∈Lq+v for q∣c, it follows from (3.1) that x+v∈Λp(L)q+pjv for q=p. For q=p, since n is represented by Λp(L)p+pjv=Λp(L)p, p∣n by Lemma 3.8. It follows that p∣Q(x+v) and so x+v∈Λp(L)p by Lemma 3.8 again. Hence x+v∈Λp(L)p+pjv by (3.1). Thus x+v∈Λp(L)q+pjv for each prime q and so x+v∈Λp(L)+pjv by Lemma 3.5. Therefore, Λp(L)+pjv is regular.
Since scaling of Λp(L) preserves the conductor and the regularity, λp(L)+pjv is of conductor c and regular. Suppose that n(Λp(L))=pin(L) for some i∈{1,2}. By the definition of λp, λp(L) is a Z-lattice on the quadratic space (V,Q′), where Q′(x)=p−iQ(x). By (3.1),
[TABLE]
and hence n(λp(L)+pjv)=Z, showing the primitivity.
∎
Given a primitive regular Z-coset L+v and an odd prime p∤c for which p2∣d(L) and Lp is not split by H, we are able to iteratively obtain primitive regular Z-cosets of conductor c until p2∤d(L) or Lp is split by H by applying Lemma 3.9 repeatedly, say λpℓ(L)+ptv, where t>0, ℓ∈N, and pt≡1(modc). Also, d(λpℓ(L))∣d(L) by Lemma 3.6. We define the successive operations above by τp(L+v):=λpℓ(L)+ptv.
Lemma 3.10**.**
Let m≥3 be a fixed integer. Given a primitive regular ternary m-gonal form △m,(a,b,c) associated with Gm(a,b,c)=Bm(a,b,c), there exists a primitive regular form △m,(a′,b′,c′) such that a′b′c′∣abc and Gm(a′,b′,c′)=Bm(a′,b′,c′).
Proof.
Fix m≥3, clearly a ternary m-gonal form △m,(a,b,c) represents n if and only if φm,(a,b,c) represents 8(m−2)n+(m−4)2(a+b+c). Then we associate φm,(a,b,c) with a Z-coset L+v on the quadratic space (Q,V) over Q; that is
L≅⟨c(m−4)a/2,c(m−4)b/2,c(m−4)c/2⟩ under the standard basis {e1,e2,e3} and v=−d(e1+e2+e3)/c∈V,
where d=(m−4)/gcd(m−4,2(m−2)). One can check that φm,(a,b,c) represents 8(m−2)n+(m−4)2(a+b+c) if and only if L+v represents
[TABLE]
where the conductor c of L+v is given by
[TABLE]
Hence we always have c∣2(m−2).
Assume that △m,(a,b,c) is primitive and regular. Then by the relation above, we see that φm,(a,b,c) is primitive and (8(m−2),(m−4)2(a+b+c))-regular (and so it is regular by Remark 3.2). Hence L+v is primitive and regular. We let
[TABLE]
Then for i=1,…,s, pi∤c follows from pi∤2(m−2). Applying the operation τ:=τp1∘⋯∘τps to L+v, Lemma 3.9 implies that we obtain a primitive regular Z-coset K+u of conductor c. Also, d(K)∣d(L) and p2∤d(K) or Kp is split by H for each p∈Gm(a,b,c)\Bm(a,b,c). The idea is to relate the regularity of K+u to the regularity of another △m,(a′,b′,c′).
In order to obtain a connection with regularity of a form △m,(a′,b′,c′), we first need to show that representation by K+u corresponds to representation by some φm,(a′,b′,c′). For this, let
[TABLE]
and u=ℓv, where ℓ=p1t1⋯psts and piti≡1(modc). Then ℓ≡1(modc). Put ℓ=1+cℓ0 and thus u=cℓ0v+v. Hence the quadratic polynomial QK+u associated with the Z-coset K+u is given by
[TABLE]
Consider the quadratic polynomial φm,(a′,b′,c′), which satisfies
[TABLE]
Since ℓ0d∈Z, QK+u(x,y,z)=φm,(a′,b′,c′)(x−ℓ0d,y−ℓ0d,z−ℓ0d) and so they are equivalent. Since QK+u is primitive and regular, so is φm,(a′,b′,c′). Again using Remark 3.2, we see that φm,(a′,b′,c′) is (8(m−2),(m−4)2(a+b+c))-regular in particular and so △m,(a′,b′,c′) is regular. Also, △m,(a′,b′,c′) is clearly primitive.
Define the corresponding set Gm(a′,b′,c′) for φm,(a′,b′,c′). Next, we show
[TABLE]
Let p∈Gm(a′,b′,c′). Then p∤2(m−2) and so p∤c. One can check that
[TABLE]
This implies that p2∣abc and so p∈Gm(a,b,c), and hence Gm(a′,b′,c′)⊆Gm(a,b,c). Note that Bm(a,b,c)⊆Bm(a′,b′,c′) by [3, Lemma 2.6] (or [3, Lemma 2.7] with β=0 and γ≥2). Combining these, we have
[TABLE]
showing (3.2). Now suppose p∈Gm(a′,b′,c′)\Bm(a′,b′,c′)=∅. Since p∈Gm(a′,b′,c′), p must divide two of a′, b′, and c′, and hence p2∣d(K), while p∈/Bm(a′,b′,c′) implies that Kp is not split by H. But on the other hand, p∈Gm(a,b,c)\Bm(a,b,c) from the containment (3.2) and so p2∤d(K) or Kp is split by H by the construction of K+u, which is impossible. Thus Gm(a′,b′,c′)\Bm(a′,b′,c′) must be empty and hence Gm(a′,b′,c′)=Bm(a′,b′,c′).
∎
Remark 3.11*.*
Reordering the coefficients of the form △m,(a′,b′,c′) obtained by Lemma 3.10, say △m,(a′′,b′′,c′′), it is not difficult to see that △m,(a′′,b′′,c′′) is still primitive and regular. Also, a′′b′′c′′∣abc and Gm(a′′,b′′,c′′)=Bm(a′′,b′′,c′′). Hence we may require a′≤b′≤c′ in Lemma 3.10 further.
3.3. Representation by φm,(a,b,c)
Based on the study of Dickson [8], Jones [12] and Chan and B.-K. Oh [4], we build sufficient conditions for a positive integer to be represented by φm,(a,b,c) over Zp. Note that for any n∈N, we have nR→φm,(a,b,c) if a,b and c are positive integers, so we may suppose p=∞. We require two well-known lemmas [2, Theorem 1 and Theorem 3, p. 41–42] in order to determine necessary conditions for solvability over Zp to occur.
Lemma 3.12**.**
Let F(x1,⋯,xℓ)∈Z[x1,⋯,xℓ]. Then F(x1,⋯,xℓ)≡0(modpt) is solvable for all t≥1 if and only if the equation F(x1,⋯,xℓ)=0 is solvable in Zp.
Lemma 3.13**.**
Let F(x1,⋯,xℓ)∈Z[x1,⋯,xℓ]. If ω1,…,ωℓ∈Zp is a solution of the following system of congruences
[TABLE]
for some i (1≤i≤ℓ), where t is a nonnegative integer and Fx:=∂x∂F denotes the derivative with respect to x, then the equation F(x1,⋯,xℓ)=0 is solvable in Zp.
Proposition 3.14**.**
Let m,a,b,c,n be positive integers and p be prime.
(i).
Assume gcd(a,b,c)=1. If p∈P(21−δ(m−2)), then 8(m−2)n+(m−4)2(a+b+c)Zp→φm,(a,b,c); If \mboxord2(m)≥2 and n≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8), then 8(m−2)n+(m−4)2(a+b+c)Z2→φm,(a,b,c).
2. (ii).
If p∈/P(2(m−2)) and p∤abc, then nZp→φm,(a,b,c).
3. (iii).
If p∈/P(2(m−2)), p∣c but p∤abn, then nZp→φm,(a,b,c).
4. (iv).
If p∈/P(2(m−2)), p∣c, p∣n, but p∤ab, and if p∈/P(−4ab), then nZp→φm,(a,b,c).
Remark 3.15*.*
For p∈Pm(a,b,c), if p∤n, then nZp→φm,(a,b,c) by Proposition 3.14 (ii) and (iii).
Proof.
(i). Let p∈P(2(m−2)). Since gcd(a,b,c)=1, we may assume without loss of generality that p∤a. We split into cases based on \mboxord2(m).
We first consider the case \mboxord2(m)=0. Define the polynomial
[TABLE]
in Z[x,y,z] and then Fx(x,y,z)=2a(m−2)x−a(m−4). Also, gcd(m−2,m−4)=1 and 2∤(m−2)(m−4). Take
[TABLE]
where the inverse is taken in Zp. One can compute F(x0,y0,z0)≡0(modp) but Fx(x0,y0,z0)≡−a(m−4)≡0(modp). Hence F(x,y,z)=0 is solvable in Zp by Lemma 3.13, and thus there also exists a solution over Zp to the equation △m,(a,b,c)(x,y,z)=n.
We next assume that \mboxord2(m)≥1. In this case, we define the polynomial
[TABLE]
in Z[x,y,z]. Then Fx(x,y,z)=a(2tm′−2)x−a(2t−1m′−2), where m′=m/2t with t≥1 and 2∤m′. Take
[TABLE]
One can see that F(x0,y0,z0)≡0(modp) while Fx(x0,y0,z0)≡−2a(2t−2m′−1)≡0(modp). Hence △m,(a,b,c)(x,y,z)=n is solvable in Zp for p∣Pm−2 and p=2, \mboxord2(m)=1 by Lemma 3.13.
We finally consider the case \mboxord2(m)≥2 and p=2. If \mboxord2(m)=2 and n≡a+b+c(mod8), then one can put (x1,y1,z1)=(1,1,1) and check that
[TABLE]
while Fx(x1,y1,z1)=2t−1m′a=2m′a≡2(mod4). Hence △m,(a,b,c)(x,y,z)=n is solvable in Z2 by Lemma 3.13 (taking t=1). If \mboxord2(m)>2 and n≡0(mod8), then 2∤2t−2m′−1. Put (x2,y2,z2)=(0,0,0) in this case. One can see that F(x2,y2,z2)=−n≡0(mod8) while Fx(x2,y2,z2)=−2a(2t−2m′−1)≡2(mod4). Hence △m,(a,b,c)(x,y,z)=n is solvable in Z2 by Lemma 3.13.
(ii)–(iv). Assume p∈/P(2(m−2)). Since the linear map x↦2(m−2)x−(m−4) is a bijection in Zp, the statements (ii)-(iv) follow immediately from Dickson’s results [8, Lemma 3-5, p. 107] and Lemma 3.12.
∎
For given positive integers a,b and c with gcd(a,b,c)=1, we may assume Gm(a,b,c)=Bm(a,b,c) by Lemma 3.10 and a≤b≤c by Remark 3.11. Then 8(m−2)n+(m−4)2(a+b+c)Zp→φm,(a,b,c) for p∈P(21−δ(m−2)) and p∤abc by Proposition 3.14 (i) and (ii). For p∈Gm(a,b,c)=Bm(a,b,c), 8(m−2)n+(m−4)2(a+b+c)Zp→φm,(a,b,c) by Remark 3.7. For p∣abc, without loss of generality, let p∣a. If (−4bc/p)=1, then 8(m−2)n+(m−4)2(a+b+c)Zp→φm,(a,b,c) by Proposition 3.14 (iii) if p∤8(m−2)n+(m−4)2(a+b+c) and by Proposition 3.14 (iv) if p∣8(m−2)n+(m−4)2(a+b+c). Therefore, to check whether 8(m−2)n+(m−4)2(a+b+c) is locally represented by φm,(a,b,c) or not, it is sufficient to consider the local representation over Zp for p∈Pm(a,b,c)∪{2}δ. In other words, to show Theorem 1.2, it is enough to consider the case of Gm(a,b,c)=Bm(a,b,c).
4. Bounding the coefficients a, b and c
For each fixed integer m>3, we always assume that q0 is the smallest prime in the set P(−4ab)\(P(m−2)∪Pm(c,ab)) (the existence follows from Lemma 4.4 (i) below) and denote by {qi}ab,m (i=1,2,…) the sequence of all primes in P(−4ab)\P(q0(m−2)) in ascending order for brevity. The following useful proposition may be found in [14, Lemma 3.5].
Proposition 4.1** (B. M. Kim, M.-H. Kim, and B.-K. Oh [14]).**
Let T be a finite set of primes. Set P:=∏p∈Tp and let ℓ be an integer relatively prime to P. Then for any integer u, the number of integers in the set
[TABLE]
that are relatively prime to P is at least
nϕ(P)/P−2ω(P)+1.
We next give an upper bound on the product ab for a regular ternary m-gonal form by using Proposition 4.1.
Lemma 4.2**.**
Let a≤b≤c be positive integers for which gcd(a,b,c)=1 and Gm(a,b,c)=Bm(a,b,c). If △m,(a,b,c) is regular, then
[TABLE]
Hence
ab<C2Pabcρ(Pabc)2/ϕ(Pabc),
where C2:=29⋅11 is a constant.
Remark 4.3*.*
At first glance, it is not obvious whether the right-hand sides of the inequalities in Lemma 4.2 grow faster or slower than the left-hand sides. However, since ϕ(n)n=O(nε), the right-hand side grows like Pabcε≪(abc)ε. Thus if c may be bounded as a function of a and b slower than (ab)1/ε, then such a bound may be combined with Lemma 4.2 to obtain a restriction on the possible choices of a, b, and c for which the form △m,(a,b,c) may be regular.
Proof.
If \mboxord2(m)<2, we take w0=1; if \mboxord2(m)≥2, we choose w0 such that 0<w0≤8 and w0≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8). Clearly, for every v∈N, if \mboxord2(m)≥2, then the integer n=8δv+w0 is congruent to ⌊2/\mboxord2(m)⌋(a+b+c) modulo 8.
We next construct a pair of integers N0 and Nˉ0 which are locally represented by φm,(a,b,c) and then use the regularity to obtain upper bounds for a and b. We do so in a series of steps (a)–(c) below, first constructing them in (a), showing that they are locally represented in (b), and finally obtaining the bounds for a and b in (c).
(a) We first construct the integers N0 and Nˉ0 such that gcd(N0,Pabc)=1=gcd(Nˉ0,Pabc), N0≡(m−4)2(a+b+c)(mod8(m−2)) and there exists some positive integer n0ˉ such that N0ˉ=8(m−2)n0ˉ+(m−4)2(a+b+c) and nˉ0Z→△m,a.
Write u=8(m−2)w0+(m−4)2(a+b+c). Since gcd(2(m−2),Pabc)=1, when v=⌊ρ(Pabc)⌋, vϕ(Pabc)/Pabc−2ω(Pabc)+1>0. By Proposition 4.1, there exists at least one integer 0≤v0≤v−1 such that N0:=8δ+1(m−2)v0+u is relatively prime to Pabc and N0≡u≡(m−4)2(a+b+c)(mod8(m−2)).
We now construct Nˉ0. We claim that for v≥2, there are at most 22⋅8δ(v+1)+1/4 integers between 1 and v represented by (△m,a(x)−w0)/8δ. Indeed, solving the inequality
[TABLE]
for x, we see that x−≤x≤x+, where
[TABLE]
Since a≥1, m>3 and w0≤8δ, we have
[TABLE]
yielding the claim. Taking v=⌊11⋅8δPabcρ(Pabc)/ϕ(Pabc)⌋, one can compute
[TABLE]
Since gcd(2(m−2),Pabc)=1, Proposition 4.1 implies that there exists an integer 0≤vˉ0≤v−1 for which Nˉ0:=8δ+1(m−2)vˉ0+u is relatively prime to Pabc. Furthermore, vˉ0 is not represented by (△m,a(x)−w0)/8δ;
namely, n0ˉ:=8δvˉ0+w0 is not represented by △m,a(x) as desired.
(b) For each prime p, we have N0Zp→φm,(a,b,c) and Nˉ0Zp→φm,(a,b,c).
By the construction in (a), we see that N0 can be rewritten as 8(m−2)n0+(m−4)2(a+b+c),
where n0=8δv0+w0 is a positive integer.
Clearly, when \mboxord2(m)≤1, by the first part of Proposition 3.14 (i), N0Z2→φm,(a,b,c); when \mboxord2(m)≥2, n0≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8) and by the second part of Proposition 3.14 (i) we conclude that N0Z2→φm,(a,b,c).
For p∈Pm(a,b,c), since gcd(N0,Pabc)=1, by Remark 3.15 we further see that
N0Zp→φm,(a,b,c). Hence N0Zp→φm,(a,b,c) for each prime p. Note that Nˉ0 can be also rewritten as 8(m−2)nˉ0+(m−4)2(a+b+c), where nˉ0=8δvˉ0+w0>0. Repeating the above argument, we deduce that Nˉ0Zp→φm,(a,b,c) for each prime p.
(c) We finally use N0 and Nˉ0 to bound a and b.
From (b), we see that N0 is locally represented by φm,(a,b,c). Since φm,(a,b,c) is regular, N0 is globally represented by φm,(a,b,c). It follows that n0Z→△m,(a,b,c) from Remark 3.2. Hence
[TABLE]
From the construction in (a), we see that Nˉ0 can be written as 8(m−2)nˉ0+(m−4)2(a+b+c), where nˉ0=8δvˉ0+w0 is not represented by △m,a. By (b), Nˉ0 is also locally represented by φm,(a,b,c) and so also globally represented from the regularity of φm,(a,b,c). This implies nˉ0Z→△m,(a,b,c).
Since nˉ0Z→△m,a, it must be the case that
[TABLE]
Lemma 4.4**.**
Let a,b,c be positive integers. Set C1:=29/5C06/5, where C0 is the constant defined as in Lemma 2.6. Fix an integer m≥4. Then the following hold.
(i).
We have q0<42/3C0(Pm−2Pc)1/6(ab)2/3.
2. (ii).
We have
q1<C1Pm−21/6(q03K(a,b,c))1/5(ab)4/5.
3. (iii).
Assume that qi0+1 is the least prime in {qi}ab,m greater than
[TABLE]
Then the inequality
[TABLE]
holds for i≥i0.
Proof.
Let t be the least positive integer such that Pm−2/t is prime to 4ab. Then Pm−2=tu, where u≥1 and gcd(2ab,u)=1.
(i). Take D=−4ab and M=Pcu in Lemma 2.6. We see that there exists some prime q∈P(−4ab)\(P(m−2)∪Pm(c,ab)) such that
[TABLE]
(ii). Taking D=−4ab and M=q0u in Lemma 2.6, we see that there exists some prime q∈P(−4ab)\P(q0(m−2)) such that
[TABLE]
(iii). Now suppose that q1q2⋯qj≤K(a,b,c)q02qj+1/(m−3) for some j≥i0. Taking D=−4ab and M=q1q2⋯qjq0u in Lemma 2.6, one deduces that there exists some prime q′∈P(−4ab)\P(q0(m−2)) different from q1,q2,⋯,qj such that
[TABLE]
where in the last line we bounded Pm−2/(m−3)≤(m−2)/(m−3)≤2 for m≥4. It follows that
[TABLE]
which contradicts the assumption that j≥i0.
∎
Lemma 4.5**.**
Let a,b,c be positive integers. Then for each qi in the prime sequence {qi}ab,m (i=1,2,⋯), there exists some Ni such that \mboxordqiNi=1, Ni≡(m−4)2(a+b)(mod8(m−2)), Ni≡8(m−2)c+q0(modq02) and gcd(Ni/qi,Pc)=1.
Hence there exists a positive integer ni such that Ni=8(m−2)ni+(m−4)2(a+b). Also, ni≤K(a,b,c)q02qi. If moreover \mboxord2(m)≥2, then ni≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8).
Proof.
Write Pab=Pab′s, where gcd(s,Pab′)=1. For i=1,2,⋯, observe that
gcd(sqi,2(m−2)q0)=1 from 2∤Pab, and gcd(qi,2(m−2)q0ab)=1. Also, gcd(q0,8(m−2))=1 and moreover, when \mboxord2(m)≥2, gcd(8(m−2),128)=16∣8(m−2)⌊2/\mboxord2(m)⌋(a+b+c).
By the Chinese remainder theorem, the system of congruences
[TABLE]
is solvable (in terms of u) for each m>3. Then we take its solution, say ui, in the range
[TABLE]
Since 8δ+1(m−2)q02v+ui is also a solution for any v∈N (note that 128∣8δ+1(m−2) when \mboxord2(m)≥2) and gcd(2(m−2)q02,qiPcPab′)=1, choosing v=⌊ρ(qiPcPab′)⌋, we see by Proposition 4.1 that there exists at least one integer 0≤vi≤v−1 for which wi:=8δ+1(m−2)q02vi+ui satisfies gcd(wi,qiPcPab′)=1. Take Ni:=sqiwi. Since gcd(swi,qi)=1, \mboxordqiNi=1. Also, since gcd(Pabwi,Pc)=1, it follows that gcd(Ni/qi,Pc)=gcd(swi,Pc)=1. Moreover,
[TABLE]
as ρ(qi)≤3, s≤Pab and Pab′∣Pab.
For the second part, from the last two congruences, we have
[TABLE]
for some positive integer ti, as Ni>(m−4)2(a+b) by construction. Take ni:=8δti. Then ni>0. Since Ni≤24⋅8δ(m−2)q02Pabρ(Pabc)qi+(m−4)2(a+b) by (4.2), we have
[TABLE]
When \mboxord2(m)≥2, we also have
[TABLE]
which implies ni≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8).
∎
Lemma 4.6**.**
Let a,b,c be positive integers with a≤b≤c and gcd(a,b,c)=1. Let Ni be the integers as defined in Lemma 4.5, i=1,2,⋯. Then the following hold.
(i).
We have NiZ→φm,(a,b); if Ni−8(m−2)c>0, then Ni−8(m−2)cZ→φm,(a,b).
2. (ii).
If p∈Pm(a,bc)∪Pm(b,ac)∪{2}δ, then Ni+(m−4)2cZp→φm,(a,b,c).
3. (iii).
If p∈Pm(c,ab) and qi∤c, then Ni+(m−4)2cZp→φm,(a,b,c).
4. (iv).
If qi∤c and a, b, and c satisfy Gm(a,b,c)=Bm(a,b,c), then Ni+(m−4)2cZp→φm,(a,b,c) for each prime p.
Proof.
Let Ni=sqiwi be as constructed in Lemma 4.5 for i=1,2,⋯.
(i). If Ni is represented by φm,(a,b), then the equation Ni=ax2+by2 is solvable (in Z) and so is 4aNi=x2+4aby2. Hence x2+4aby2≡0(modqi). But (−4ab/qi)=−1, which implies that x≡y≡0(modqi). So qi2∣x2+4aby2=4aNi
and hence qi2∣Ni, which contradicts \mboxordqiNi=1.
If Ni−8(m−2)c>0, then since Ni≡8(m−2)c+q0(modq02), we conclude that \mboxordq0(Ni−8(m−2)c)=1. If Ni−8(m−2)c is represented by φm,(a,b) over Z, then since (−4ab/q0)=−1, we again conclude by a similar argument that q02∣Ni−8(m−2)c, yielding a contradiction.
(ii). Let p∈Pm(a,bc)∪Pm(b,ac) be given, from which we conclude that p∣Pab but p∤c. We again write Pab=Pab′s, with gcd(Pab′,s)=1, so that either p∣s or p∣Pab′. If p∣s, then p∤m−4 and it follows that gcd(Ni+(m−4)2c,p)=gcd((m−4)c,p)=1 and so Ni+(m−4)2cZp→φm,(a,b,c) by Remark 3.15. On the other hand, if p∣Pab′, then p∣m−4 and p∤s and hence
[TABLE]
We then note that p=qi because gcd(qi,ab)=1 and gcd(wi,Pab′)=1, from which we conclude that gcd(qiwi,p)=1. So Ni+(m−4)2cZp→φm,(a,b,c) by Remark 3.15.
Now consider p=2 and assume without loss of generality that \mboxord2(m)≥2, since otherwise this case is covered above. Observe that Ni+(m−4)2c=8(m−2)ni+(m−4)2(a+b+c) and ni≡⌊2/\mboxord2(m)⌋(a+b+c)(mod8), where ni is constructed as in the proof of Lemma 4.6. By the second part of Proposition 3.14 (i), we have Ni+(m−4)2cZ2→φm,(a,b,c).
(iii). Let p∈Pm(c,ab) with qi∤c be given. Then p∣Pc but p∤ab. Since p∣c, s∣Pab, gcd(wi,Pc)=1, and qi∤c by construction (and hence qi=p), we have
[TABLE]
Therefore Ni+(m−4)2cZp→φm,(a,b,c) by Remark 3.15.
(iv). The statement follows immediately from parts (ii) and (iii).
∎
Recall from Remark 4.3 that one obtains a bound for a and b in a regular △m,(a,b,c) which is non-trivial when a, b, and c satisfy Gm(a,b,c)=Bm(a,b,c). It was then explained that obtaining a bound for c in terms of a and b would lead to a bound for the possible choices of a, b, and c for which △m,(a,b,c) is regular. Following Dickson’s proofs of [7, Theorem 5, 6], we deduce such a bound for c.
Lemma 4.7**.**
Let a≤b≤c be positive integers for which gcd(a,b,c)=1 and Gm(a,b,c)=Bm(a,b,c). If △m,(a,b,c) is regular, then
[TABLE]
with the constant C3:=426/15C013/5C1, where C0 and C1 are defined as in Lemma 2.6 and 4.4, respectively.
Proof.
Consider the prime sequence {qi}ab,m. By Lemma 4.4 (ii), we see that
[TABLE]
for some i0≥1. For each i, we take Ni=sqiwi=8(m−2)ni+(m−4)2(a+b) as constructed in Lemma 4.5.
If qj∤c for some 1≤j≤i0, then Nj+(m−4)2cZp→φm,(a,b,c) for every prime p by Lemma 4.6 (iv). But then njZp→△m,(a,b,c) for each prime p and △m,(a,b,c) is regular so njZ→△m,(a,b,c). Hence Nj+(m−4)2cZ→φm,(a,b,c). Namely,
[TABLE]
for some x0,y0,z0∈Z. Since NjZ→φm,(a,b) by Lemma 4.6 (i), it follows that z0=0. Moreover, since (4.3) is increasing as a function of z0 for z0>0 and decreasing for z0<0, we have Nj−8(m−2)c≥φm,(a,b)(x0,y0).
Since Nj−8(m−2)c≡q0(modq02) from Lemma 4.5, we have Nj−8(m−2)c>0 and so Nj−8(m−2)cZ→φm,(a,b) by Lemma 4.6 (i), which in turn implies that z0=1. Hence Nj≥φm,(a,b)(x0,y0)+8c(m−2)(m−3); that is
nj≥△m,(a,b)(x0,y0)+c(m−3) and so, using the fact that nj≤K(a,b,c)q02qj by Lemma 4.5 and qj≤qi0,
Hence if there exists some j≤i0 for which qj∤c, then we have (4.4), which implies the claim.
On the other hand, if no such j exists, then we have q1q2⋯qi0∣Pc. We claim that for every i≥i0 we have q1q2⋯qi∣Pc, leading to a contradiction because c is finite. The case i=i0 is assumed, and we proceed by induction. Suppose that i≥i0 and q1⋯qi∣Pc. If qi+1∤c, then we again have ni+1≤K(a,b,c)q0qi+1 by Lemma 4.5 and repeating the above argument we obtain
[TABLE]
which contradicts the inequality (4.1) in Lemma 4.4. We conclude that j must exist, and therefore (4.4) follows.
∎
Proof of Theorem 1.2.
First suppose that △m,(a,b,c) is primitive and regular and Gm(a,b,c)=Bm(a,b,c). By Lemma 3.10, there exists another primitive regular form △m,(a′,b′,c′) with Gm(a′,b′,c′)=Bm(a′,b′,c′). It thus suffices to prove that there do not exist any primitive regular forms with Gm(a,b,c)=Bm(a,b,c) for m sufficiently large.
Assume that a≤b≤c, gcd(a,b,c)=1 and Gm(a,b,c)=Bm(a,b,c). Note that Pab≤ab and Pc≤c. Also, when m≥6, Pm−23/5<(m−3)4/5. By Lemma 4.7,
[TABLE]
Therefore, (m−3)2/5c<C32(24ρ(Pabc))12/5(ab)112/15.
By Lemma 4.2, we deduce that
[TABLE]
and so
(m−3)2/5c<C4ρ(Pabc)18(Pabc/ϕ(Pabc))112/15, where C4=2412/5C2112/15C32. Since Pabc/ϕ(Pabc)≪Pabcε and 2ω(Pabc)≪Pabcε,
[TABLE]
This implies that (m−3)6/5abc≪Pabcε, where the implied constant only depends on ε, but not on m. Since 1≤a≤b≤c, this leads to a contradiction for m sufficiently large.
Acknowledgments
The authors would like to thank Yuk-Kam Lau for helpful discussion and the referee for his/her useful comments and suggestions.
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