Tetranomial Thue equations

Journal of Number Theory 133 (2013) 4140–4174

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Journal of Number Theory www.elsevier.com/locate/jnt

Tetranomial Thue equations Helen G. Grundman a,∗ , Daniel P. Wisniewski b a

Department of Mathematics, Bryn Mawr College, Bryn Mawr, PA 19010, United States Department of Mathematics/Computer Science, DeSales University, Center Valley, PA 18034, United States

b

a r t i c l e

i n f o

Article history: Received 29 April 2013 Revised 19 June 2013 Accepted 19 June 2013 Available online 30 August 2013 Communicated by David Goss MSC: 11D41 11D45

a b s t r a c t Let F (x, y) be an irreducible binary form of degree n  6 with exactly four nonzero terms. Assuming certain conditions relating the coefficients and the degrees of the different terms are satisfied, we prove upper bounds on the number of equivalent pairs of nontrivial solutions of the Thue equation |F (x, y)| = 1. Improved bounds are provided for a variety of cases, where more information about the form is known. © 2013 Elsevier Inc. All rights reserved.

Keywords: Thue equations Tetranomial

1. Introduction Let F (x, y) be a Thue form, an irreducible binary form of degree n  3 with integer coefficients. In 1909, Thue [22] proved that the equation |F (x, y)| = 1 has at most finitely many integer solutions. Since that time, much attention has been given to finding upper bounds for the finite number of solutions. See, for example, [5,6,9,10,12,19,20,23], and of * Corresponding author. E-mail addresses: [email protected] (H.G. Grundman), [email protected] (D.P. Wisniewski). 0022-314X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jnt.2013.06.005

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recent note, [1–4,16,24]. Of interest here are bounds for when F (x, y) has few coefficients, such as those in [13–15,18]. In particular, in 2000, Thomas [21] proved that when F (x, y) has exactly three nonzero terms and degree at least 38, the Thue equation |F (x, y)| = 1 has at most 20 equivalent pairs of nontrivial solutions. He gave additional bounds for degrees at least 5, and improved these to smaller bounds in the case when the degree of F (x, y) is odd. His work inspired this paper. A standard method for bounding the numbers of solutions of Thue equations is, for each root, to bound the number of solutions for which it is the “closest” root of F (x, 1). In contrast, in his work with trinomial Thue equations, Thomas considered what he called the exceptional points of F (x, 1) and defined, for an arbitrary solution, to which exceptional point the solution “belongs.” He then, for each exceptional point, bounded the number of possible solutions that can belong to it. Following convention, for a chosen number Y , Thomas classified a solution (p, q) as either large (q > Y ) or small (1  q  Y ) and then bounded the number of each type of solution. To bound the number of large solutions, he applied the Thue–Siegel principle as developed in [7–9]; and, to bound the number of small solutions, Thomas used a gap principle argument. Finally, by bounding the number of possible exceptional points, he proved his results. In this paper, we use Thomas’s ideas to investigate the problem of finding bounds in the case where the Thue forms have exactly four nonzero terms. In his paper, Thomas suggested that his approach could be applied to the tetranomial case, but until now this possibility had not been formally studied. Our results, given in Theorems 1.1–1.4, are very similar to those found by Thomas, yet not as all-encompassing in that we need an added restriction on the form. Specifically, if F (x, y) = axn + rxm y n−m − sxk y n−k + ty n and εn is as defined below, we require that    an  −1    rm  > (1 − ε)

and

   tn  −1    s(n − k)  > (1 − ε) ,

(1)

with εn  ε < 1. The need for this restriction appears to be fundamental to this method of approach, as we explain at the relevant points in our argument. The key is that Thomas frequently exploited the fact that for trinomials, the derivative of F (x, 1) has only two terms, something that, obviously, does not extend to the tetranomial case. Further, the additional term in the tetranomial allows for a wider variation in shapes of the graphs of y = F (x, 1), resulting in a need for some new arguments in proving the bounds. Following standard terminology, we say that integer solutions (p, q) and (−p, −q) to a Thue equation |F (x, y)| = 1 are equivalent. Further, they are trivial if p, q ∈ {−1, 0, 1}. We let NF denote the number of equivalent pairs of nontrivial integer solutions to |F (x, y)| = 1. Note that NF is also the number of regular solutions to |F (x, y)| = 1: integer solutions (p, q) such that p = 0,

|p| = q,

and q > 0.

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Table 1 Values for wn and dn . n

9

10

11

12

13–14

15–18

19–21

22–59

 60

wn dn

19 149

17 3

14 1

11 1

9 1

8 1

7 1

7 0

6 0

For n  9, let wn and dn be defined as in Table 1, let P = Pn be defined by ⎧ 23, ⎪ ⎪ ⎨ 4, Pn = ⎪ 3, ⎪ ⎩ 2,

if if if if

n = 9, n = 10, 11  n  21, n  22,

(2)

and let 

εn =

4(n − 1) . nP n

(3)

Our first result gives numerical bounds on NF for tetranomial Thue equations satisfying conditions (1), with ε = εn , based on the degree of the form, and with possible improvements based on the degrees of x in the different terms. Theorem 1.1. Let F (x, y) ∈ Z[x, y] be an irreducible form, F (x, y) = axn + rxm y n−m − sxk y n−k + ty n , with n > m > k > 0, a, r, s, t = 0, and n  9, satisfying condition (1) with ε = εn . Further, let ⎧ ⎨ 4, vF = 5, ⎩ 6,

if n is even and m or k is odd, if n is odd, if n is even and both m and k are even.

(4)

Then, the number of regular solutions to |F (x, y)| = 1 is bounded: NF  wn · vF + 4dn . For example, if F (x, y) is of degree 60, then NF  36 and if, in addition, either m or k is odd, then NF  24. If F (x, y) is of degree 25, then NF  35 and if the degree is 10, then NF  114. The upper bounds on NF can be improved for small values of n by increasing the required bounds in (1). In the following theorem, we provide two such improvements. With these new bounds, we obtain results for tetranomial Thue forms of degree 6 and greater. As will become evident, similar results can be computed for other values of ε in condition (1).

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Table 2 Values for wn,ε and dn,ε for ε = 0.01 and ε = 0.5. (a) ε = 0.01 n

6

7

8

9

10

11

12

13–15

16–17

18–32

 33

wn,0.01 dn,0.01

22 359

18 31

15 11

11 9

11 3

11 1

10 1

8 1

7 1

7 0

6 0

(b) ε = 0.5 n

6

7

8

9

wn,0.5 dn,0.5

16 9

17 3

11 3

10 3

10

11–12

13–17

18–28

 29

9 1

8 1

7 1

7 0

6 0

Theorem 1.2. Let F (x, y) ∈ Z[x, y] be an irreducible form, F (x, y) = axn + rxm y n−m − sxk y n−k + ty n , with n > m > k > 0, a, r, s, t = 0, and n  6, satisfying (1) with ε = 0.01 or ε = 0.5. Then, the number of regular solutions to |F (x, y)| = 1 is bounded: NF  wn,ε · vF + 4dn,ε , where vF is defined in Eq. (4) and the values of wn,ε and dn,ε are given in Table 2. For example, if F (x, y) is of degree is 10 and ε = 0.5, then NF  55. Let g(x) be a nonconstant polynomial with real coefficients. We reserve the term critical point for real numbers μ such that g  (μ) = 0. We call μ ∈ R a proper critical point of g(x) if g  (μ) = 0, g(μ) = 0, and there exists a punctured neighborhood U of μ such that for all x ∈ U , g  (x)g(x) > 0. Hence, near a proper critical point μ, the graph of y = g(x) is concave up and above the x-axis, or concave down and below the x-axis. Given an irreducible tetranomial form F (x, y), let RF be the number of real roots of F (x, 1) and let CF be the number of proper critical points of F (x, 1). In Section 9, we prove Theorems 1.1 and 1.2 along with the following two theorems, which provide tighter bounds on NF in cases when more information about the form is available. Theorem 1.3. Given F (x, y) as in Theorem 1.1, let wn and dn be as given in Table 1 and wn as in Table 3(a). The number of regular solutions to |F (x, y)| = 1 is bounded: NF  wn RF + wn CF + 4dn . So if F (x, y) is of degree is 11 and F (x, 1) has exactly one real root and one critical point, then NF  24. Again, we have parallel results for larger choices of ε. Theorem 1.4. Given F (x, y) as in Theorem 1.2 with ε = 0.01 or 0.5, let wn,ε and dn,ε be as given in Table 2 and wn,ε as in Table 3(b). The number of regular solutions to

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|F (x, y)| = 1 is bounded: NF  wn,ε RF + wn,ε CF + 4dn,ε . Table 3 Values of wn and wn,ε . (a) wn n

9

10

11

12

13–14

15–17

18

19–54

55–59

 60

wn

10

8

6

4

4

4

3

3

2

2

(b) wn,ε n

6

7

8

9–11

12–14

15–21

22–32

 33

wn,0.01 wn,0.5

8 6

5 6

6 4

4 4

4 3

3 3

3 2

2 2

In Section 2, we establish the values of vF in Eq. (4) by considering bounds on the numbers of real roots and critical points of nonconstant tetranomials. In Section 3, we define what it means for a solution to belong to a particular exceptional point of f (x), and in Sections 4–8, we define and determine bounds on the number of special solutions that belong to an exceptional point τ > 1. Finally, in Section 9, we complete the proofs of Theorems 1.1–1.4. 2. Exceptional points In this section, we define exceptional points of a polynomial and use their properties to prove bounds on the numbers of real roots and proper critical points of a nonconstant tetranomial. Then, in Corollary 2.6, we prove that for F (x, y) as in Theorem 1.1, the number of exceptional points of at least one of F (x, 1) or F (1, x) is at most vF . Again, let g(x) be a nonconstant real polynomial. A real number, τ , is an exceptional point of g(x) if τ is either a real root or a proper critical point of g(x). Let E(g) denote the set of all exceptional points of g(x). Since g(x) is a nonconstant polynomial, E(g) is nonempty and finite. We say that τ1 , τ2 ∈ E(g) with τ1 < τ2 are neighboring exceptional points if there does not exist τ ∈ E(g) such that τ1 < τ < τ2 . We call η ∈ R an improper critical point of g(x) if g  (η) = 0, g(η) = 0, and there exists a punctured neighborhood U of η such that for all x ∈ U , g  (x)g(x) < 0. The following lemma is from Thomas [21, Lemma 2.1]. Lemma 2.1 (Thomas). Let g(x) ∈ R[x] be nonconstant. Between any two neighboring exceptional points of g(x), there exists an improper critical point of g(x). We now restrict our focus to tetranomials. Fix g(x) = axn + rxm − sxk + t ∈ R[x], with n > m > k > 0 and a, r, s, t ∈ R − {0}.

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Proposition 2.2. The tetranomial g(x) has at most two positive critical points and at most two negative critical points, for a total of at most five critical points. If n − m is odd, then g(x) has at most three nonzero critical points. Proof. Suppose that g(x) has at least three positive critical points. Then g  (x) = xk−1 (anxn−k + rmxm−k − sk) has at least three positive roots and so gˆ(x) = anxn−k + rmxm−k − sk has at least three positive roots. Then, by Lemma 2.1, gˆ(x) must have at least two positive improper critical points. But since gˆ (x) = xm−k−1 (a(n − k)nxn−m + r(m − k)m) has at most one positive root, this is a contradiction. So g(x) has at most two positive critical points. Similarly, g(x) has at most two negative critical points. If n − m is odd, then gˆ (x) has at most one nonzero real root and so gˆ(x) has at most two critical points. Thus, by Lemma 2.1, gˆ(x) has at most three nonzero real roots and so g(x) has at most three nonzero critical points. 2 The first of the following two corollaries is immediate from Lemma 2.1 and Proposition 2.2. Corollary 2.3. The tetranomial g(x) has at most six exceptional points, at most three positive real roots, at most three negative real roots, at most one positive proper critical point, and at most one negative proper critical point. If n − m is odd, then g(x) has at most five exceptional points. Corollary 2.4. If n is odd, then g(x) has at most five real roots. If n is even and m is odd, then g(x) has at most four real roots. Proof. Suppose that n is odd and g(x) has more than five real roots. By Corollary 2.3, g(x) has exactly six real roots. By Lemma 2.1 and Proposition 2.2, g(x) has exactly five critical points, each of which is improper and, therefore, is not a root. So the sign of g(x) changes at each of the real roots. Since there are exactly six real roots, there are exactly six sign changes, and therefore lim g(x) = lim g(x) = ∞. Hence, g(x) must be of x→−∞

x→+∞

even degree, a contradiction. Now, suppose that n is even, m is odd, and g(x) has more than four real roots. Then, by Corollary 2.3, g(x) has exactly five real roots. Using the same reasoning as above, the sign of g(x) changes at each of the five real roots and thus g(x) is of odd degree, a contradiction. 2 Proposition 2.5. With the above notation, ⎧ 4,   ⎨ E(g)  5, ⎩ 6,

if n is even and m is odd, if n is odd, if n and m are both even.

Proof. By Corollary 2.3, |E(g)|  6. So, if n is even and m is even, we are done.

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If n is odd, then by Corollary 2.4, there are at most 5 real roots. Suppose that |E(g)| = 6. Then there must be at least one proper critical point. So, by Proposition 2.2, there are at most four improper critical points. Thus, by Lemma 2.1, g(x) has at most five exceptional points, contradicting the assumption that |E(g)| = 6. If n is even and m is odd, then by Corollary 2.4, there are at most 4 real roots. Suppose that |E(g)|  5 and obtain a contradiction as above. 2 Finally, note that given F (x, y) as in Theorem 1.1 or Theorem 1.2, the form G(x, y) = F (y, x) also satisfies the hypotheses of the theorem. Further, |F (x, y)| = 1 and |G(x, y)| = 1 have the same number of regular solutions. This allows us, in Section 9, to use the following corollary to complete the proofs of Theorems 1.1 and 1.2. Corollary 2.6. Let F (x, y) = axn +rxm y n−m −sxk y n−k +ty n ∈ R[x], with n > m > k > 0, a, r, s, t = 0. Then vF , as defined in Eq. (4), is an upper bound for the number of exceptional points of at least one of F (x, 1) or F (1, x). The proof is immediate, considering the degrees of the terms in F (1, x). 3. Regular solutions belonging to exceptional points For the remainder of this paper, we fix an irreducible binary form, F (x, y) = a0 xn + r0 xm y n−m − s0 xk y n−k + t0 y n ∈ Z[x, y],

(5)

with n > m > k > 0, a0 > 0, r0 , s0 , t0 = 0, and n  6, satisfying    a0 n  −1    r0 m  > (1 − ε)

   t0 n   > (1 − ε)−1 , and  s0 (n − k) 

(6)

for some εn  ε < 1, with εn as given in Eq. (3) where, for now, we assume only that P is an integer greater than or equal to 2. Note that since |F (x, y)| = |−F (x, y)|, adding the assumption that a0 > 0 to those in Theorem 1.1 or 1.2 results in no loss of generality. We also fix f (x) = F (x, 1). Let E(f ) = {τ1 , . . . , τ|E(f )| } with τ1 < · · · < τ|E(f )| and let η1 , . . . , η|E(f )|−1 be improper critical points, the existence of which is guaranteed by Lemma 2.1, such that τ1 < η1 < τ2 < · · · < η|E(f )|−1 < τ|E(f )| . (It is easy to see that these improper critical points are uniquely determined, but we do not need this fact here.) For convenience, let η0 = −∞ and η|E(f )| = +∞. Given ρ ∈ R, we say that ρ belongs to τi and τi belongs to ρ, if ηi−1  ρ < ηi . So for each ρ ∈ R, there is a unique τ ∈ E(f ) such that ρ belongs to τ . If (p, q) is a regular solution to |F (x, y)| = 1 and pq belongs to τ ∈ E(f ), we say that (p, q) belongs to τ .

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Again, following Thomas, we partition R − {−1, 0, 1} into four disjoint subsets: I(1) = (−∞, −1),

I(2) = (−1, 0),

I(3) = (0, 1),

I(4) = (1, +∞),

and define three new Thue forms: F1 (x, y) = ±F (−x, y),

F2 (x, y) = ±F (−y, x),

F3 (x, y) = ±F (y, x),

where in each case the sign is chosen so that the new form has positive leading coefficient. Set F0 (x, y) = F (x, y). Note that for 0  i  3, since Fi (x, y) and fi (x) = Fi (x, 1) satisfy the assumptions made about F (x, y) and g(x) in Section 2, the results of that section hold for Fi (x, y) and fi (x). Recall that if (p, q) is a regular solution, then by definition, pq ∈ / {−1, 0, 1}. For 0  i  3 and 1  j  4, let Si (j) denote the set of regular solutions (p, q) to the equation |Fi (x, y)| = 1 such that pq ∈ I(j). Note that the exceptional point to which (p, q) ∈ Si (j) belongs is also in I(j) [21, Lemma 2.4]. Let Ni (j) = |Si (j)|. Then NF =

4 

N0 (j).

(7)

j=1

The following lemma, a reformulation of [21, Corollary 2.2], allows us to bound N0 (j) for each 0  j  3 via the bounds on the number of regular solutions of |Fi (x, y)| = 1 in the interval I(4). Lemma 3.1. For 1  j  3, N0 (j) = Nj (4). Proof. Define ψ1 : S0 (1) → S1 (4) by ψ1 ((p, q)) = (−p, q). Note that pq ∈ I(1) if and only if −p q ∈ I(4). Since |F (p, q)| = |F1 (−p, q)| and (p, q) is regular if and only if (−p, q) is regular, ψ1 defines a one-to-one correspondence. Hence, N0 (1) = |S0 (1)| = |S1 (4)| = N1 (4). Similarly, ψ2 : S0 (2) → S2 (4) defined by ψ2 ((p, q)) = (q, −p) is a one-to-one correspondence, implying that N0 (2) = |S0 (2)| = |S2 (4)| = N2 (4). And ψ3 : S0 (3) → S3 (4) defined by ψ3 ((p, q)) = (q, p) is a one-to-one correspondence, yielding N0 (3) = N3 (4). 2 Still following Thomas [21], we distinguish between solutions that belong to real roots and those that belong to proper critical points. We introduce the following notation. For 0  i  3 and 1  j  4, let Ri (j) denote the number of real roots of fi (x) in I(j), and let Ci (j) denote the number of proper critical points of fi (x) in I(j). Further, let NiR (j) denote the number of regular solutions (p, q) to |Fi (x, y)| = 1 such that p C q ∈ I(j) belongs to a real root and let Ni (j) denote the number of regular solutions (p, q) to |Fi (x, y)| = 1 such that pq ∈ I(j) belongs to a proper critical point. So Ni (j) = NiR (j) + NiC (j).

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The remainder of this section is a series of lemmas concerning the values of the functions just defined. Thomas’s analogous results [21, Lemmas 2.3 & 2.5] do not hold for tetranomials, and so have been modified considerably. For example, for tetranomials, it is not always true that Ri (j) · Ci (j) = 0, nor that N0C (i) = NiC (4), for 1  i  3. Lemma 3.2. Let 0  i  3 and 1  j  4. Then 0  Ri (j)  3 and 0  Ci (j)  1. Further, if Ri (j) > 1, then Ci (j) = 0. Proof. Fix i and j. The first statement is Corollary 2.3 applied to fi (x). If Ri (j) > 1 and Ci (j) = 1, then, there are at least 3 exceptional points in I(j). By Lemma 2.1, there are at least two improper critical points in I(j). Hence, at least 3 critical points lie in I(j), contradicting Proposition 2.2. 2 The next lemma, which is used in proving Lemma 3.4, gives an important relationship between the number of solutions and the number of real roots in the interval I(4). This is the first instance in which we need the restrictions in (6) to guarantee that the result holds. As is frequently the case, however, the only condition on ε that is needed is that 0 < ε < 1. Lemma 3.3. For 0  i  3, if NiC (4) = 0, then Ri (4) = 0. Proof. We prove the case i = 0. The other cases are similar, noting that (6) holds for F0 if and only if it holds for each Fi , 1  i  3. Suppose N0C (4) = 0 and R0 (4) > 0. Then there exists a proper critical point μ > 1 and a root λ > 1 of fi (x) = f (x). By Lemma 2.1, there exists an improper critical point η1 > 1 between μ and λ. So μ and η1 are both positive roots of f  (x) = xk−1 (a0 nxn−k + r0 mxm−k − s0 k) and therefore of the trinomial fˆ(x) = a0 nxn−k + r0 mxm−k − s0 k. Thus, by Lemma 2.1, there exists an improper critical point η2 > 1 of fˆ(x) between μ and η1 . Now, a0 n(n − k)η2n−m + r0 m(m − k) = fˆ (η2 ) = 0 and thus, since |r0 |m < (1−ε)a0 n < a0 n, η2n−m < 1. But this contradicts that η2 > 1. 2 Note that Lemma 3.3 need not hold, if the restrictions in (6) are not satisfied. For example, consider F (x, y) = x6 − 1215x2 y 4 + 4668xy 5 − 4541y 6 . Here, N0C (4) = 0, since (2, 1) is a solution to |F (x, y)| = 1 and 2 is a proper critical point of F (x, 1). Yet, R0 (4) > 0, since F (2, 1) = −1 and F (5, 1) > 0. Lemma 3.4. For 1  i  3, (a) R0 (i) = Ri (4). (b) If N0C (i) = 0, then Ci (4)  C0 (i).

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(c) If R0 (i) = 0 or C0 (i) = 0, then N0R (i) = NiR (4) and N0C (i) = NiC (4). (d) If R0 (i) = C0 (i) = 1, then N0R (i) + N0C (i) = NiR (4). Table 4 Values for Pn,ε for ε = 0.01 and ε = 0.5. n

6

7

8

9

10

11–17

 18

Pn,0.01 Pn,0.5

35 6

11 4

7 4

6 4

4 3

3 3

2 2

Proof. Let 1  i  3. (a) Immediate using correspondences as in the proof of Lemma 3.1. (b) If N0C (i) = 0, then C0 (i) = 0. Hence, by Lemma 3.2, Ci (4)  1 = C0 (i). (c) If R0 (i) = 0, then N0R (i) = 0. By part (a), we have Ri (4) = 0 and therefore, R Ni (4) = 0. Hence, N0R (i) = NiR (4) and, since, by Lemma 3.1, N0 (i) = Ni (4), N0C (i) = NiC (4). On the other hand, if R0 (i) = 0, then by hypothesis, C0 (i) = 0 and so N0C (i) = 0. By part (a), Ri (4) = R0 (i) = 0, and so by Lemma 3.3, NiC (4) = 0. Thus, N0C (i) = NiC (4) and hence N0R (i) = NiR (4). (d) If R0 (i) = C0 (i) = 1, then by part (a), Ri (4) = R0 (i) = 1. Therefore, by Lemma 3.3, NiC (4) = 0. By Lemma 3.1, N0 (i) = Ni (4), and therefore, N0R (i) + N0C (i) = NiR (4) + NiC (4) = NiR (4). 2 4. Special solutions For the remainder of this work, we assume that we are in one of two situations. In the first, used for proving Theorems 1.1 and 1.3, we assume that n  9, condition (6) holds with ε = εn , and P = Pn as defined in Eq. (2). In the second, used for proving Theorems 1.2 and 1.4, we assume that n  6, condition (6) holds with ε = 0.01 or 0.5, and P = Pn,ε as defined in Table 4. We say that a nontrivial regular solution (p, q) to |F (x, y)| = 1 is a special solution if p > 1 and p  P . Notice that the first condition is equivalent to the solution’s belonging q to an exceptional point in I(4). We first bound the number of regular solutions that belong to an exceptional point in I(4), but are not special solutions. When P = 2, every such regular solution is a special solution. When P = 3, there is at most one regular solution that is not a special solution, namely (2, 1). For P = 4, the only possible regular solutions that are not special are (3, 1), (3, 2), and, again, (2, 1). For larger values of P , we note that if |F (p, q)| = 1, then gcd(p, q) = 1. Thus, the number of possible values of q for a given value of p is at most ϕ(p). So the number of regular solutions belonging to an exceptional point in I(4) that are not special is bounded by

d(P ) =

P −1  i=2

ϕ(i).

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The following results reduce the proofs of our main theorems to verifying bounds on the numbers of solutions belonging to real roots and proper critical points. Here, again, the difficulties caused by the additional term in the tetranomial require a more complicated argument than that given for trinomials in [21]. Theorem 4.1. Assume that, independent of i, w ∈ R+ is an upper bound on the number of special solutions that can belong to a real root of fi (x) in I(4) and w ∈ R+ is an upper bound on the number of special solutions that can belong to a proper critical point of fi (x) in I(4). Let d = d(P ) as defined above. Then NF  wRF + wCF + 4d. Proof. We begin by showing that for each 1  i  4, N0 (i)  wR0 (i) + wC0 (i) + d.

(8)

First note that N0 (4) is the number of regular solutions of F (x, y) belonging to exceptional points in I(4), wR0 (4) is a bound on the number of special solutions of F (x, y) belonging to real roots in I(4), wC0 (4) is a bound on the number of special solutions of F (x, y) belonging to proper critical points in I(4), and d is a bound on the number of regular solutions in I(4) that are not special. Thus, inequality (8) holds for i = 4. Now fix 1  i  3. By the same reasoning as above, NiR (4)  wRi (4) + d and C Ni (4)  wCi (4) + d. Suppose that R0 (i) = 0. Then N0R (i) = 0 and so N0 (i) = N0C (i). If N0C (i) = 0, then the lemma clearly holds. If N0C (i) = 0, then by Lemma 3.4(b), Ci (4)  C0 (i). Thus, from Lemma 3.4(c), N0 (i) = N0C (i) = NiC (4)  wCi (4) + d  wC0 (i) + d and inequality (8) follows. Suppose that C0 (i) = 0. Then N0C (i) = 0 and so N0 (i) = N0R (i). By Lemma 3.4(c) and then Lemma 3.4(a), N0 (i) = N0R (i) = NiR (4)  wRi (4) + d = wR0 (i) + d and inequality (8) follows. Finally, suppose that R0 (i) > 0 and C0 (i) > 0. By Lemma 3.2, R0 (i) = C0 (i) = 1. By Lemma 3.4(d), N0 (i) = N0R (i) + N0C (i) = NiR (4)  wRi (4) + d = wR0 (i) + d, by Lemma 3.4(a), and, again, inequality (8) follows. Thus, combining Eq. (7) with inequality (8),

NF =

4 

N0 (i) 

i=1

=w

4  i=1

as desired.

2

4 

wR0 (i) + wC0 (i) + d

i=1

R0 (i) + w

4  i=1

C0 (i) + 4d  wRF + wCF + 4d,

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Table 5 Values for κn and κn,ε . n

6

7

8

9

10

11–12

13–15

16–17

18–21

 22

κn κn,0.01 κn,0.5

– 208 5

– 84 4

– 84 4

177 50 4

22 22 4

6 6 4

3 3 3

2 2 2

2 2 2

n−1 2 n−1 2 n−1 2

It remains, in Sections 5–8, to establish both the existence of and methods of calculation of bounds needed to apply Theorem 4.1. In Section 9, we then prove Theorems 1.1–1.4. 5. Bounding a key distance Retaining the notation and assumptions from previous sections, fix an arbitrary special solution (p, q) to |F (x, y)| = 1 and set ρ=

p >1 q

and α = f (ρ) = ±

1 . qn

Note that since (p, q) is a special solution, ρ  P |α|1/n . The goal of this section is to find an upper bound for the distance between ρ and the exceptional point to which it belongs. To do this, we follow Thomas’s approach of first defining a family of polynomials of which f (x) is a member and then solving the problem, in general, for polynomials in the family. Because of the additional term in the tetranomial, the bounds and the arguments are more complicated than those of Thomas. For j  1, define ωj (x) = xj − ρj ∈ R[x] and for each s ∈ R, define gs (x) = a0 ωn (x) + r0 ωm (x) − sωk (x) + α. Let F = {gs (x) | s ∈ R} and note that f (x) = a0 ωn (x) + r0 ωm (x) − s0 ωk (x) + α = gs0 (x) ∈ F. For each s ∈ R, let τ (s) be the exceptional point of gs (x) belonging to ρ. Set  n−2 2(1 + κ−1 ) and Δ = , n ˆ= 2 a0 n(n − m)

(9)

where, as defined in Table 5, we let κ = κn for ε = εn , and we let κ = κn,ε for ε = 0.01 or 0.5. The introduction of the variable κ, which does not appear in Thomas’s work, enables the extension of our arguments to tetranomials of small degree. At the end of this section, we prove that for every s ∈ R − {0}, |τ (s) − ρ| < pΔ n ˆ q . As a  first step, we solve the equations gs (x) = 0 and gs (x) = 0 for the variable s and examine the behaviors of the resulting functions.

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Define R : [0, ρ) ∪ (ρ, ∞) → R by R(x) =

a0 ωn (x) + r0 ωm (x) + α ωk (x)

φ(x) =

a0 n n−k r0 m m−k x x + . k k

and φ : [0, ∞) → R by

For s, λ ∈ R and μ ∈ R − {0}, it is easily verified that gs (λ) = 0 if and only if R(λ) = s and gs (μ) = 0

if and only if φ(μ) = s.

Here we see a major difference from Thomas’s work with trinomials. In that work, φ(x) has only one term and R(x) is also less complicated; essentially, Thomas considered 0n the case of r0 = 0. In this work, we are considering 0 < |r0 | < (1−ε)a . For r0 outside m of these bounds, the overall approach fails. It is easy to see that the largest value at which φ(x) = 0 is given by ⎧ ⎪ ⎨ 0,   1 x˙ =  r0 m  n−m   ⎪ , ⎩ a0 n 

if r0 > 0, if r0 < 0.

Notice that 0  x˙ < 1 and φ is positive and increasing on (x, ˙ ∞). By direct calculation, ω  (x) we have that R (x) = ωkk (x) (φ(x) − R(x)) and so, for x ∈ (x, ˙ ρ) ∪ (ρ, ∞), R (x) = 0 if and only if R(x) = φ(x). We focus first on the case α > 0 and consider the graphs of s = R(x) and s = φ(x) for x  x. ˙ Detailed analysis using elementary methods yields the following lemma. Lemma 5.1. If α > 0, then R(x) = φ(x) at exactly two values greater than x: ˙ σ1 ∈ (x, ˙ ρ) and σ2 ∈ (ρ, ∞). Further, R has a unique maximum on (x, ˙ ρ) at σ1 and a unique minimum on (ρ, ∞) at σ2 . Let s1 = R(σ1 ), s2 = R(σ2 ), and s∗ = φ(ρ). In Fig. 1(a), we give a sketch of the graphs of R and φ for α > 0 and x  x. ˙ Although the graphs resemble the analogous graphs in Thomas’s paper [21, Fig. 1], this is only because we have restricted the domains. In fact, on the interval 0 < x < x, ˙ the behaviors of R and φ vary a great deal depending on the coefficients of F (x, y), particularly on the sign of r0 . This is one of the reasons for the restrictions given in (6), specifically, to guarantee that the interval I(4) ⊆ (x, ˙ ∞).

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Fig. 1. Graphs of R and φ.

Recall that for s ∈ R, we let τ (s) be the unique exceptional point of gs (x) ∈ F that belongs to ρ = pq . For α > 0, we see that this defines an exceptional point function, τ : R → [σ1 , σ2 ], given by ⎧ −1 ⎨ (R|[σ1 ,ρ) ) (s), τ (s) = (φ|(σ1 ,σ2 ) )−1 (s), ⎩ (R|(ρ,σ2 ] )−1 (s),

s ∈ (−∞, s1 ], s ∈ (s1 , s2 ), s ∈ [s2 , ∞).

Note that for all s < s∗ , σ1  τ (s) < ρ. So |τ (s) − ρ|  |σ1 − ρ|. Similarly, for all s > s∗ , |τ (s) − ρ|  |σ2 − ρ|. Therefore, we now bound the distance between σi and ρ, for i = 1, 2. Let A1 (x) = −(a0 ωn (x) + r0 ωm (x) − φ(x)ωk (x) + α) and note that A1 (x) = 0 if and only if R(x) = φ(x). It follows from Lemma 5.1 that σ1 and σ2 are the only roots of ˙ The next proposition bounds the values of yi = |σiρ−ρ| , for A1 (x) = 0 greater than x. i = 1, 2 and α > 0. Proposition 5.2. If α > 0, then the equation A1 (ρ(1 + y)) = 0 has two solutions −y1 and y2 such that for i = 1, 2, 

0 < yi < Δ

α . ρn

Proof. We apply [21, Theorem 9.1] with d = 2. Let B(y) = A1 (ρ(1 + y)) and for 0  (j) n A (ρ)ρj j  n, let Bj = 1 j! . So B(y) = j=0 Bj y j and for 1  j  n,

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Bj =

   n−1 n−k−1 − j−1 j−1

    r0 m(m − k) m m − 1 m−k−1 ρ + − . kj j−1 j−1

a0 n(n − k) n ρ kj



Note that B0 = A1 (ρ) = −α < 0, B1 = 0, and Bn = (1 − ε)a0 n > |r0 |m and ρ > 1, B2 =

a0 (n−k) n ρ k

= 0. Since a0 n >

a0 n(n − k) n r0 m(m − k) m a0 n(n − m) n ρ + ρ > ρ > 0. 2 2 2

Let Φ = 23 (n − 1)(n − 2) > 0 and let c = κ > 1. Since ρ  P α1/n , nρn nP n a0 n(n − m)ρn B2   . > −B0 2α 2α 2 Using direct calculation and elementary bounds, we obtain

2

nP n  2(1 + c)Φ 1 + c−1 . 2

To verify that the hypotheses of [21, Theorem 9.1] are satisfied, it remains to show |B | that, for 1  j  n − 2, Bj+2  Φj . 2 For 0  j  n − 1, let

    a0 n(n − k) n n − 1 n−k−1 ρ Cj = − , k(j + 1) j j and r0 m(m − k) m ρ Dj = k(j + 1)

    m−1 m−k−1 − , j j

so that, for 1  j  n, Bj = Cj−1 + Dj−1 . By [21, Corollary 10.1], for 1  j  n − 2, D

Cj+1 C1

 ( 2n−k−3 )j and, if r0 > 0, then for 3

1  j  n − 2, Dj+1  ( 2m−k−3 )j . 3 1 If r0 > 0, then Bj+2 = Cj+1 + Dj+1 > 0 and so, for all 1  j  n − 2, Cj+1 Dj+1 Cj+1 C1 Dj+1 D1 |Bj+2 | = + = · + · . B2 B2 B2 C1 B2 D1 B2 Now,

C1 B2

=

C1 C1 +D1

< 1 and, similarly,

< 1. Thus, for 1  j  n − 2,

j  j 2n − k − 3 2m − k − 3 + 3 3  j  j 2n − k − 3 2n − k − 3 <2  (n − k) . 3 3

|Bj+2 | < B2



D1 B2

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If r0 < 0, then since a0 n > |r0 |m, we have, for 0  j  n − 1, Cj > −Dj > 0. Thus, for 1  j  n − 2, Cj+1 + Dj+1 Cj+1 C1 Cj+1 |Bj+2 | = < = · . B2 B2 B2 B2 C1 Now, a0 n > |r0 |m implies that

C1 B2

<

n−k n−m

 n − k. Thus, for all 1  j  n − 2,

 j |Bj+2 | 2n − k − 3 < (n − k) . B2 3

Therefore, regardless of the sign of r0 , since k  1, for 1  j  n − 2,  j  j |Bj+2 | 2n − k − 3 2 (n − 1)(n − 2) = Φj . < (n − k)  B2 3 3

Hence, by [21, Theorem 9.1], the equation A1 (ρ(1 + y)) = 0 has two solutions −y1 and y2 such that, for i = 1, 2, 

0 < yi <

−B0 (1 + c−1 ) = B2



2α(1 + κ−1 ) <Δ a0 n(n − k)ρn



α . ρn

2

We now consider the case α < 0. Note that some of our notation differs from that in the case α > 0. Again, we start with a basic, but important lemma. ˙ σ1 ∈ (x, ˙ ρ) Lemma 5.3. If α < 0, then R(x) = φ(ρ) at exactly two values greater than x: ˙ ρ) and on (ρ, ∞). and σ2 ∈ (ρ, ∞). Further, R is increasing on (x, ˙ Again, let s∗ = φ(ρ). Fig. 1(b) is a sketch of the graphs of R and φ for α < 0 and x  x. Again, because we have restricted the domains, these graphs resemble the analogous graphs in Thomas’s work [21, Fig. 2], but the behaviors of R and φ for 0 < x < x˙ vary significantly with the coefficients and the degrees of the terms of F (x, y). For the case α < 0, the exceptional point function τ : R → [σ1 , σ2 ] is given by 

τ (s) =

(R|(σ1 ,ρ) )−1 (s), (R|(ρ,σ2 ] )−1 (s),

s ∈ (s∗ , ∞), s ∈ (−∞, s∗ ].

Note that for s  s∗ , ρ < τ (s)  σ2 and so |τ (s) − ρ|  |σ2 − ρ|. Similarly, for s > s∗ , |τ (s) − ρ|  |σ1 − ρ|. Let A2 (x) = gs∗ (x) = a0 ωn (x) + r0 ωm (x) − φ(ρ)ωk (x) + α, so A2 (x) = 0 if and only if R(x) = φ(ρ). By Lemma 5.3, σ1 and σ2 are the only roots of A2 (x) = 0 greater that x. ˙ Again, we bound the values of yi = |σiρ−ρ| , for i = 1, 2.

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Proposition 5.4. If α < 0, then the equation A2 (ρ(1 + y)) = 0 has two solutions −y1 and y2 such that for i = 1, 2, 

0

2|α|(1 − κ−1 ) < yi < Δ a0 n(n − k)ρn + r0 m(m − k)ρm



|α| . ρn

Proof. Again, we apply [21, Theorem 9.1] with d = 2. Let B(y) = A2 (ρ(1 + y)), and for (j) A (ρ)ρj 0  j  n, let Bj = 2 j! . So B(y) = nj=0 Bj y j and, for 1  j  n, Bj =

a0 n n ρ j

   

    r0 m m m − 1 n−1 k−1 k−1 ρ − + − . j j−1 j−1 j−1 j−1

Now B0 = A2 (ρ) = α < 0, B1 = 0, and Bn = a0 ρn = 0. Since a0 n > (1 − ε)a0 n > |r0 |m and ρ > 1, B2 > 0. Again we let Φ = 23 (n−1)(n−2), and c = κ and, following the proof of Proposition 5.2, B2 obtain −B > (2(1 + c)Φ)2 (1 + c−1 ). 0 For 0  j  n − 1, let Cj =

a0 n n ρ j+1

Dj =

r0 m m ρ j+1



   n−1 k−1 − , j j

and



   m−1 k−1 − , j j

so that, again, for 1  j  n, we have Bj = Cj−1 + Dj−1 . As in the proof of Proposition 5.2, the hypotheses of [21, Corollary 10.1] are satisfied and the desired bound follows. 2 Combining Propositions 5.2 and 5.4 yields the following corollary. Corollary 5.5. Regardless of the sign of α, for i = 1, 2, 

σi − ρ = (−1) ρyi i

with 0 < yi < Δ

Δ |α| = n/2 . ρn p

We now state and prove the main theorem of this section. ˆ and Δ be as defined in (9). Then Theorem 5.6. Let s ∈ R − {0} and let n   τ (s) − ρ < Δ . pnˆ q

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Δ Proof. By Corollary 5.5, |σi −ρ| = ρyi < pq pn/2 = we have σ1 − ρ  τ (s) − ρ  σ2 − ρ, and so

Δ ˆq , pn

4157

for i = 1, 2. Since σ1  τ (s)  σ2 ,

  τ (s) − ρ  max |σi − ρ| < Δ . i=1,2 pnˆ q

2

The following corollary, used in Sections 7 and 8, is immediate from the inequalities n  6, n − m  1, a0  1, κ > 1, and p  P  2. Corollary 5.7. For s ∈ R − {0}, |τ (s) − ρ| <

1 2q .

6. Further bounds For f (x) = a0 xn + r0 xm − s0 xk + t0 , let the height of f (x) be defined by   H = H(f ) = max |a0 |, |r0 |, |s0 |, |t0 | .

In this section, we refine the bound on |τ (s) − ρ|, restricting to the specific case of s = s0 , introducing H and ε into the bound, while removing the dependencies on a0 and m. This new bound will enable us, in Section 8, to bound the number of special solutions belonging to each exceptional point. We continue to follow the ideas in Thomas’s work [21], though the details are quite different, including our introduction of σ3 and σ4 , which are needed in proving our bounds in the case where α < 0. Retaining our notation from the previous sections, let 

Un = 2 2ε−1 1 + κ−1 ,



Vn =

97 32



1/n

,

and Wn =

2(1 + κ−1 ) . nP n

Note that since p  P , Corollary 5.5 implies that yi < Wn , for i = 1, 2. For 0  b  n ˆ, define

b Kn (b) = Un Vn (1 + Wn ) .

The goal of this section is to prove the following theorem. ˆ , and set c = n ˆ − b. If (p, q) is a special solution Theorem 6.1. Let b ∈ R, with 0  b  n to |F (x, y)| = 1 belonging to an exceptional point τ = τ (s0 ), then     Kn (b) τ − p  < .   b/n q 2H pc q b+1

We begin with a series of technical lemmas leading to Proposition 6.7, in which we bound |s|j |τ (s) − ρ|, for s = 0 and 0  j < 12 . For ease in notation, let h(s) = |τ (s) − ρ| and, for 0  j < 12 , define ξj : R − {0} → R by ξj (s) = |s|j h(s). Let |s| = υs s, with

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υ0 = 1, and let h(s) = νs · (τ (s) − ρ), with νs∗ = 1. Recall that for each gs (x) ∈ F, s is the coefficient of the term of degree k. Using this fixed k, define P (x, y) =

k−1 

xi y k−1−i .

i=0

Then for all s ∈ R, ωk (τ (s)) = (τ (s))k − ρk = νs h(s)P (τ (s), ρ) and ωk (τ (s)) = P (τ (s), τ (s)). Finally, define γs ∈ R by P (τ (s), ρ) = (1 − νs γs )P (τ (s), τ (s)). In the following two lemmas, we first bound the value of γs , and then determine the d sign of ds ξj (s) for certain values of s. Lemma 6.2. For s = 0, 0  γs < 16 . Proof. If k = 1, then P (τ (s), ρ) = P (τ (s), τ (s)) = 1 and so γs = 0. Now assume that k > 1. If τ (s) < ρ, then νs = −1 and P (τ (s), τ (s)) < P (τ (s), ρ), implying that γs > 0. Similarly, if τ (s)  ρ, then νs = 1 and P (τ (s), τ (s))  P (τ (s), ρ), implying that γs  0. Thus in either case, γs  0. Assume that τ (s) < ρ, so σ1  τ (s) < ρ. Define γ  > 0 by γ =



−1 kρk−1 P (ρ, ρ) −1= − 1 = (1 − y1 )−(k−1) − 1 < 1 − (k − 1)y1 − 1, k−1 P (σ1 , σ1 ) kσ1

using Corollary 5.5 and basic calculus. Then γs < γ  , since 1 + γs =

P (ρ, ρ) P (τ (s), ρ) < = 1 + γ. P (τ (s), τ (s)) P (σ1 , σ1 )

√ Now, we have y1 < Wn , κ > 1, and P  2, and so (k−1)y1 < (n−3)Wn < 2(n−3)/ n2n . For n  10, noting that the latter expression is a decreasing function of n, we have √ (k − 1)y1 < 2(7)/(32 10 ) < 17 . And by direct calculation, for 6  n  9, (k − 1)y1 < (n − 3)Wn < 17 . Hence, for n  6, γs < γ   (1 − (k − 1)y1 )−1 − 1 < (1 − 17 )−1 − 1 = 16 . Next, assume that ρ < τ (s). So ρ < τ (s)  σ2 . For this case, define γ  > 0 by γ = 1 −

kρk−1 P (ρ, ρ) = 1 − k−1 = 1 − (1 + y2 )−(k−1) . P (σ2 , σ2 ) kσ2

Then γs < γ  , since 1 − γ =

P (τ (s), ρ) P (ρ, ρ) < = 1 − γs . P (σ2 , σ2 ) P (τ (s), τ (s))

√ Since y2 < Wn , κ > 1, P  2, and n  6, (1+y2 )k−1 < (1+Wn )n−3 < (1+2/ n2n )n−3 < (1 + 2−n/2 )n . Now, again using basic calculus, n log(1 + 2−n/2 ) < n2−n/2 , a decreasing −n/2 function of n, since n  6. Hence, for n  14, (1 + y2 )k−1 < (1 + 2−n/2 )n < en2 

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−7

4159

e14·2 < 65 . By direct calculation, for 6  n  13, (1 + y2 )k−1 < (1 + Wn )n−3 < Therefore, for n  6, γs < γ  < 1 − 56 = 16 . 2

6 5.

If α < 0, let σ3 > ρ be the unique real number such that R(σ3 ) = −2φ(σ3 ). Lemma 6.3. Let 0  j < (a) (b) (c) (d)

If If If If

α>0 α>0 α<0 α<0

and and and and

1 2

and s = 0.

d s > s2 , then ds ξj (s) < 0. d ∗ s < −s , then ds ξj (s) > 0. d ∗ s > s , then ds ξj (s) < 0. d s < R(σ3 ), then ds ξj (s) > 0.

Proof. Note that, in all four cases, on the given interval, τ (s) = R−1 (s). So, within each ω  (x) interval, since R (x) = ωkk (x) (φ(x) − R(x)), ωk (τ (s)) νs h(s)P (τ (s), ρ) d τ (s) =  = . ds ωk (τ (s))(φ(τ (s)) − s) P (τ (s), τ (s))(φ(τ (s)) − s) It follows that 

d  h(s)(1 − νs γs ) h(s)P (τ (s), ρ) d h(s) = = , τ (s) − ρ = ds ds P (τ (s), τ (s))(φ(τ (s)) − s) φ(τ (s)) − s

and therefore, that

d j |s|j h(s)(1 − νs γs ) d ξj (s) = |s| h(s) = υs j|s|j−1 h(s) + ds ds φ(τ (s)) − s

=





h(s)|s|j−1 υs jφ τ (s) + |s|(1 − j − νs γs ) . φ(τ (s)) − s

(10)

In parts (a) and (c), s > 0 and so υs = 1. Further, s > φ(τ (s)) > 0. Thus, the first factor of (10) is negative and, using Lemma 6.2, both terms in the second factor are d positive. Therefore, in parts (a) and (c), ds ξj (s) < 0. In parts (b) and (d), s < 0 and so υs = −1. Further, φ(τ (s)) > 0 > s. Thus, the first factor of (10) is positive. So it suffices to prove that for these parts, the second factor is also positive, that is, that |s|(1 − j − νs γs ) > jφ(τ (s)). In part (b), τ (s) < ρ and so νs = −1 and, by Lemma 6.2, γs  0. Thus, |s|(1 − j − νs γs ) > 12 |s| > js∗ > jφ(τ (s)). In part (d), τ (s) > ρ and so νs = 1 and, by Lemma 6.2, γs < 16 . Also, |s| > |R(σ3 )| = 2φ(σ3 ) > 2φ(τ (s)). Thus |s|(1 − j − νs γs ) > 13 |s| > 23 φ(τ (s)) > jφ(τ (s)). Therefore, in d parts (b) and (d), ds ξj (s) > 0, completing the proof. 2 The following two lemmas deal specifically with the case where α < 0, which, compared to α > 0, requires a much more careful analysis. If α < 0, let σ4 be the unique

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real number greater than ρ such that R(σ4 ) = −s∗ . Note that ρ < σ3 < σ4 < σ2 . Define y¯ by σ4 = ρ(1 + y¯). Lemma 6.4. If α < 0, then, |α| . a0 nρn + r0 mρm

0 < y¯ <

Proof. We apply [21, Theorem 9.1] with d = 1 to the equation B(y) = A3 (ρ(1 + y)) = 0, where A3 (x) = g−s∗ (x) = a0 ωn (x) + r0 ωm (x) + φ(ρ)ωk (x) + α. Since σ4 is the unique root of g−s∗ (x) = 0 greater than ρ, y¯ is the unique root of B(y) = 0 greater than zero. Letting B(y) = nj=0 Bj y j , we have B0 = A3 (ρ) = α < 0 and for 1  j  n, a0 n n ρ Bj = j

   

    r0 m m m − 1 n−1 k−1 k−1 ρ + + + . j j−1 j−1 j−1 j−1

So, Bn = a0 ρn = 0, and since (1 − ε)a0 n > |r0 |m and ρ > 1, B1 = 2(a0 nρn + r0 mρm ) > 2εa0 nρn > 0. 1/n Let Φ = n−1 and ε  εn , ε > 0 and c = 1. Then, since ρ  P |α|  



2εa0 nρn B1 n −1 4(n − 1)  2εnP  2ε > nP n  2(1 + c)Φ 1 + c−1 . −B0 |α| nP n

For 1  j  n − 1, we have, if r0 > 0, since m < n, |Bj+1 | < B1



(a0 nρn +r0 mρm ) n−1

[ j + k−1 j ] j+1 n m 2(a0 nρ + r0 mρ )

<

  n−1  (n − 1)j < Φj , j

and if r0 < 0, |Bj+1 | < B1





a0 n n n−1 j+1 ρ [ j

+

k−1

j

2εa0 nρn

The lemma now follows from [21, Theorem 9.1].

]

  1 n−1 < < Φj . ε j

2

Lemma 6.5. If α < 0, then |σ4 − ρ| < 12 |σ2 − ρ|. Proof. By Proposition 5.4 and Lemma 6.4, we have |α|2 a0 n(n − k)ρn + r0 m(m − k)ρm y¯2 < · 2 n m 2 y2 (a0 nρ + r0 mρ ) 2|α|(1 − κ−1 )  |α|

a0 n(n − k)ρn + r0 m(m − k)ρm , (a0 nρn + r0 mρm )2

since κ  2. If r0 > 0, then, since ε2 < 1  a0 ,

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n−k n−k n−k y¯2 < |α| < |α| < |α| 2 n , y22 a0 nρn + r0 mρm a0 nρn ε nρ and if r0 < 0, then, since r0 m = −|r0 m| > −(1 − ε)a0 n, a0 n(n − k)ρn n−k y¯2 < |α| < |α| 2 n . 2 n 2 y2 (εa0 nρ ) ε nρ 4(n−1) nP n ,

Since P n |α|  ρn and ε2 

we have

1 n−k n−k y¯2  . < |α| 2 n < 2 y2 ε nρ 4(n − 1) 4 Thus y¯ < 12 y2 and therefore |σ4 − ρ| < 12 |σ2 − ρ|.

2

We now establish two important bounds of s2 , in terms of Wn . Lemma 6.6. (a) If r0 > 0, then s2 < (1 + Wn )n s∗ . (b) If r0 < 0, then s2 < ε−1 (1 + Wn )n s∗ . Proof. First if r0 > 0, then a0 nσ2n−m ρn−m + r0 mρn−m < a0 nσ2n−m ρn−m + r0 mσ2n−m and so σ2n−m a0 nσ2n−m + r0 m < = a0 nρn−m + r0 m ρn−m



σ2 ρ

n−m

.

Thus, φ(σ2 ) s2 = = ∗ s φ(ρ)



a0 nσ2n−m + r0 m a0 nρn−m + r0 m



σ2 ρ

m−k



<

σ2 ρ

n−k

.

Alternatively, if r0 < 0, then, since r0 m = −|r0 m| > −(1 − ε)a0 n, s2 < s∗



a0 n εa0 n



σ2 ρ

n−k

= ε−1



σ2 ρ

n−k

.

The theorem now follows from 

σ2 ρ

n−k

= (1 + y2 )n−k < (1 + Wn )n−k < (1 + Wn )n .

2

We use the above lemmas with Theorem 5.6 to bound the value of ξj (s). These bounds are key to the proof of Theorem 6.1, at the end of this section.

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Proposition 6.7. Let 0  j <

1 2

and s ∈ R − {0}. Then 

ξj (s) = |s|j h(s) <

(1 + Wn )n s∗ ε

j

Δ . pnˆ q

Further, if r0 > 0, then ξj (s) = |s|j h(s) < ((1 + Wn )n s∗ )j pΔ n ˆq . Proof. If 0 < |s|  s∗ , then, since (1 + Wn )n > 1, ξj (s) = |s|j h(s)  (s∗ )j h(s) < ((1 + Wn )n s∗ )j h(s) < ((1 + Wn )n s∗ )j pΔ n ˆ q , by Theorem 5.6. ∗ Assume now that α > 0 and |s| > s . If s > s2 , then, by Lemma 6.3(a), ξ(s) < ξ(s2 ) = |s2 |j h(s2 ). Applying Theorem 5.6 and Lemma 6.6(a), the result follows immediately. If 0 < s∗ < s  s2 , then ξj (s) = |s|j h(s)  |s2 |j h(s) and the result follows as before. And if s < −s∗ , then by Lemma 6.3(b), ξj (s) < ξj (−s∗ ) = |−s∗ |j h(−s∗ ) < sj2 h(−s∗ ). Once more, the result follows from Theorem 5.6 and Lemma 6.6(a). Next, assume that α < 0 and s > s∗ . By Lemma 6.3(c),   j j ξj (s)  lim∗+ ξj (t) = lim∗+ |t|j τ (t) − ρ = s∗ |σ1 − ρ| = s∗ h(s1 ). t→s

t→s

Since (1 + Wn )n > 1, the result follows from Theorem 5.6. Finally, assume that α < 0 and s < −s∗ . If R(σ3 )  s < −s∗ = R(σ4 ), then ρ < τ (s) < σ4 and so h(s) = |τ (s) − ρ| < |σ4 − ρ|. Hence, ξj (s) = |s|j h(s) < |R(σ3 )|j |σ4 − ρ|. Alternatively, if s < R(σ3 ), then, by Lemma 6.3(d), j  j

 ξj (s)  ξj R(σ3 ) = R(σ3 ) |σ3 − ρ| < R(σ3 ) |σ4 − ρ|.

Note that since 1 < σ3 < σ2 and φ is increasing for x > 1, we have |R(σ3 )| = 2φ(σ3 ) < 2φ(σ2 ) = 2s2 . So in either case, we apply Lemma 6.5 to get ξj (s) < |R(σ3 )|j |σ4 − ρ| < (2s2 )j ( 12 |σ2 − ρ|) < sj2 h(σ2 ), since j < 12 . The result now follows from Theorem 5.6 and Lemma 6.6(b). 2 Let J = max{Jn , Jm , Jk }, where Jn = a0 ρn , Jm = |r0 |ρm , Jk = |s0 |ρk . We need one more inequality before proving Theorem 6.1. Lemma 6.8. H < Vnn J. Proof. Since ρ > 1, if H ∈ {|a0 |, |r0 |, |s0 |}, then H < J < Vnn J. So assume that H = |t0 |. 1 If q = 1, then ρ = pq  2 and so J  a0 ρn  2n > 25 , implying that |α| = 1 < 32 J. If 1 1 1 n m q  2, then, since J > 1, |α| = qn < 25  32 J. Thus H = |t0 | = |−a0 ρ − r0 ρ + s0 ρk + n α|  3J + |α|  97 32 J = Vn J. 2 Note that, since k  1 and n − m  1, 

a0 n εk

1/2



Δ=

2(1 + κ−1 ) εk(n − m)

1/2



Un . 2

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Thus, for 0  j < 12 , 

a0 n εk

j

<

Un . 2Δ

(11)

We now prove the main result of this section. Proof of Theorem 6.1. First, assume that J ∈ {Jn , Jm }. Then we have J = 0n p n max{a0 ρn , |r0 |ρm }  max{a0 ρn , |r0 |ρn } < max{ a0kn ρn , |r0k|m ρn } = a0kn ρn < aεk (q) . So, by Lemma 6.8 and (11), 

H b/n < Vnb J b/n < Vnb

a0 n εk

b/n  b   b p Un p < Vnb . q 2Δ q

Thus, by Theorem 5.6,     b     p Un Vnb Kn (b) τ − p H b/n = τ (s0 ) − ρH b/n < Δ · Vnb Un = < c b+1 ,   n ˆ c b+1 q p q 2Δ q 2p q 2p q Kn (b) since Wn > 0. So |τ − pq | < 2H b/n . pc q b+1 Now assume that J = Jk = |s0 |ρk . By Lemma 6.8,

H b/n < Vnb J b/n = Vnb |s0 |b/n ρbk/n .

(12)

0 n n−k Recall that s∗ = φ(ρ) = a0kn ρn−k + r0km ρm−k . Thus, if r0 > 0, then s∗ < (2−ε)a ρ < k a0 n n−k a0 n n−k ∗ , and if r0 < 0, then s < k ρ . Combining this with Proposition 6.7 and εk ρ then using (11),



|s0 |

b/n

h(s0 ) < <

(1 + Wn )n a0 n εk

b/n

ρb(n−k)/n

Δ pnˆ q

Un (1 + Wn )b b−bk/n ρ . 2pnˆ q

Thus, by inequality (12), H b/n h(s0 ) < |s0 |b/n h(s0 )Vnb ρbk/n < So |τ − pq | = h(s0 ) <

Kn (b) . 2H b/n pc q b+1

2

Un (Vn (1 + Wn ))b b Kn (b) ρ = c b+1 . n ˆ 2p q 2p q

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7. Bounding the number of large special solutions In this section, following standard practice, we use a fixed number, Y = Y (F ), to classify solutions to |F (x, y)| = 1, with F (x, y) as in (5), as being either small or large. We then apply the Thue–Siegel principle to bound the number of large special solutions belonging to a fixed exceptional point. In Section 8, we bound the number of small special solutions, obtaining bounds on the total number of special solutions belonging to a fixed exceptional point. Let a1 and a2 be as given in Table 6(a) for ε = εn or ε = 0.01, and let a1 and a2 be as given in Table 6(b) for ε = 0.5. Let 

u=

2 , n + a21

w = a2 u,

and  =

2 . u−w

(13)

Recalling that H is the height of f (x), define Y by log Y = χn log H + πn ,

(14)

where 

χn =

n n−

n,

+

 , a21 (n−)

if 6  n  32, if n  33,

and ⎧ ( n + 2 ) log(n + 1) + ⎪ ⎪ ⎨ 2(n−) 2a1 (n−) n n , + 2(n−) log n + 2a2 (n−) πn = 1 ⎪ ⎪ ⎩ log 2 + n+4 2 log n + n log 5, n+4

n+1+2 n−

log 2 if 6  n  32, if n  33.

Note that for n  33, Y = 2n 2 (5H)n . A direct calculation shows that, in fact, for n+4 n  6, Y  2n 2 (5H)n . We say that a special solution (p, q) is a small special solution if 1  q  Y , and that (p, q) is a large special solution if q > Y . To bound the number of large special solutions belonging to a fixed exceptional point, we use an application of the Thue–Siegel principle developed by Bombieri and Schmidt [9, Lemma 2]. The following lemmas allow us to apply this result, despite significant differences between the hypotheses in that work and those herein. The first difference is that, given a large special solution (p, q), we consider to which exceptional point it belongs, while Bombieri and Schmidt consider to which root of f (x) p q is closest. Lemma 7.1 provides that the exceptional point belonging to a large solution (p, q) is, in fact, a real root of f (x) and Lemma 7.2 provides that the real root is, of all roots of f (x), the closest to pq .

(a) ε = εn or ε = 0.01 n

6

7

8–9

10

11–16

17

18

19–20

21

22–25

26–32

 33

a1 a2

0.4 0.41

0.4 0.45

0.44 0.47

0.44 0.46

0.4 0.48

0.4 0.51

0.39 0.52

0.35 0.53

0.28 0.54

0.28 0.56

0.22 0.58

0.46 0.5

(b) ε = 0.5 n

6

7–10

11–12

13–16

17

18

19

20

21–27

28

29–30

31–32

 33

a1 a2

0.2 0.3

0.3 0.45

0.35 0.46

0.37 0.46

0.37 0.48

0.37 0.51

0.37 0.53

0.37 0.55

0.26 0.57

0.25 0.57

0.24 0.58

0.22 0.58

0.46 0.5

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Table 6 Values for a1 and a2 .

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Lemma 7.1. If the special solution (p, q) belongs to a proper critical point, then q < 2n−1 25nH n and (p, q) is a small special solution. Proof. Let (p, q) belong to a proper critical point τ . Since τ is not a real root of f (x), we must have α > 0 and so f ( pq ) = q1n > 0. Thus, since pq belongs to τ , f (τ ) > 0. Because τ is a proper critical point, f  (τ ) > 0 and therefore f (x) has a local minimum 1 at τ . Hence q1n = f ( pq )  f (τ ). By a result of Rump [17, Lemma 3], f (τ ) > nn (S+1) 2n−1 , 1 1 1 where S = S(f ) = |a0 | + |r0 | + |s0 | + |t0 |  4H. Thus qn > nn (S+1)2n−1 > nn (5H)2n−1 2n−1

2n−1

and so q < n(5H) n < 25nH n . 2n−1 n+4 Finally, for all n  6, q < 25nH n < 2n 2 (5H)n  Y , and thus (p, q) is a small special solution. 2 Thus if (p, q) is a large special solution, it must belong to a real root of f (x). We now show that this real root is the root of f (x) that is closest to pq . We note that, although Thomas [21] used the analogous result for trinomials, he did not explicitly state or verify it in his paper. Lemma 7.2. If a large special solution (p, q) belongs to a real root τ , then for any root β of f (x), |τ − pq |  |β − pq |. Proof. Suppose, to the contrary, that there exists a root β of f (x) such that |β − pq | < 1 |τ − pq |. Then, by Corollary 5.7, |β − pq | < |τ − pq | < 2q . Thus, since (p, q) is a large special solution,      1 p  p 1 1 |τ − β|  τ −  + β −  < < < , n+4 n+4 n q q q 2 2 2n (5H) 2n (S + 1)n where, again, S = |a0 |+|r0 |+|s0 |+|t0 |. But this contradicts the minimum root separation result of Rump [17, Theorem 4]. Hence, τ is the root of f (x) closest to pq . 2 The remaining key difference between Bombieri and Schmidt’s hypotheses and ours is the definition of large solution, in which we use the value Y , defined in (14), and they use the value Y0 , defined below. In their definition, Bombieri and Schmidt allow for any values of a1 and a2 , with 0 < a1 < a2 < 1. For our purposes, we choose these values as in our definition of Y . Using the notation in (13), set   n A = a−2 log M (f ) + , 1 2

1

n C = 2n 2 M (f ) ,

 1 Y0 = (2C) n− 4eA n− ,

and (15)

where M (f ) denotes the Mahler measure of f (x) (in [9], this is called the Mahler height). The following lemma provides that, for each value of n, we have Y > Y0 . Therefore each large solution in this work is also a large solution under Bombieri and Schmidt’s

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definition. This justifies the application of their bound in the proof of Theorem 7.4. The need for our definition of Y also to differ from that used by Thomas [21, Eq. (5.4)], specifically our definition of πn , is seen in the following proof. Lemma 7.3. Let a1 and a2 be as given in Table 6(a), for ε = εn or ε = 0.01, and let a1 and a2 be as given in Table 6(b), for ε = 0.5. Let Y and Y0 be defined as in (14) and (15), respectively. Then, Y > Y0 . 1

Proof. Using the fact that M (f )  (n + 1) 2 H (see [11, p. 343]), we have 

log Y0 

  n + 2 log H n −  a1 (n − )    n + log(n + 1) + 2(n − ) 2a21 (n − )

+

n + 1 + 2 n n log 2 + log n + 2 . n− 2(n − ) 2a1 (n − )

(16)

Hence, if 6  n  32, we have Y0  Y . Now, viewing u, w, and  as functions of n and noting that, for n  33, we have a1 = 0.46 and a2 = 0.5, the function G, defined for n  33 by G(n) =

 n + , n −  a21 (n − )

is a decreasing function of n. Since G(33) < 7 < 33, we have G(n) < 7 < n, for all n  33. Thus, for all n  33,  n +
and

n  7 + < . 2(n − ) 2a21 (n − ) 2

(17)

Using basic calculus, we also have that, for n  33, n + 1 + 2 < 4, n−

n < 1, 2(n − )

and

 2a21 (n

− )

<

7 . 3

Thus, combining Eqs. (16)–(18), for n  33, log Y0 < n log H +

7 7 log(n + 1) + 4 log 2 + log n + n. 2 3

Verification that, for n  33, 7 n+4 7 log(n + 1) + 4 log 2 + log n + n < log 2 + log n + n log 5 2 3 2 completes the proof. 2

(18)

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We now bound the number of large special solutions belonging to a fixed real root (recalling that, by Lemma 7.1, no large special solutions belong to any proper critical point). Theorem 7.4. Let τ > 1 be a real root of f (x). For n  9, there are at most Ln large special solutions belonging to τ , where 

Ln =

4, 3,

if 9  n  12, if n  13.

(19)

Further, if ε = 0.01 or 0.5, then there are at most Ln,ε large special solutions, where

Ln,0.01

⎧ ⎨ 5, = 4, ⎩ 3,

if n = 6, if 7  n  12, if n  13,



and

Ln,0.5 =

4, 3,

if 6  n  7, if n  8.

Proof. Again let a1 and a2 be as given in Table 6(a), for ε = εn or ε = 0.01, and let a1 and a2 be as given in Table 6(b), for ε = 0.5. Let Y and Y0 be defined as in (14) and (15), using the values of u, w, and  in (13). Let Sτ be the set of large special solutions (p, q) that belong to τ . Let (p, q) ∈ Sτ . Then, since (p, q) is a large special solution, p > q > Y and by Lemma 7.2, τ is the root of f (x) closest to pq . By Lemma 7.3, Y > Y0 . Therefore, we can apply the bound from Bombieri and Schmidt [9, Lemma 2], to get 

 log(δ −1 ( − 2)−1 ) |Sτ |  + 2, log(n − 1) +w −2 where δ = nu n−1 . For n  32, direct calculation of individual cases now yields the desired bounds. For n  33, notice that 2

2

0 < δ −1 ( − 2)−1 =

and thus,

log(δ −1 (−2)−1 ) log(n−1)

(n + a21 )(n − 1)  < (n − 1)2 , (1 − 4a21 )( 2(n + a21 ) − 1)

< 2, giving us |Sτ |  1 + 2 = 3, as desired. 2

8. Bounding the total number of special solutions Fix an exceptional point τ of F (x, 1), with F (x, y) as in (5). In this section, we use the results from Sections 5–7 to determine upper bounds on the total number of special solutions to |F (x, y)| = 1 that belong to τ . Specifically, we prove that for τ a real root, the number of special solutions belonging to τ is at most wn , as in Table 1, and for τ a proper critical point, the number is at most wn , as in Table 3. Further, with the added

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assumptions in Theorem 1.2, we prove smaller bounds for 6  n  59. In Section 9, we use these bounds to complete the proofs of Theorems 1.1–1.4. Again using notation set by Thomas [21], we fix t ∈ Z such that there are t + 2 small special solutions that belong to τ . Since our bounds are all at least 2, we assume that t  0. Letting {(p0 , q0 ), (p1 , q1 ), . . . , (pt+1 , qt+1 )} be the set of distinct small special solutions that belong to τ , labeled so that 1  q0  q1  · · ·  qt+1  Y, it follows from Corollary 5.7 that 2  p0 < p1 < · · · < pt+1 . The following lemma provides a lower bound on each qi , 1  i  t+1, paralleling similar results of Thomas [21, Theorem 4.1, Eq. (4.4)]. For i  1, define Ei (x) =

xi − 1 . x−1

Lemma 8.1. For b, b0 ∈ R, with 2  b  n ˆ and 0  b0  n ˆ − 0.5, let c = n ˆ − b, and c0 = n ˆ − b0 . If 1  i  t, then i

qi+1 >

i

H Bb Qb1 , K Ei (b)

where B=

1 + b0 , n

Q1 =

pc00 , Kn (b0 )

and

K=

Kn (b) . (Q1 + 1)c

(20)

Proof. Let (pi , qi ) and (pj , qj ) be small special solutions that belong to τ with 0  i < j  t + 1. By Theorem 6.1, we have      pi   pj  Kn (b) Kn (b) Kn (b) 1  τ −  + τ −  < + < , qi qj qi qj 2H b/n pci qib+1 2H b/n pcj qjb+1 H b/n pci qib+1

and thus, qj >

H b/n pci qib . Kn (b)

(21)

Applying Theorem 6.1 as above to (p0 , q0 ) and (p1 , q1 ) with b0 in place of b, we get q1 >

H b0 /n pc00 H b0 /n pc00 q0b0  = H b0 /n Q1 . Kn (b0 ) Kn (b0 )

(22)

Now, since p1 and q1 are positive integers, p1  q1 + 1 and so p1 > H b0 /n Q1 + 1. Because H  1, we have

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p1 > Q1 + 1. Next, by applying inequality (21) to (p1 , q1 ) and (p2 , q2 ), then using inequality (22), we have q2 >

H b/n (Q1 + 1)c (H b0 /n Q1 )b H Bb Qb1 H b/n pc1 q1b > > . Kn (b) Kn (b) K

The result now follows from an inductive argument using inequality (21). 2 Lemma 8.2. Let Q2 = then

Q1 K .

Under the hypotheses of Lemma 8.1, if K > 1 and Q2 > 1,

log qi+1 > Bbi log H + bi log Q2 . Proof. Let K > 1 and Q2 > 1. Since b  2, we have Ei (b) = Lemma 8.1, i

qi+1 >

i

i

bi −1 b−1

< bi . Therefore, by

i

i i H Bb Qb1 H Bb Qb1 > = H Bb Qb2 . E (b) K i K bi

Taking logarithms, the lemma follows.

2

We now use Lemma 8.1 to bound the number of small special solutions and, therefrom, the total number of special solutions, in the case n  60, regardless of the value of ε  εn . Theorem 8.3. Assume n  60. If τ is a real root, then there are at most wn = 6 special solutions belonging to τ . If τ is a proper critical point, then there are at most wn = 2 special solutions belonging to τ . Proof. Let n  60. Using the notation of Lemma 8.1, set b =  n ˆ − 1 = 0.5n  − 2 and n2n n −1 n ˆ b0 = 0.13ˆ n. Then Bb > 2 and Bb2 > n. Since κ = n−1 and ε = = 2 2 4(n−1) n−1 <   n+1 2nˆ +0.02 , we have Un = 2 2ε−1 (1 + κ−1 ) < 20.5ˆn+1.51 n−1 < 20.25n+1.04 . Now, (1 + Wn )b decreases as n  60 increases and so (1 + Wn )b  (1 + W60 )28 < 20.01 . Since Vnb < 2, we have Kn (b) = Un (Vn (1 + Wn ))b < 20.25n+2.05 < 20.3n . From p0  2, c0 = 0.87ˆ n > 0.42n, and Kn (b0 ) < Kn (b), it follows that Q1 =

pc00 > 20.42n−0.3n = 20.12n Kn (b0 )

and so, since K < Kn (b) and b  28, Qb1 > 2(0.12n)(28)−0.3n > 23n . K

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Suppose that τ is a real root and that there are at least 4 small special solutions belonging to τ . Setting i = 2 in Lemma 8.1 and using the above,  b b 2 2 2 H Bb Qb1 1 n Q1 > H n · 23n(0.5n−2)−0.3n > H n · 21.3n . q3 > >H K K K E2 (b) n+4

0.3n Since 2n > n > n+4 > 10, q3 > H n (21.3n )n > H n n 2 ·2·5n = Y , contradicting 2 and 2 that (p3 , q3 ) is a small special solution. Thus, there are at most 3 small special solutions belonging to τ . Combining this with Theorem 7.4 yields that there are at most 6 special solutions belonging to τ . Now suppose that τ is a proper critical point and that there are at least 3 small special solutions belonging to τ . Setting i = 1 in Lemma 8.1,

q2 > H Bb

2n−1 Qb1 > H 2 · 23n > 25nH n , K

contradicting Lemma 7.1. Thus, there are at most 2 small special solutions belonging to τ , and by Lemma 7.1, there are no other special solutions belonging to τ . 2 The following two theorems provide bounds for the cases with n  59. Unfortunately, Lemma 8.1 and Lemma 8.2 are required for this method and, with ε = εn , they do not apply for values of n less than 9. Indeed, for n  5, n ˆ = n−2  1.5, so the condition 2 2bn ˆ , in the hypothesis of Lemma 8.1, is impossible to satisfy. And for 6  n  8, we have QK1 < 1 (independent of choices of P  2 and κ  2), contradicting the hypothesis of Lemma 8.2. With the larger values of ε assumed in Theorems 1.2 and 1.4, however, the process goes through and we determine bounds for all n  6. For the remainder of this section, assume the notation defined in Lemma 8.1 and Lemma 8.2. Theorem 8.4. Assume that τ is a proper critical point. For 9  n  59, there are at most wn special solutions belonging to τ . Further, if ε = 0.01 or 0.5, for 6  n  59, there are at most wn,ε . Proof. Recall that there are t + 2 small special solutions belonging to τ and that, by Lemma 7.1, there are no large special solutions belonging to τ . Also, by Lemma 7.1, log qt+1 <

2n − 1 log H + log(25n). n

Assuming first only that ε  εn , suppose, for a contradiction, that t + 2 > wn . Letting b0 and b be as given in Table 7(a), K > 1 and Q2 > 1. Then Lemma 8.2 applies and so log qt+1 > Bbt log H + bt log Q2 . But, direct calculation verifies that 2n − 1 < (1 + b0 )bwn −1  (1 + b0 )bt and log(25n) < bwn −1 log Q2  bt log Q2 , yielding a contradiction, 0 since B = 1+b n .

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Table 7 Values for b and b0 . (a) ε  εn n

9–11

12–18

19–23

24–26

27–32

33–49

50–59

b b0

2 0

n ˆ−2 0

n ˆ−2 0.5

n ˆ−2 1

n ˆ − 1.5 1

n ˆ−1 1

n ˆ−1 0.125ˆ n

(b) ε = 0.01 n

6–7

8–9

10–11

12–15

16–17

18–20

21–23

24–32

33–49

50–59

b b0

2 0

n ˆ − 0.5 0

3 0

n ˆ−1 0

n ˆ − 0.5 0.5

n ˆ − 1.5 0.25

n ˆ−1 0.5

n ˆ − 0.5 1

n ˆ 3.7

n ˆ 0.45ˆ n

(c) ε = 0.5 n

6–12

13–19

20–32

33–59

b b0

n ˆ 0

n ˆ 1

n ˆ 0.33ˆ n

n ˆ 0.6ˆ n

If ε = 0.01 or 0.5, let b0 and b be as given in Table 7(b) or 7(c) and suppose for a contradiction that t+2 > wn,ε . Verifying the above inequalities completes the proof. 2 We now bound the total number of special solutions that belong to a real root, for n  59. Theorem 8.5. Assume that τ is a real root. For 9  n  59, there are at most wn special solutions belonging to τ . Further, if ε = 0.01 or 0.5, for 6  n  59, there are at most wn,ε . Proof. First, consider ε = εn . Using the values of a1 and a2 given in Table 6(a), let χn and πn be as defined in Section 7. Using the values of b and b0 given in Table 7(a), set Tn =



log(χn ) − log(B) log(b)



and Tn =



 log(πn ) − log log(Q2 ) . log(b)

(23)

Recalling that t + 2 is the number of small solutions belonging to τ , suppose that t > max{Tn , Tn }. Then t  Tn + 1 >

log(χn ) − log(B) , log(b)

implies that Bbt > χn , and t  Tn + 1 >

log(πn ) − log log(Q2 ) , log(b)

implies that bt log Q2 > πn . Since, for 9  n  59, K > 1 and Q2 > 1, we can apply Lemma 8.2 to i = t obtaining

H.G. Grundman, D.P. Wisniewski / Journal of Number Theory 133 (2013) 4140–4174

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log qt+1 > Bbt log H + bt log Q2  χn logH + πn = log Y, contradicting the definition of a small special solution. Thus, t  max{Tn , Tn }, with the number of special solutions belonging to τ at most t + 2 + Ln . Direct calculation yields that, for each n,   t + 2 + Ln  max Tn , Tn + 2 + Ln = wn .

For ε = 0.01 or 0.5, let a1 , a2 , b and b0 be as defined in Tables 6 and 7 and de  fine Tn,ε and Tn,ε analogous to definition (23). Verifying that, for each 6  n  59,   max{Tn,ε , Tn,ε } + 2 + Ln,ε  wn,ε completes the proof. 2 9. Proofs of the main theorems We now have all of the pieces in place for proving the theorems in Section 1. We begin by assuming that F (x, y) satisfies the hypotheses of Theorems 1.1 and 1.3 and, without loss of generality, that a > 0. Let τ ∈ I(4) be an exceptional point of fi (x), for some 1  i  4. If τ is a real root, then by Theorems 8.3 and 8.5, the number of special solutions belonging to τ is bounded by wn . If τ is a proper critical point, then by Theorems 8.3 and 8.4, the number of special solutions belonging to τ is bounded by wn . Verifying that, for each n, dn = d(Pn ) as defined in Section 4, Theorem 1.3 follows immediately from Theorem 4.1. From Tables 1 and 3, we see that, regardless of the value of n, wn  wn . Thus, Theorem 1.3 implies that   NF  wn RF + wn CF + 4dn  wn (RF + CF ) + 4d  wn E(f ) + 4d.

As noted in Section 2, the form G(x, y) = F (y, x) also satisfies the hypotheses of Theorem 1.1 and the equations |F (x, y)| = 1 and |G(x, y)| = 1 have the same number of regular solutions. Thus, without loss of generality, by Corollary 2.6, NF  wn |E(f )| + 4d  wn vF + 4d, completing the proof of Theorem 1.1. Next, in addition, assume that F (x, y) satisfies condition (6) with ε = 0.01 or 0.5. As above, Theorems 4.1, and 8.3–8.5, and verification that dn,ε = d(Pn,ε ), for each n and ε, yields Theorem 1.4. And finally, verifying from Tables 2 and 3(b) that, wn,ε  wn,ε , for each n and ε, Theorem 1.2 follows from Theorem 1.4 and Corollary 2.6. Acknowledgments The authors would like to thank the anonymous referee of an earlier version of this paper for many helpful suggestions.

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