Max–min of polynomials and exponential diophantine equations (II)

Journal of Number Theory 157 (2015) 397–423

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

Max–min of polynomials and exponential diophantine equations (II) Shih Ping Tung 1 Department of Applied Mathematics, Chung Yuan Christian University, Chung Li, 32023 Taiwan, ROC

a r t i c l e

i n f o

Article history: Received 15 November 2013 Received in revised form 7 May 2015 Accepted 7 May 2015 Available online 7 July 2015 Communicated by David Goss Keywords: Polynomials Exponential diophantine equations Max–min values

a b s t r a c t Based on the estimate of max–min values of polynomials over integers, we study exponential diophantine equations with parameters. We proved the case for exponential equations with two terms. In this paper, we extend our previous results to general exponential equations. In particular, we have the  Gk (x,y) following: Let F (x, y) = m , where fk (x, y) k=0 fk (x, y)b and Gk (x, y) are in Z[x, y], b is a positive integer, and Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z for each 1 ≤ i = j ≤ m. Then for every integer α there is an integer c such that F (α, c) = 0 if and only if for every integer α there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(α) ∈ Z. This result can be extended to higher orders of exponential diophantine equations. © 2015 Elsevier Inc. All rights reserved.

1. Introduction Let F (x1 , x2 , . . . , xn ) be a polynomial over Z. Many questions arise from the equation F (x1 , x2 , . . . , xn ) = 0. For example, what can we say about the polynomial E-mail address: [email protected]. Supported by a grant from the National Science Council of the Republic of China, NSC 102-2115-M-033-001. 1

http://dx.doi.org/10.1016/j.jnt.2015.05.020 0022-314X/© 2015 Elsevier Inc. All rights reserved.

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F (x1 , x2 , . . . , xn ) if F (x1 , x2 , . . . , xn ) = 0 has sufficiently many integer solutions? Let F (x1 , x2 , . . . , xn , y) ∈ Z[x1 , x2 , . . . , xn , y]. For any integers a1 , a2 , . . . , an , does there exist an integer b such that F (a1 , a2 , . . . , an , b) = 0? Equivalently, is ∀x1 ∀x2 . . . ∀xn ∃y F (x1 , x2 , . . . , xn , y) = 0 true in Z? We give a necessary and sufficient condition that ∀x1 ∀x2 . . . ∀xn ∃y F (x1 , x2 , . . . , xn , y) = 0 is true in Z [5]. In fact, the decision problem of determining for an arbitrarily given F (x1 , x2 , . . . , xn , y) ∈ Z[x1 , x2 , . . . , xn , y] whether ∀x1 ∀x2 . . . ∀xn ∃y F (x1 , x2 , . . . , xn , y) = 0 is true in Z or not is co-NP-complete [6]. This kind of problems and their generalizations are called diophantine equations with parameters [1,3], and Schinzel devoted almost all of Chapter 5 in [4] to this problem. Previously, these problems were considered about polynomials, but in [8] we considered a similar problem for exponential polynomials. In order to solve that problem, we first studied the maximum–minimum of a polynomial over Z. Let C and D be arbitrary non-empty sets, and let F be a function from C × D to [−∞, +∞]. For each u ∈ C one can take the minimum of F (u, v) over v ∈ D and then take the maximum of this minimum as a function on C. The quantity so obtained is maxu∈C minv∈D F (u, v). We use the properties of the max–min values of polynomials over Z to solve our problem. Let A and B be subsets of Z, and we may ask what is the value of maxx∈A miny∈B |F (x, y)|? This should also be an interesting topic on its own. In a certain sense, this is another generalization of the problem of diophantine equations with parameters. Sentence ∀x∃y F (x, y) = 0 is true in Z if and only if maxx∈Z miny∈Z |F (x, y)| = 0. Let A ⊆ Z be a set with upper asymptotic density, and A(T ) = {x ∈ A : |x| ≤ T }. In [2], it is proved that if max min |F (x, y)| = o(T 1/2 )

x∈A(T ) y∈Z

then there is an h(x) ∈ Q[x] such that F (x, h(x)) ≡ c for a c ∈ Q. Based on this result, we study exponential diophantine equations with parameters having two terms in [8]. In this paper we extend our results to more general exponential equations. In particular, m we have the following: Let F (x, y) = k=1 fk (x, y)bGk (x,y) , where fk (x, y) and Gk (x, y) are in Z[x, y], b is a positive integer, and

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Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z for each 1 ≤ i = j ≤ m. Then for every integer α there is an integer c such that F (α, c) = 0 if and only if for every integer α there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(α) ∈ Z. 2. Preliminaries and notation Let f (x, y) be a polynomial over Q. We write degx (f (x, y)) and degy (f (x, y)) to denote the degrees of the variables x and y in f (x, y), respectively. If f (x) is a polynomial with a single variable, then we simply write deg(f (x)) to denote the degree of the polynomial f (x). If D is the least common multiple of the denominators of the coefficients of f (x, y), then D is called the denominator of the polynomial f (x, y) and denoted by Den(f (x, y)). Hence, Den(f (x, y)) · f (x, y) ∈ Z[x, y]. We use f (x, y) to denote the maximum of the absolute values of the coefficients of the polynomial Den(f (x, y))·f (x, y). For every subset A of Z, |A| denotes the number of the elements of A and A(T ) = {a ∈ A, |a| ≤ T }. The set A has density δ if and only if d(A) = lim

T →∞

|A(T )| = δ. 2T

¯ The upper asymptotic density (abbreviated UAD) d(A) of a set A is |A(T )| ¯ . d(A) = lim sup 2T T →∞ Similarly, we define the lower asymptotic density (abbreviated LAD) d(A) of a set A to be d(A) = lim inf T →∞

|A(T )| . 2T

¯ If d(A) = d(A) = d, then d(A) = d. We have the following results. Theorem A. (See [2].) Let F (X, Y ) ∈ Q[X, Y ]. If A is a set of positive UAD, and y(a), a ∈ A, are integers such that |F (a, y(a))| = o(a1/2 ), then there exists a polynomial f (X) ∈ Q[X] such that F (X, f (X)) is a constant. Now, we give some definitions. Definition. Let F (x, y) be a polynomial over Q, the solvable set SF of F (x, y) over Z is SF = {t ∈ Z | F (t, y) = 0 is solvable in Z}.

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Definition. Let F (x, y) be a polynomial over Q, and A a subset of Z. Then SA,F (T ) = max min |F (x, y)|. x∈A(T ) y∈Z

Note that for any x ∈ SF we may choose a y such that F (x, y) = 0. Therefore, SSF ,F (T ) = 0. Also, if F (x) ∈ Q[x], and F (x) may be viewed as an element of Q[x, y], then SF is simply the set of roots in Z of F (x) = 0. Hence, if F (x) ∈ Q[x] and SF is infinite, then F (x) ≡ 0. Definition. A polynomial f (x, y) is said to be with polynomial growth on an infinite set A if SA,f (T ) = o(T 1/2 ). 3. Max–min values and near solutions In this section we give the properties of near solutions of polynomials and their relationship with the max–min values of the corresponding polynomials. We first studied the near solutions of polynomials in [7], and though we did not use the name there, and we will use their properties in this paper repeatedly. For the sake of convenience, we restate the results in [7,8] with the name “near solution.” Here x ¯ denotes (x1 , x2 , . . . , xn ) for n ≥ 1. In this paper we do not need the multivariate form of near solutions. But because their properties may be used in some other places, we also write these results in the more general multivariate forms. We first define the near solutions of a polynomial. x, y) ∈ Q[¯ x, y]. Then H(¯ x) ∈ Q[¯ x] is a near solution of F (¯ x, y) if Definition. Let F (¯ F (¯ x, H(¯ x)) ≡ c ∈ Q. x, y) ∈ Q[¯ x, y] and degy (F (¯ x, y)) = n. Theorem 3.1. (See [7, Theorem 2.5].) Let F (¯ 1. If F (¯ x, y) has only a finite number of near solutions, then F (¯ x, y) has at most n near solutions. 2. F (¯ x, y) has infinitely many near solutions if and only if there is a unique polynomial g(¯ x) ∈ Q[¯ x], with its constant term equal to zero, such that F (¯ x, y) =

n 

cj (y − g(¯ x))j

j=0

for some rational numbers cj , and h(¯ x) is a near solution of F (¯ x, y) if and only if h(¯ x) = g(¯ x) + a for an a ∈ Q. Lemma 3.2. (See [7, Proposition 2.1].) Let F (¯ x, y) = (y − g(¯ x))G(¯ x, y) + a, where a ∈ Q, F (¯ x, y), g(¯ x) and G(¯ x, y) are polynomials over Q. Let h(¯ x) ∈ Q[¯ x] and h(¯ x) = g(¯ x). If h(¯ x) is a near solution of F (¯ x, y), then h(¯ x) is a near solution of G(¯ x, y). Moreover, if F (¯ x, g(¯ x)) = F (¯ x, h(¯ x)), then g(¯ x) − h(¯ x) is rational number.

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In this paper we study the polynomials over Z. The following lemma allows us to transform the polynomials having infinitely many near solutions to polynomials over Z, which will significantly simplify the proofs. Here, f (¯ x, y) ∈ Q[g(¯ x, y)] means that there exist ci ∈ Q, 1 ≤ i ≤ m, such that f (¯ x, y) =

m 

ci (g(¯ x, y))i .

i=0

Lemma 3.3. Suppose that g(¯ x) ∈ Q[¯ x]. Let G(¯ x) = g(¯ x) − g(¯0) and c be the least common multiple of the denominators of the coefficients of G(¯ x). Let h(¯ x) = c · G(¯ x). Then Z[¯ x, y] ∩ Q[y − g(¯ x)] = Z[c · y − h(¯ x)]. Proof. First, Q[y − g(¯ x)] = Q[cy − h(¯ x)] ⊇ Z[cy − h(¯ x)], so it will suffice to prove that any f (¯ x, y) ∈ Z[¯ x, y] ∩ Q[cy − h(¯ x)] belongs to Z[cy − h(¯ x)]. This we do by induction on the y degree of f (¯ x, y). The degree 0 case is trivial. There exist di ∈ Q such that s f (¯ x, y) = i=0 di (cy − h(¯ x))i . Since h(¯0) = 0, substituting (¯ x, y) = (¯0, 0) shows that f (¯0, 0) = d0 , so d0 ∈ Z. The polynomial f (¯ x, y) − d0 ∈ Z[¯ x, y] is divisible by cy − h(¯ x) in Q[¯ x, y], so by Gauss’ Lemma, f (¯ x, y) −d0 = (cy −h(¯ x))F (¯ x, y) for some F (¯ x, y) ∈ Z[¯ x, y]. In addition, F (¯ x, y) ∈ Q[cy − h(¯ x)]. By induction hypothesis, F (¯ x, y) ∈ Z[cy − h(¯ x)]. Hence, f (¯ x, y) = (cy − h(¯ x))F (¯ x, y) + d0 ∈ Z[cy − h(¯ x)]. 2 Near solutions are closely related to the max–min values of polynomials. We now restate Theorem A in the previous section with our newly given definitions. Theorem A. (See [2].) Let F (X, Y ) ∈ Q[X, Y ] and A ⊆ Z be a set of positive UAD. If F (X, Y ) is not with polynomial growth on A, then F (X, Y ) has a near solution in Q[x]. We also have the following. Theorem 3.4. (See [8, Theorem 2.5].) If F (x, y) ∈ Q[x, y] does not have polynomial growth on a positive UAD set A, then for each a ∈ A, except a set of zero UAD, there exists a near solution g(x) of F (x, y) such that g(a) ∈ Z. Corollary 3.5. (See [8].) Let F (x, y) ∈ Q[x, y] and A ⊆ Z with positive UAD. If for each a ∈ A, F (a, y) = 0 is solvable in Z, then for each a ∈ A, except a set of zero UAD, there exists a g(x) ∈ Q[x] with F (x, g(x)) ≡ 0 and g(a) ∈ Z. Corollary 3.6. (See [8].) Let F (x, y) ∈ Q[x, y]. For every integer a, F (a, y) = 0 is solvable in Z if and only if for every integer a there exists a g(x) ∈ Q[x] with F (x, g(x)) ≡ 0 and g(a) ∈ Z. Theorem 3.7. (See [8, Theorem 2.6].) Let F (x, y) be a polynomial over Q. The following statements are equivalent:

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(i) SZ,F (T ) = o(T 1/2 ) as T → ∞. (ii) There is a positive integer B such that SZ,F (T ) ≤ B as T → ∞. (iii) There is a positive integer B, integers {a1 , . . . , an } with their absolute values less than or equal to B, and near solutions {g1 (x), . . . , gn (x)} over Q such that for every integer α there is an i, where 1 ≤ i ≤ n, F (x, gi (x)) = ai and gi (α) is in Z. (iv) There exist integers bi , di , primitive polynomials Hi (x) ∈ Z[x] for 1 ≤ i ≤ n, and G(x, y) ∈ Z[x, y] such that F (x, y) = (b1 y − H1 (x)){(b2 y − H2 (x))[(b3 y − H3 (x))(· · · (G(x, y) · · ·) + d3 ] + d2 } + d1 , where either di = 0 or Hi (x)/bi − H1 (x)/b1 = ci ∈ Q for 2 ≤ i ≤ n, and for every integer α there is an i such that Hi (α)/bi is an integer. From (iv) of this theorem we may observe that for each i, gi (x) = Hi (x)/bi is a near solution of F (x, y). 4. Exponential diophantine equations This section considers what may be called exponential diophantine equations with parameters. We first prove some lemmas to simplify the proof of our main theorem. m Lemma 4.1. Let F (x) = i=1 fi (x)bgi (x) ≡ 0, where fi (x) and gi (x) are in Z[x], b is a positive integer. There are at most finitely many integers a such that F (a) = 0. Proof. If b = 1, F (x) =

m 

fi (x)bgi (x) =

i=1

m 

fi (x) ≡ 0

i=1

is a polynomial over Z. There are at most finitely many integers a such that F (a) = 0. n Thus, we assume that b > 1. Let g(x) = i=0 axi , where n is an even integer greater than all the degrees of gi (x), and a is a positive integer greater than all gi (x) for 1 ≤ i ≤ m. We may multiply F (x) = 0 with the term bg(x) to obtain the equation F (x)bg(x) =

m 

fi (x)bgi (x)+g(x) = 0.

i=1

Clearly, every gi (x) + g(x) has even degree and all of its coefficients are positive. For any integer x, bg(x) = 0, we have F (x)bg(x) = 0 if and only if F (x) = 0. Thus, we may further assume that the degree of gi (x), 1 ≤ i ≤ m, is an even number greater than 0, and all the

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coefficients of each gi (x) are also positive. By doing so, we are only dealing with integers in our proof. Now, let gi (x) = Gi (x) + ci , where Gi (x) has no constant term and ci ∈ Z is positive, then fi (x)bgi (x) = (fi (x)bci )bGi (x) for each i, where fi (x)bci is a polynomial over Z. Hence, we may further assume that each gi (x) has no constant term. Finally, after we have done all the above simplifications, we also assume that gi (x) = gj (x) if 1 ≤ i = j ≤ m. For if gi (x) = gj (x), then fi (x)bgi (x) + fj (x)bgj (x) = (fi (x) + fj (x))bgi (x) . We may combine these two exponential terms into one. We now summarize our assumptions on the exponential polynomial F (x) as F (x) =

m 

fi (x)bgi (x) ,

i=1

where fi (x) and gi (x) are polynomials over Z, and b > 1 is a positive integer. For every i, 1 ≤ i ≤ m, gi (x) is a polynomial with even degree, all coefficients nonnegative, and without constant term. Also, gi (x) = gj (x) if i = j. Let a be an integer with |a| sufficiently large such that gi (a) > 0 for every i, gi (a) = gj (a) for 1 ≤ i = j ≤ m, and F (a) =

m 

fi (a)bgi (a) = 0.

i=1

We may rearrange the indices if necessary, and we assume that gm (a) > gi (a) for every 1 ≤ i < m. Since gm (x) − gi (x) ∈ Z[x] is not identically equal to 0, and has no constant term, for |a| sufficiently large we have that gm (a) − gi (a) − |a| ≥ 0. Thus, bgi (a)−gm (a)+|a| ≤ 1. For any integer a, either fm (a) = 0 or fm (a) = 0. The number of roots of fm (x) = 0 is less than or equal to deg(fm (x)). Now, assume that fm (a) = 0. Then, |fm (a)| ≥ 1. We have   m−1  m−1  m−1      gi (a)−gm (a)+|a|  fi (a)b |fi (a)| .  ≤ fi (a)bgi (a)−gm (a)+|a|  ≤   i=1

i=1

i=1

Since m m   gi (a) −gm (a)+|a| ( fi (a)b )b = fi (a)bgi (a)−gm (a)+|a| = 0, i=1

i=1

then m−1         gi (a)−gm (a)+|a|  fi (a)b   = fm (a)bgm (a)−gm (a)+|a|  .   i=1

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We obtain that m−1 

|fi (a)| ≥ |fm (a)b|a| | ≥ b|a| .

i=1

Thus, for every integer x with |x| sufficiently large and F (x) = 0, after rearranging the indices we will have that m−1 

|fi (x)| ≥ b|x| .

i=1

However, the left-hand side of this inequality grows polynomially and the right-hand side grows exponentially as |x| approaches infinity. But there are only finitely many integers m−1 a such that i=1 |fi (a)| ≥ b|a| . Therefore, there are at most finitely many integers a such that F (a) = 0. 2 m Lemma 4.2. Let F (x, y) = i=1 fi (x, y)bgi (x,y) ≡ 0, where fi (x, y) and gi (x, y) are polynomials over Z, b is a positive integer. There are at most finitely many integers c such that F (x, c) ≡ 0. Proof. We may multiply F (x, y) = 0 with the term b−g1 (x,y) to obtain the equation F (x, y)b−g1 (x,y) = 0. For any integer c, we have F (c, y)b−g1 (c,y) ≡ 0 if and only if F (c, y) ≡ 0. Thus, without loss of generality, we may assume that g1 (x, y) ≡ 0. As we did in the proof of Lemma 4.1, we may also assume that for each i, gi (x, y) has no constant term, and gi (x, y) = gj (x, y) if 1 ≤ i = j ≤ m. We first prove that there is an integer α such that F (α, y) ≡ 0. For each i, let fi (x, y) =

mi 

fi,j (x)y j ,

j=0

where fi,j (x) ∈ Z[x], and gi (x, y) =

ni 

gi,j (x)y j ,

j=0

where gi,j (x) ∈ Z[x], respectively. Let T be the set of indices such that k ∈ T if and only if gk (x, y) = gk (x) ∈ Z[x]. Note that 1 ∈ T . For each i ∈ / T there is a j > 0 such that gi,j (x) ≡ 0. Thus, there is a finite set S1 such that for each a ∈ Z − S1 and each i ∈ / T, gi (a, y) ∈ (Z[y] − Z). Let H(x, y) =

 i∈T

fi (x, y)bgi (x)

S.P. Tung / Journal of Number Theory 157 (2015) 397–423

=

405

mi  ( fi,j (x)y j )bgi (x) i∈T j=0

=

M   ( fi,j (x)bgi (x) )y j j=0 i∈T

=

M 

Hj (x)y j ,

j=0

 where Hj (x) = i∈T fi,j (x)bgi (x) , and M is the maximum of mi for 1 ≤ i ≤ m. Since f1 (x, y) ≡ 0, there is a k, 0 ≤ k ≤ m1 ≤ M , such that f1,k (x) ≡ 0. Thus Hk (x) ≡ 0, and Hk (x) = 0 has only finitely many solutions by Lemma 4.1. Let S2 be the set of solutions of Hk (x) = 0. For any integer α ∈ Z − S2 ,

H(α, y) =

M 

Hj (α)y j ≡ 0.

j=0

Now, let α ∈ Z − (S1 ∪ S2 ). Then, F (α, y) =

 i∈T

fi (α, y)bgi (α) +



fi (α, y)bgi (α,y) = H(α, y) +

i∈T /



fi (α, y)bgi (α,y) .

i∈T /

For each i ∈ / T , gi (α, y) ∈ (Z[y] − Z). Thus, F (α, y) ≡ 0. It is assumed that F (x, c) ≡ 0. Hence, F (α, c) = 0 for any integer α ∈ Z − (S1 ∪ S2 ). This implies that c is a solution of the exponential equation

F (α, y) =

m 

fi (α, y)bgi (α,y) = 0.

i=1

There are only finitely many such solutions by Lemma 4.1. Therefore, there are at most finitely many integers c such that F (x, c) ≡ 0. 2 m Lemma 4.3. Let F (x, y) = i=1 fi (x, y)bgi (x,y) ≡ 0, where fi (x, y) and gi (x, y) are polynomials over Z, b is a positive integer. There are at most finitely many h(x) ∈ Q[x] such that h(a) ∈ Z for some integer a and F (x, h(x)) ≡ 0. Proof. If b = 1, then F (x, y) ∈ Z[x, y]. There are at most finitely many h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0. Hence, we assume that b > 1. Similar to the arguments of Lemma 4.1, we may assume that

F (x, y) =

m  i=1

fi (x, y)bgi (x,y) ,

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where fi (x, y) and gi (x, y) are polynomials over Z, and b > 1 is a positive integer. Each gi (x, y) is a polynomial with no constant term, and gi (x, y) = gj (x, y) if i = j. m For any exponential polynomial G(x) = i=1 fi (x)bgi (x) over Q with the assumption that there is a k, 1 ≤ k ≤ m, such that fk (x) ≡ 0 and gk (x) ≡ c for a c ∈ Q. If G(x) ≡ 0, then there must exist a j, 1 ≤ j = k ≤ m, such that gj (x) − gk (x) is a rational number. This implies that if F (x, h(x)) ≡ 0, then either f1 (x, h(x)) ≡ 0, or there exists a j, 1 < j ≤ m, such that g1 (x, h(x)) − gj (x, h(x)) = c for a rational number c, i.e., h(x) is a near solution of g1 (x, y) − gj (x, y). Now, let h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0. There are only finitely many h(x) ∈ Q[x] such that f1 (x, h(x)) ≡ 0. Thus we assume that f1 (x, h(x)) ≡ 0. Then h(x) is a near solution of g1 (x, y) − gj (x, y) for some j, 1 < j ≤ m. If g1 (x, y) − gj (x, y) has only a finite number of near solutions, then we are finished. But suppose that g1 (x, y) − gj (x, y) has infinitely many near solutions for a j where 1 < j ≤ m. Then there exists a unique g(x) ∈ Q[x], with its constant term equal to zero, and ci ∈ Q such that g1 (x, y) − gj (x, y) =

p 

ci (y − g(x))i

i=0

by Theorem 3.1. Moreover, h(x) is a near solution of g1 (x, y) − gj (x, y) if and only if there exists a c ∈ Q such that h(x) = g(x) + c by Theorem 3.1 again. Suppose that there is another index k such that k = 1 and k = j. Following the same argument we only need to handle the case that there is another index l = k and a unique r(x) ∈ Q[x], with its constant term equal to zero, and di ∈ Q such that gk (x, y) − gl (x, y) =

q 

di (y − r(x))i .

i=0

Also if F (x, h(x)) ≡ 0, then h(x) = r(x) + d for a d ∈ Q. Since h(x) = g(x) + c = r(x) + d, where g(x), r(x) have no constant terms and c, d are rational numbers, we obtain that g(x) = r(x). Notice that we may have that l = 1 or l = j. If l = 1, then g1 (x, y) − gk (x, y) =

q 

−di (y − g(x))i .

i=0

Suppose that l = j. Then, g1 (x, y) − gk (x, y) = (g1 (x, y) − gj (x, y)) − (gk (x, y) − gj (x, y)) =

p  j=0

ci (y − g(x))i −

q  i=0

di (y − g(x))i

S.P. Tung / Journal of Number Theory 157 (2015) 397–423

=

r 

407

(ci − di )(y − g(x))i ,

j=0

where r is the maximum of p and q. This implies that there is a g(x) ∈ Q[x], with its constant term equal to zero, such that for any two indices 1 ≤ i = j ≤ m, if gi (x, y) − gj (x, y) has near solutions, then there exist ci,j,k ∈ Q such that 

mi,j,k

gi (x, y) − gj (x, y) =

ci,j,k (y − g(x))k .

k=0

Therefore, there is a partition P = {P1 , P2 , . . . , Pn } of the set of exponents {g1 (x, y), g2 (x, y), . . . , gm (x, y)} appeared in F (x, y) such that if gi (x, y) and gj (x, y) are in the same set Pk ∈ P if and only if gi (x, y) − gj (x, y) has near solutions, and if and only if there exist ci,j,k ∈ Q such that 

mi,j,k

gi (x, y) − gj (x, y) =

ci,j,k (y − g(x))k .

k=0

Let D = Den(g(x)), and G(x) = D · g(x) ∈ Z[x]. From Lemma 3.3, we may rewrite that 

mi,j,k

gi (x, y) − gj (x, y) =

di,j,k (Dy − G(x))k ,

k=0

where di,j,k ∈ Z. Choose one pj (x, y) from each Pj for 1 ≤ j ≤ n. We then have that if s = t, then for any c ∈ Q ps (x, g(x) + c) − pt (x, g(x) + c) ∈ (Q[x] − Q). Then we may rewrite F (x, y) such that F (x, y) =

m 

fi (x, y)bgi (x,y)

i=1

=

n  

fi (x, y)bgi (x,y)

j=1 i∈Pj

=

n   ( fi (x, y)bgi (x,y)−pj (x,y) )bpj (x,y) j=1 i∈Pj

=

n   mi,j,k k ( fi (x, y)b k=0 di,j,k (Dy−G(x)) )bpj (x,y) , j=1 i∈Pj

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where di,j,k ∈ Z. Let h(x) ∈ Q[x], such that F (x, h(x)) ≡ 0. With the assumption that h(a) ∈ Z for some integer a, we then have that h(x) = g(x) + c/D for some c ∈ Z. Then F (x, g(x) + z/D) =

n   mi,j,k k ( fi (x, g(x) + z/D)b k=0 di,j,k (z) )bpj (x,g(x)+z/D) ≡ 0. j=1 i∈Pj

Let Hj (x, z) =



mi,j,k

fi (x, g(x) + z/D)b

k=0

di,j,k (z)k

,

i∈Pj

then F (x, g(x) + z/D) =

n 

Hj (x, z)bpj (x,g(x)+z/D)

j=1

and F (x, h(x)) = F (x, g(x) + c/D) =

n 

Hj (x, c)bpj (x,g(x)+c/D) ≡ 0.

j=1

Since for every 1 ≤ i = j ≤ n, pj (h(x)) − pi (h(x)) ∈ Q[x] − Q, this implies that Hj (x, c) ≡ 0 for every j, 1 ≤ j ≤ n. Let r be a positive integer greater than all degy (fi (x, y)), then for each i, Dr · fi (x, g(x) + z/D) ∈ Z[x, z]. From Lemma 4.2, there are only finitely many integers c such that Dr · Hj (x, c) ≡ 0. Thus there are only finitely many h(x) = g(x) + c/D such that h(a) ∈ Z for some integer a and F (x, h(x)) ≡ 0. For any j = 1, no matter whether g1 (x) − gj (x) has finitely many or infinitely many near solutions, there are only finitely many near solutions h(x) of g1 (x) − gj (x) such that h(a) ∈ Z for some integer a and F (x, h(x)) ≡ 0. There are finitely many possible choices for j. Therefore, there are only finitely many h(x) ∈ Q[x] such that h(a) ∈ Z for some integer a and F (x, h(x)) ≡ 0. 2 To simplify the proof of the main theorem, we show a special case first. m Lemma 4.4. Let F (x, y) = i=1 fi (x, y)bGi (y) , where fi (x, y) and Gi (y) are polynomials over Z, and b is a positive integer. Let A ⊂ Z with positive UAD. If for every α ∈ A there is a y in Z such that F (α, y) = 0, then there is a polynomial h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0.

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m m Gi (y) Proof. If b = 1, F (x, y) = = i=1 fi (x, y)b i=1 fi (x, y) which is a polynomial over Z, then by Corollary 3.5 this lemma is true. Similar to the arguments of Lemma 4.1, m we assume that F (x, y) = i=1 fi (x, y)bGi (y) , where fi (x, y) and Gi (y) are polynomials over Z, and b > 1 is a positive integer. For every i, 1 ≤ i ≤ m, Gi (y) is a polynomial with even degree, positive leading coefficient, and without constant term. In addition, Gi (y) = Gj (y) if 1 ≤ i = j ≤ m. We assume that there is no β in Q such that F (x, β) ≡ 0; otherwise, simply let h(x) = β and we obtain the conclusion. m Now, assume that F (x, y) = i=1 fi (x, y)bGi (y) has at least one exponential term. For every a ∈ A there is an integer y such that F (a, y) = 0. Let Ap and An be the subsets of A such that a ∈ Ap if F (a, y) = 0 has a positive integer solution, and c ∈ An if F (c, y) = 0 has a negative integer solution, respectively. Either Ap or An must have positive UAD. We may rearrange the terms of F (x, y) as follows: 1. Ap is of positive UAD. Then for each i, 1 ≤ i < m, Gi+1 (y) − Gi (y) is with positive leading coefficient. 2. An is of positive UAD. Then for each i, 1 ≤ i < m, Gi+1 (y) − Gi (y) is with positive leading coefficient if deg(Gi+1 (y)−Gi (y)) is even, and Gi+1 (y)−Gi (y) is with negative leading coefficient if deg(Gi+1 (y) − Gi (y)) is odd. In doing so, for every integer c, where |c| is sufficiently large, G1 (c), G2 (c), . . . , Gm (c) is an increasing sequence. (In fact, in our proof we only require that Gi (c) < Gm−1 (c) < Gm (c) for 1 ≤ i < m − 1.) We then have that bGj (c)−Gi (c) > 1 for 1 ≤ i < j ≤ m. The proofs for these two cases are almost the same. We demonstrate only the case that Ap is of positive UAD. Let B ⊆ Ap be the set such that α ∈ B if and only if there is a positive y in Z such that F (α, y) = 0 and fm (α, y) = 0. Thus, m−1 

fi (α, y)bGi (y) = 0.

i=1

¯ ¯ p − B) > 0. We will show that this leads to a Suppose that d(B) = 0, then d(A contradiction. Let a ∈ Ap − B and c be a positive integer such that

F (a, c) =

m  i=1

fi (a, c)bGi (c) = 0.

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Then,   m−1       Gi (c)  fi (a, c)b   = fm (a, c)bGm (c)  .   i=1

Since a ∈ Ap − B, fm (a, c) = 0. Hence, |fm (a, c)| ≥ 1. For any integer c, equation ¯ p − B) > 0, as |a| grows larger F (x, c) = 0 has only finitely integer solutions. Since d(A and larger so does |c|. Thus,     m−1   m−1     Gi (c)  Gi (c)−Gm−1 (c) Gm−1 (c)  fi (a, c)b fi (a, c)b )·b   = (      i=1 i=1   m−1    fi (a, c)bGi (c)−Gm−1 (c)  · bGm−1 (c) =   i=1

m−1 

≤(

|fi (a, c)bGi (c)−Gm−1 (c) |) · bGm−1 (c)

i=1 m−1 

<(

|fi (a, c)|) · bGm−1 (c) .

i=1

  m−1 Then, ( i=1 |fi (a, c)|) · bGm−1 (c) ≥ fm (a, c)bGm (c) . Hence, m−1 

|fi (a, c)| ≥ |fm (a, c)|bGm (c)−Gm−1 (c) ≥ b(Gm −Gm−1 )(c) .

i=1

Now, let a be an integer in Z such that |a| > fi (x, y) for each i, and |a| > (2(Gm − Gm−1 )(y))2 . Then (Gm − Gm−1 )(c) > |c|1/2 if |c| > |a|. Let fi (x, y) =

mi 

ai,j (y)xj

j=0

and let r be a positive integer larger than each degy (fi (x, y)) for 1 ≤ i ≤ m. Then for any y in Z    mi  mi mi     |fi (a, y)| =  ai,j (y)aj  ≤ |ai,j (y)||a|j ≤ |a||y|r |a|j < |y|r |a|mi +2 .  j=0  j=0 j=0 Let E be a fixed positive integer greater than m and each mi , 1 ≤ i < m, but less than |a|. For any c in Z, |c| |a| r

E+3

> E · |c| |a| r

E+2

≥

m−1  i=1

|fi (a, c)| ≥ b(Gm −Gm−1 )(c) .

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Suppose that a is sufficiently large and |c| > a, we then have that b|c|

1/2

< b(Gm −Gm−1 )(c) <

m−1 

|fi (a, c)| < |c|r aE+3 < |c|r+E+3 .

i=1

Since r and E are constants and b > 1, this is impossible when a, and hence |c|, are sufficiently large. This implies that for any a ∈ Z sufficiently large and F (a, c) = 0 for some c, we should have that |c| < a. This then implies that |fi (a, c)| < |c|r |a|E+2 < |a|E+2+r for each i. From F (a, c) = 0 and |fm (a, c)| = 0, we obtain that b|c|

1/2

< b(Gm −Gm−1 )(|c|) < |a|r+E+3 .

We then have that |c| < ((r + E + 3) logb (|a|))2 . This means that if F (a, c) = 0 and |a| is large enough, then |c| is much smaller than |a|. For every β in Z, it is assumed that F (x, β) ≡ 0, then there are at most M = max{mi : 1 ≤ i ≤ m} many α’s in Z such that

F (α, β) =

m 

fi (α, β)bGi (β) = 0.

i=1

Let S ∈ Ap be an integer large enough to ensure the condition that if F (S, β) = 0 for some β, then |β| < ((r + E + 3) logb (|S|))2 ¯ p ) = δ. We also may require an integer T large enough such as derived above. Let d(A that |Ap (T )|/2T > δ/2, S/T < δ/4 and T > (2/δ)((r + E + 3) logb (T ))2 · M.

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Then |{x : |x| ≤ T, ∃y F (x, y) = 0}| 2T Ap (S) |{x : S < |x| ≤ T, ∃y F (x, y) = 0}| + = 2T 2T

|Ap (T )|/2T =

< δ/4 +

((r + E + 3) logb (T ))2 · M 2T

< δ/4 +

((r + E + 3) logb (T ))2 · M (4/δ)((r + E + 3) logb (T ))2 · M

= δ/4 + δ/4 = δ/2. ¯ p − B) = 0. Therefore, d(B) ¯ This contradiction shows that d(A > 0. ¯ If d(B) > 0, then there is an h(x) ∈ Q[x] such that fm (x, h(x)) ≡ 0 by Corollary 3.5. m−1 If i=1 fi (x, h(x))bGi (h(x)) ≡ 0, let B1 be the set of roots of the equation m−1 

fi (x, h(x))bGi (h(x)) = 0.

i=1

B1 is a finite set by Lemma 4.1. Moreover, since fm (x, h(x)) ≡ 0, we may write that fm (x, y) = (y − h(x))r · Hm (x, y) ¯ −B1 ) > 0, there is a y in Z such that fm (α, y) = 0 with r > 0. For every α ∈ B −B1 , d(B and m−1 

fi (α, y)bGi (y) = 0.

i=1

Since α is not in B1 , m−1 

fi (α, h(α))bGi (h(α)) = 0.

i=1

This implies that y = h(α). Therefore, Hm (α, y) = 0. We then obtain that if α ∈ B − B1 , then there is a y in Z such that Hm (α, y) = 0 and m−1  i=1

fi (α, y)bGi (y) = 0.

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Since degy (Hm (x, y)) < degy (fm (x, y)), by induction on degy (fm (x, y)), we obtain that there exists a polynomial h(x) ∈ Q[x] such that fm (x, h(x)) ≡ 0 and m−1 

fi (x, h(x))bGi (h(x)) ≡ 0.

i=1

Therefore F (x, h(x)) =

m−1 

fi (x, h(x))bGi (h(x)) + fm (x, h(x))bGm (h(x)) ≡ 0.

2

i=1

Remark. From the proof of Lemma 4.4 we may observe the following fact about the solutions of F (x, y) = 0. There is a set B with zero UAD such that if α ∈ (A − B) and F (α, y) = 0 for an integer y, then there is an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(α) = y. This will be stated explicitly in the next theorem. m Theorem 4.5. Let F (x, y) = k=1 fk (x, y)bGk (x,y) , where fk (x, y) and Gk (x, y) are in Z[x, y], b is a positive integer, and Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z for each ¯ 1 ≤ i = j ≤ m. Let A ⊂ Z with d(A) > 0, and for every α ∈ A there is a y in Z such that ¯ F (α, y) = 0. Then there is a set B with d(B) = 0 such that for every α ∈ A − B there exists an h(x) ∈ Q[x] with F (x, h(x)) ≡ 0 and h(α) ∈ Z. Furthermore, there is a finite set S of polynomials over Q such that for every α ∈ A − B and every y ∈ Z, F (α, y) = 0 if and only if there exists an h(x) ∈ S such that F (x, h(x)) ≡ 0 and y = h(α). Proof. As we reasoned above, we may assume that b > 1, Gi (x, y) has no constant term, and Gi (x, y) = Gj (x, y) for 1 ≤ i = j ≤ m. Let i, j be two fixed indices such that Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z. There are two possible cases for the polynomial Gi (x, y) − Gj (x, y): Case 1. Gi (x, y) − Gj (x, y) is with polynomial growth on A. Case 2. Gi (x, y) − Gj (x, y) is not with polynomial growth on A. We claim that in both cases there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0. We prove Case 1 by induction on the number of exponential terms of F (x, y). The case that F (x, y) has only two terms is proved in [8]. Let F (x, y) =

m 

fk (x, y)bGk (x,y)

k=1

with m > 1. Because Gi (x, y) − Gj (x, y) is of polynomial growth on A, this means that SA,(Gi −Gj ) = o(T 1/2 ). For any  > 0 we may choose T ∈ A to be sufficiently large and y

S.P. Tung / Journal of Number Theory 157 (2015) 397–423

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to be an integer such that F (T, y) = 0 and |Gi (T, y) − Gj (T, y)| > (T 2 − ). Without loss of generality, we assume that Gi (T, y) > Gj (T, y). Let K be the set of indices such that k ∈ K if and only if 1

Gj (T, y) ≤ Gk (T, y) ≤ Gi (T, y). We may rearrange the indices in K such that the sequence Gk (T, y) : k ∈ K is increasing. There exist two adjacent Gp (T, y) and Gq (T, y) in the sequence such that Gp (T, y) − Gq (T, y) > T 2 −2 . 1

Thus for every k, 1 ≤ k ≤ m, either Gk (T, y) ≤ Gq (T, y), or Gp (T, y) ≤ Gk (T, y). Also, Gj (T, y) ≤ Gq (T, y) < Gp (T, y) ≤ Gi (T, y). Let IM and Im be the sets of indices such that l ∈ IM if Gl (T, y) > Gp (T, y), and l ∈ Im if Gl (T, y) < Gq (T, y), respectively. Since F (T, y) =

m 

fk (T, y)bGk (T,y) = 0,

k=1

we have that           Gl (T,y) Gq (T,y)  Gl (T,y) Gp (T,y)  fl (T, y)b + fq (T, y)b fl (T, y)b + fp (T, y)b  = .     l∈Im

l∈IM

Now, let Fm (x, y) = fq (x, y) +



fl (x, y)bGl (x,y)−Gq (x,y)

l∈Im

and FM (x, y) = fp (x, y) +



fl (x, y)bGl (x,y)−Gp (x,y) .

l∈IM

Thus,         Fm (T, y)bGq (T,y)  = FM (T, y)bGp (T,y)  . Note that, Fm (T, y) = 0 if and only if FM (T, y) = 0.

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Assume that FM (T, y) = 0. Since for each i ∈ IM we have Gi (T, y) > Gp (T, y), b is an integer. Then, |FM (T, y)| ≥ 1. For each i ∈ Im , Gi (T, y) − Gq (T, y) < 0, we then have that Gi (T,y)−Gp (T,y)

0 < bGi (T,y)−Gq (T,y) < 1 and |fi (T, y)bGi (T,y)−Gq (T,y) | ≤ |fi (T, y)|. Therefore       Gl (T,y)−Gq (T,y)  |Fm (T, y)| ≤ |fq (T, y)| +  fl (T, y)b |fi (T, y)|.  ≤ |fq (T, y)| +   l∈Im

l∈Im

Hence bGp (T,y) ≤ (|fq (T, y)| +



|fi (T, y)|)bGq (T,y) ,

i∈Im

or bGp (T,y)−Gq (T,y) ≤ |fq (T, y)| +



|fi (T, y)|.

i∈Im

Since Gp (T, y) − Gq (T, y) > T 2 −2 , we obtain that 1

1 −2

bT 2

≤ |fq (T, y)| +



|fi (T, y))|.

i∈Im

For any integer y, the left-hand side of the inequality grows faster than the right-hand side as T approaches infinity. To satisfy this inequality, |y| needs to be much larger than 1 −3

T as T is sufficiently large. We can take T to be large enough so that |y| > bT 2 1 −4

(Gp − Gq )(T, y) = Gp (T, y) − Gq (T, y) > bT 2

and

.

Thus, from the fact that Gp (T, y) − Gq (T, y) > T 2 − , we will get 1

1 −4

(Gp − Gq )(T, y) = Gp (T, y) − Gq (T, y) > bT 2

.

This means Gp (T, y) − Gq (T, y) needs to become larger and larger, as does the value of |y|; but this is impossible. We call this “ascending argument.” Thus, there is an upper bound U such that if F (T, y) = 0 and FM (T, y) = 0 for some integer y, then T < U . There are only finitely many possible combinations for the indices. We then obtain that

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if Gi (x, y) − Gj (x, y) is of polynomial growth on A, then there exists a set of indices L with |L| < m and a positive UAD set C ⊂ Z such that for every a ∈ C there is a y, 

fi (a, y)bGi (a,y) = 0,

and



fi (a, y)bGi (a,y) = 0.

i∈L /

i∈L

From our induction hypothesis, there exist two non-empty but finite sets of polynomials SM and Sm such that h(x) ∈ SM if and only if 

fi (x, h(x))bgi (x,h(x)) ≡ 0

i∈L

and g(x) ∈ Sm if and only if 

fi (x, g(x))bgi (x,g(x)) ≡ 0,

i∈L /

respectively. Also, from our induction hypothesis, there exist two sets BM , Bm with zero UAD such that for every a ∈ C − (BM ∪ Bm ) and an integer y, where  i∈L

fi (a, y)bGi (a,y) = 0 (then



fi (a, y)bGi (a,y) = 0).

i∈L /

Then there exist h1 (x) ∈ SM and h2 (x) ∈ Sm , respectively, such that h1 (a) = y = h2 (a). For any two polynomials f (x) ≡ g(x) there are only finitely numbers a such that f (a) = g(a). We may delete the polynomials h(x) from SM and Sm , respectively, if h(x) is not in SM ∩ Sm . We then obtain that, except with a zero UAD set B, for every a ∈ C − B there is an h(x) such that FM (x, h(x)) ≡ 0, Fm (x, h(x)) ≡ 0, and h(a) is an integer. Note that if FM (x, h(x)) ≡ 0 and Fm (x, h(x)) ≡ 0, then F (x, h(x)) ≡ 0. We also obtain that for every α ∈ C − B and every y ∈ Z, F (α, y) = 0 if and only if there exists an h(x) ∈ S, F (x, h(x)) ≡ 0, and y = h(α). This finishes the proof of our claim for Case 1. Now, we prove the claim for Case 2. We assume that Gi (x, y) − Gj (x, y) is not of polynomial growth on A. Without loss of generality, we may take i = 2 and j = 1. Then G2 (x, y) − G1 (x, y) must have near solutions by Theorem 3.4. There are two possible subcases. Case 2.1. G2 (x, y) − G1 (x, y) has finitely many near solutions. Case 2.2. G2 (x, y) − G1 (x, y) has infinitely many near solutions. We prove Case 2.1 by induction on the number of near solutions of G2 (x, y) −G1 (x, y). Let h(x) be a near solution of G2 (x, y) − G1 (x, y), and G2 (x, h(x)) − G1 (x, h(x)) = a

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for an a in Q. We may write the polynomial G2 (x, y) − G1 (x, y) = (y − h(x))q(x, y) + a, where q(x, y) and h(x) are polynomials over Q. We want to show that if F (x, h(x)) ≡ 0, then G2 (x, y) − G1 (x, y) must have another near solution h1 (x) = h(x). Suppose that F (x, h(x)) ≡ 0. Then equation F (x, h(x)) = 0 has only finitely many solutions in Z by ¯ − B) > 0. Lemma 4.1. Let B be the set of solutions of F (x, h(x)) = 0. Then d(A Suppose that q(x, y) is of polynomial growth on A −B; then there is a positive constant C such that miny∈Z |q(T, y)| > T C for infinitely many T in A − B. We may require that this |T | is greater than |2a|1/C and all the roots of F (x, h(x)) = 0. If F (T, β) = 0, then β = h(T ). We obtain that |(G2 − G1 )(T, β)| = |(β − h(T ))q(T, β) + a| > (1/(d + 1))T C , where d = Den(h(x)) for T sufficiently large. With the same “ascending argument” as above we may prove the claim. Thus, we assume that q(x, y) is not of polynomial growth on A − B. This implies that q(x, y) has a near solution h1 (x). Let q(x, y) = (y − h1 (x))q1 (x, y) + c for a q1 (x, y) in Q[x, y] and a c in Q. Hence G2 (x, y) − G1 (x, y) = (y − h(x))[(y − h1 (x))q1 (x, y) + c] + a. We want to show that G2 (x, y) − G1 (x, y) has another near solution. For every α ∈ A − B let bα ∈ Z such that F (α, bα ) = 0. Let 0 < C < 1/2 be a constant. If there are infinitely many (α, bα ) ∈ Z2 such that (G2 − G1 )(α, bα ) > |α|C , then with the “ascending argument” we may obtain our conclusion. Hence, we assume that there are only finitely many α ∈ A − B, (G2 − G1 )(α, bα ) > |α|C . Then except a set with zero UAD, for every α ∈ A − B G2 (α, bα ) − G1 (α, bα ) = (bα − h(α))[(bα − h1 (α))q1 (α, bα ) + c] + a < |α|C . If there is a positive UAD set D ⊆ (A − B) such that for every α ∈ D, q(α, bα ) = (bα − h1 (α))q1 (α, bα ) + c = 0, then there is r(x) ∈ Q[x] such that q(x, r(x)) ≡ 0 by Corollary 3.5. Then

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G2 (x, r(x)) − G1 (x, r(x)) = (r(x) − h(x))q(x, r(x)) + a = a, and r(x) is another near solution of G2 (x, y) − G1 (x, y). Thus, without loss of generality, we may assume that, except a set of zero UAD, for every α ∈ A − B, (bα − h1 (α))q1 (α, bα ) + c = 0. There are two possible cases: (1) There is a set D ⊆ A − B with positive UAD such that q1 (α, bα ) = 0 if α ∈ D. By Corollary 3.6, there exists a finite set of polynomials gk , 1 ≤ k ≤ m such that q1 (x, gk (x)) ≡ 0 and for every α ∈ D, except for a set of zero UAD, there is gk (x) such that bα = gk (α). We then obtain that G2 (α, bα ) − G1 (α, bα ) = G2 (α, gk (α)) − G1 (α, gk (α)) = (gk (α) − h(α))[(gk (α) − h1 (α))q1 (α, gk (α)) + c] + a = (gk (α) − h(α))[c] + a < |α|C as α approaches infinity. This implies that gk (x) − h(x) = dk is a constant. Thus G2 (x, gk (x)) − G1 (x, gk (x)) = (gk (x) − h(x))[(gk (x) − h1 (x))q1 (x, gk (x)) + c] + a = dk · c + a is a constant in Q. This means that gk (x) is a near solution of G2 (x, y) − G1 (x, y). (2) There is a set E ⊆ A − B with positive UAD such that q1 (α, bα ) = 0 if α ∈ E. Let α ∈ E, then α ∈ / B. Thus, bα = h(α). Unless h(x) − h1 (x) is a constant, h(x) − h1 (x) = O(x). Then for any integer y, either y − h(x) = O(x) or y − h1 (x) = O(x) as x approaches infinity. From the fact that G2 (α, bα ) − G1 (α, bα ) = (bα − h(α))[(bα − h1 (α))q1 (α, bα ) + c] + a < |α|C for a C < 1/2, we obtain that h(x) − h1 (x) = d is a constant in Q. Then G2 (x, h1 (x)) − G1 (x, h1 (x)) = (h1 (x) − h(x))[(h1 (x) − h1 (x))q1 (x, gk (x)) + c] + a = d · c + a. Thus h1 (x) is a near solution of G2 (x, y) − G1 (x, y) too. Hence if F (x, h(x)) ≡ 0, then G2 (x, y) − G1 (x, y) must have another near solution h1 (x).

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We may write the polynomial G2 (x, y) − G1 (x, y) in a nested parenthesis form of near solutions, as given in Theorem 3.7. If F (x, h1 (x)) ≡ 0, then we may repeat the arguments above to show that G2 (x, y) −G1 (x, y) must still have another near solution. It is assumed that G2 (x, y) −G1 (x, y) has only finite number of near solutions. By induction, we obtain that G2 (x, y) − G1 (x, y) has a near solution h(x) and F (x, h(x)) ≡ 0. From this proof we also can see that, except with a finite set, for all (a, c) ∈ Z2 , F (a, c) = 0 implies that c = h(a), where h(x) is a near solution of G2 (x, y) − G1 (x, y) and F (x, h(x)) ≡ 0. Next, we prove Case 2.2. That is, G2 (x, y)−G1 (x, y) has infinitely many near solutions. Then there is a unique h(x) ∈ Q[x], with its constant term equal to zero, such that G2 (x, y) − G1 (x, y) =

m2 

c2,j (y − h(x))j ,

j=0

where c2,j ∈ Q. We then consider polynomial G3 (x, y) − G1 (x, y), and we will show that we only need to consider the situation that G3 (x, y) − G1 (x, y) =

m3 

c3,j (y − h(x))j ,

j=0

where c3,j ∈ Q. From our assumption, we have that G3 (x, y) − G1 (x, y) ∈ Z[x, y]. If G3 (x, y) − G1 (x, y) has polynomial growth on A (Case 1), or it does not have polynomial growth but has only a finite number of near solutions (Case 2.1), then we may apply the proofs above to obtain our conclusion. Thus we may also assume that G3 (x, y) − G1 (x, y) has infinitely many near solutions. Again, there is a unique g(x) ∈ Q[x], with its constant term equal to zero, such that G3 (x, y) − G1 (x, y) =

m3 

c3,j (y − g(x))j .

j=0

If g(x) = h(x), then g(x) −h(x) ∈ Q[x] −Q, then for any integer y either y −g(x) = O(x) or y − h(x) = O(x) as x approaches infinity. This then implies that there exists a set B with positive UAD such that either G3 (x, y) − G1 (x, y) = O(x) or G2 (x, y) − G1 (x, y) = O(x) as x approaches infinity. This implies that either G2 (x, y) − G1 (x, y) or G3 (x, y) − G1 (x, y) is of polynomial growth on B. With the same “ascending argument” as in the proof of Case 1 above we may obtain the conclusion. Therefore, we only need to demonstrate the case that g(x) = h(x). We may repeat the same arguments for each Gk (x, y) −G1 (x, y), 4 ≤ k ≤ m. Thus, we are finished except for the case where there is an h(x) such that for each i, 2 ≤ k ≤ m, Gk (x, y) − G1 (x, y) =

mk  j=0

ck,j (y − h(x))j ,

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where ck,j ∈ Q. Let D = Den(h(x)) and H(x) = D · h(x) ∈ Z[x]. Since for each k, 2 ≤ k ≤ m, Gk (x, y) − G1 (x, y) is a polynomial over Z, we may rewrite it as Gk (x, y) − G1 (x, y) =

mk 

dk,j (Dy − H(x))j ,

j=0

where dk,j ∈ Z by Lemma 3.3. We then have that F (x, y) =

m 

fk (x, y)bGk (x,y) = (

k=1

m 

fk (x, y)bGk (x,y)−G1 (x,y) ) · bG1 (x,y) .

k=1

Let f (x, y) =

m 

fk (x, y)bGk (x,y)−G1 (x,y) =

k=1

m 

mk

fk (x, y)b

j=0

dk,j (Dy−H(x))j

.

k=1

We have that for any (a, b) ∈ Z2 , F (a, b) = 0 if and only if f (a, b) = 0. Furthermore, for any polynomial p(x), F (x, p(x)) ≡ 0 if and only if f (x, p(x)) ≡ 0. Then y can be substituted by h(x) + z/D and we obtain for every x ∈ A, there is a z in Z such that f (x, h(x) + z/D) =

m 

mk

fk (x, h(x) + z/D)b

j=0

dk,j (z)j

= 0.

k=1

Let n be an integer larger than degy (fk (x, y)) for each k, 1 ≤ k ≤ m. Then Dn ·f (x, h(x)+ z/D) is an exponential polynomial over Z. From Lemma 4.4, we obtain that there is a K(x) ∈ Q[x] such that D · n

m 

fk (x, K(x)/D)b

mk

j=0

dk,j (K(x))j

≡ 0.

k=1

Hence f (x, h(x) + K(x)/D) ≡ 0 and F (x, h(x) + K(x)/D) ≡ 0. This completes the proof of our claim, i.e., that there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0. Let R be the set of polynomials h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(a) ∈ Z for some a ∈ Z. From Lemma 4.3, R is a finite set. Let R = {h1 (x), . . . , hn (x)} and ¯ − ∪i Si ) > 0, then with the same arguments there Si = {x ∈ Z | hi (x) ∈ Z}. If d(A must exist an h(x) over Q such that F (x, h(x)) ≡ 0. Moreover, from the proof above, no matter whether we are in Case 1 or Case 2, there exists an integer a ∈ (A − ∪i Si ) such that h(a) is an integer. Since for every i, 1 ≤ i ≤ n, hi (a) is not an integer, h(x) = hi (x). From this contradiction, (A − ∪i Si ) = B has zero UAD. Hence, for every α ∈ A − B there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0, and h(α) ∈ Z. Clearly, for any integer a if y = h(a) is an integer and F (x, h(x)) ≡ 0, then F (a, y) = 0. Therefore, for every α ∈ A − B and every y ∈ Z, F (α, y) = 0 if and only if there exists an h(x) ∈ S, F (x, h(x)) ≡ 0, and y = h(α). 2

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As explained in [8], we need the requirement that “Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z for each 1 ≤ i = j ≤ m” to obtain the conclusion that there is a polynomial h(x) ∈ Q[x] and F (x, h(x)) ≡ 0. Let F (x, y) = y 2 2x+2y + y22x+y − y22x+2y − 23x+y = (y − 2x )(y2x+2y + 22x+y ). For every integer a, F (a, 2a ) = 0. However, there is no polynomial H(x) ∈ Q[x] such that F (x, H(x)) ≡ 0. From the example in [8, p. 928] we also see that the set A being infinite does not suffice to yield our conclusion, there must be A with positive UAD to obtain the conclusion. We then have the following corollary. This corollary extends the result in [1] to exponential diophantine equations. m Corollary 4.6. Let F (x, y) = k=1 fk (x, y)bGk (x,y) , where fk (x, y) and Gk (x, y) are in Z[x, y], b is a positive integer, and Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z for each 1 ≤ i = j ≤ m. Then for every integer α there is an integer c such that F (α, c) = 0 if and only if for every integer α there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(α) ∈ Z. Proof. Assume that for every α ∈ Z there is a c in Z such that F (α, c) = 0. Let R = {h1 (x), . . . , hn (x)} be the set of polynomials h(x) ∈ Q[x] such that h(x) ∈ R if and only if F (x, h(x)) ≡ 0 and h(a) ∈ Z for some a ∈ Z. From Lemma 4.3, R is a finite set. Let Si = {x ∈ Z | hi (x) ∈ Z}. If Z − ∪i Si is non-empty, then it is of positive UAD. This contradicts Theorem 4.5 that, except with a set B with zero UAD, for every α ∈ A − B there exists an h(x) ∈ Q[x] with F (x, h(x)) ≡ 0 and h(α) ∈ Z. Therefore, Z = ∪i Si . Hence for every α ∈ Z there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0 and h(α) ∈ Z. 2 Let b be a positive integer. We use Ebn (x) to denote the function which takes the iterative exponential with base b of x n times. We define the exponential order Ob [F (x, y)] of a general exponential polynomial F (x, y) with base b as follows: 1. Ob [F (x, y)] = −∞ if F (x, y) ≡ 0. 2. Ob [F (x, y)] = 0 if F (x, y) is in Z[x, y] − {0}. 3. Ob [F (x, y)] = n + 1 if there is a general exponential polynomial G(x, y) with Ob [(G(x, y))] = n such that F (x, y) = Eb (G(x, y)) = bG(x,y) .

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4. If Ob [G1 (x, y)] ≤ n and Ob [G2 (x, y)] = n, then Ob [G1 (x, y) + G2 (x, y)] ≤ n and Ob [G1 (x, y) · G2 (x, y)] = n. Since the arguments for the general exponential polynomials are almost the same as Theorem 4.5, we only sketch its proof. m n Theorem 4.7. Let F (x, y) = k=1 fk (x, y)E[ j=1 fk,j (x, y)E[. . . E[Gk (x, y)]]], where fk (x, y), fk,j (x, y) and Gk (x, y) are in Z[x, y], and if both Gi (x, y) and Gj (x, y) are of highest exponential order of F (x, y) then Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z. Let A ⊂ Z be with positive UAD, and for every α ∈ A there is a y in Z such that F (α, y) = 0. Then there is a set B with zero UAD such that for every α ∈ A − B there exists an h(x) over Q, F (x, h(x)) ≡ 0, and h(α) ∈ Z. Furthermore, there is a finite set S of polynomials such that for every α ∈ A − B and every y ∈ Z, F (α, y) = 0 if and only if there exists an h(x) ∈ S, F (x, h(x)) ≡ 0, and y = h(α). Sketch of proof. The main ingredient in the proofs of Lemma 4.4 and Theorem 4.5 is comparing the growth of the terms. Let F (x, y) = 0 be an exponential equation in the form of Theorem 4.5. If the difference of the exponent terms of F (x, y) is of polynomial growth, then one of the terms of F (x, y) will dominate the sum of all other terms if |x| is sufficiently large. (This is our Case 1.) Thus, there are finitely many solutions for the equation F (x, y) = 0. Therefore, the difference of some exponential terms cannot have polynomial growth. Hence, we will have near solutions h(x) for some of the differences of the exponential terms. If there are finitely many near solutions (Case 2.1), then we may reduce equation F (x, y) = 0 to the form of F (x, h(x)) = 0. If there are infinitely many near solutions (Case 2.2), then we may reduce equation F (x, y) = 0 to an equation in the form that exponents have only one variable y. This special case is solved by Lemma 4.4. Similar arguments may apply to the general exponentials. We compare the growth of the terms of the highest order. Let Fi (x, y) be the term with highest order of F (x, y), and Gi (x, y) be its highest exponent. If there is no other term with the same order and Gi (x, y) is with polynomial growth, then this term will dominate all other term eventually. Therefore, we only have to handle the case that Gi (x, y) has near solution h(x). This then reduces the equation F (x, y) = 0 to the form of F (x, h(x)) = 0. Thus, if G(x, y) is with polynomial growth, then there must exist another term with the same order. Let H(x, y) be the exponent of the other term with the same order; then we only need to handle the case that G(x, y) − H(x, y) has near solutions. We may combine all the terms with the same near solutions. By induction we may obtain our conclusion. 2 Similarly, we also have the following corollary. m n Corollary 4.8. Let F (x, y) = k=1 fk (x, y)E[ j=1 fk,j (x, y)E[. . . E[Gk (x, y)]]], where fk (x, y), fk,j (x, y) and Gk (x, y) are in Z[x, y], and if both Gi (x, y) and Gj (x, y) are of highest exponential order of F (x, y) then Gi (x, y) − Gj (x, y) ∈ (Z[x, y] − Z[x]) ∪ Z. Then,

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for every α ∈ Z there is a c in Z such that F (α, c) = 0 if and only if for every integer α ∈ Z there exists an h(x) ∈ Q[x] such that F (x, h(x)) ≡ 0, and h(α) ∈ Z. Acknowledgments The author would like to thank the referees for many helpful suggestions and for pointing out a mistake in the original proof of Lemma 4.2. References [1] H. Davenport, D.J. Lewis, A. Schinzel, Polynomials of certain special types, Acta Arith. 9 (1964) 107–116. [2] R. Dvornicich, S.P. Tung, U. Zannier, On polynomials taking small values at integral arguments, II, Acta Arith. 106 (2003) 115–121. [3] A. Schinzel, Diophantine equations with parameters, in: J.V. Arimitage (Ed.), Journées Arithmetiques 1980, in: London Math. Soc. Lecture Note Ser., vol. 56, Cambridge University Press, Cambridge, 1982, pp. 211–217. [4] A. Schinzel, Selected Topics on Polynomials, The University of Michigan Press, Ann Arbor, 1982. [5] S.P. Tung, On weak number theories, Jpn. J. Math. 11 (1985) 203–232. [6] S.P. Tung, Computational complexities of diophantine equations with parameters, J. Algorithms 8 (1987) 324–336. [7] S.P. Tung, Approximate solutions of polynomial equations, J. Symbolic Comput. 33 (2002) 239–254. [8] S.P. Tung, Max–min of polynomials and exponential equations, J. Number Theory 130 (2010) 912–929.