New York Journal of Mathematics

New York J. Math. 12 (2006) 193–217.

On Pillai’s Diophantine equation

Yann Bugeaud and Florian Luca

Abstract. Let A, B, a, b and c be fixed nonzero integers. We prove several results on the number of solutions to Pillai’s Diophantine equation

Aa^{x} − Bb^{y} = c

in positive unknown integers x and y.

Contents

1. Introduction 193 2. Results 194 3. Preparations 195 4. Preliminary results 196 5. Proof of Theorem 2.1 203 6. The ABC conjecture and the equation a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}} 206 7. Comments and remarks 216 References 216

1. Introduction

Let a, b and c be nonzero integers with a ≥ 2 and b ≥ 2. As noticed by Pólya [16], it follows from a theorem of Thue that the Diophantine equation

(1) a^{x} − b^{y} = c, in positive integers x, y has only finitely many solutions. If, moreover, a and b are coprime and c is sufficiently large compared with a and b, then (1) has at most one solution. This is due to Herschfeld [9] in the case a = 2, b = 3, and to Pillai [15] in the general case. (Pillai also claimed that (1) can have at most one solution even if a and b are not coprime. This is incorrect, however, as shown by the example 6^{4} − 3^{4} = 6^{5} − 3^{8} = 1215.)

193^{ISSN 1076-9803/06}

Received December 21, 2004.

Mathematics Subject Classification. 11D61, 11D72, 11D45.

Key words and phrases. Diophantine equations, applications of linear forms in logarithms and the Subspace Theorem, ABC conjecture.

This paper was written during a visit of the second author at the Université Louis Pasteur in Strasbourg in September 2004. He warmly thanks the Mathematical Department for its hospitality. Both authors were supported in part by the joint Project France–Mexico ANUIES-ECOS M01-M02.

Further results on Equation (1) are due to Shorey [21], Le [10] (both papers are concerned with the more general equation Aa^{x} − Bb^{y} = c, in positive integers x,

y) and, more recently, to Scott and Styer [20] and to Bennett [1, 2]. We direct the reader to [23, 1] for more references.

In view of Pólya’s result, the above quoted theorem of Pillai can be rephrased as follows.

Theorem 1.1. Let a ≥ 2 and b ≥ 2 be coprime integers. Then the Diophantine equation

a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}}, (2)

in positive integers x_{1}, x_{2}, y_{1}, y_{2} with x_{1} ≠ x_{2} has at most finitely many solutions.

In (2), the bases a and b are fixed. Scott and Styer [20] allowed a to be a variable, under some additional, mild assumptions. A particular case of their Theorem 2 can be formulated as follows.

Theorem 1.2. The Diophantine equation

a^{x_{1}} − a^{x_{2}} = 2^{y_{1}} − 2^{y_{2}}, (3)

in positive integers a, x_{1}, x_{2}, y_{1}, y_{2} with x_{1} ≠ x_{2} and a prime has no solution, except for four specific cases, or unless a is a sufficiently large Wieferich prime.

Since we still do not know whether or not infinitely many Wieferich primes exist, Theorem 1.2 does not imply that (3) has only finitely many solutions. Such a result has been recently established by Luca [11]. Luca’s result is the following.

Theorem 1.3. Let b be a prime number. The Diophantine equation

a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}}, (4)

in positive integers a, x_{1}, x_{2}, y_{1}, y_{2} with a ≠ b prime and x_{1} ≠ x_{2} has only finitely many solutions.

The proof of Theorem 1.3 uses a broad variety of techniques from Diophantine approximation, ranging from Ridout’s Theorem to the theory of linear forms in logarithms.

In the present paper, our aim is to generalize Theorem 1.3 in two directions. First, we remove the assumption ‘b is prime’ and we allow b to be any fixed positive integer. Secondly, under some mild coprimality conditions, we also allow arbitrary coefficients which need not be fixed, but whose prime factors should be in a fixed finite set of prime numbers.

Acknowledgments. We thank the referee for useful suggestions. The second author also thanks Andrew Granville for enlightening conversations.

2. Results

Let P = {p_{1}, …, p_{t}} be a fixed, finite set of prime numbers. We write S = {±p_{1}^{α_{1}} … p_{t}^{α_{t}} : α_{i} ≥ 0, i = 1, …, t} for the set of all nonzero integers whose prime factors belong to P. This notation will be kept throughout this paper.

Our main result is the following extension of Theorem 1.3.

Theorem 2.1. Let b be a fixed nonzero integer. The Diophantine equation

A(a^{x_{1}} − a^{x_{2}}) = B(b^{y_{1}} − b^{y_{2}}), (5)

in positive integers A, B, a, x_{1}, x_{2}, y_{1}, y_{2} has only finitely many solutions (A, B, a, x_{1}, x_{2}, y_{1}, y_{2}) with x_{1} ≠ x_{2}, a prime, A, B ∈ S and gcd(Aa, Bb) = 1.

We display two immediate corollaries concerning Equation (1).

Corollary 2.2. Let b be a fixed positive integer. There exists a positive constant a_{0} depending only on b and S such that for any nonzero integer c, for any prime a ≥ a_{0}, and for every positive integers A, B in S coprime to c, the equation

Aa^{x} − Bb^{y} = c,

in positive integers x, y has at most one solution.

Corollary 2.3. Let b be a fixed positive integer. There exists a positive constant c_{0} depending only on b and S such that for any prime a ≥ 2, and for any integer c ≥ c_{0} coprime to a, and for every coprime integers A, B in S, the equation

Aa^{x} − Bb^{y} = c,

in positive integers x, y has at most one solution.

Besides the introduction of the coefficients A and B, the important new point in Corollary 2.2 (resp. Corollary 2.3) is that the constant a_{0} (resp. c_{0}) does not depend on c (resp. a).

The proof of Theorem 2.1 follows the same general lines as that of Theorem 1 from [11]. However, there are many additional difficulties since b is no longer prime and since the coefficients A, B are not even fixed. To overcome some of these difficulties, we are led to use the Schmidt Subspace Theorem instead of Ridout’s Theorem.

We have tried to clearly separate the different steps of the proof of Theorem 2.1 and to point out where our assumptions on a and b are needed. A short discussion on possible extensions to our theorem is given in Section 6.

Throughout this paper, we use the symbols ‘O’, ‘≪’, ‘≫’, ‘≍’ and ‘o’ with their usual meaning (we recall that A ≪ B and B ≫ A are equivalent to A = O(B) and that A ≍ B means that both A ≫ B and B ≫ A hold).

3. Preparations

In this section, we review some standard notions of Diophantine approximation. For a prime number p and a nonzero rational number x, we denote by ord_{p}(x)

the order at which p appears in the factorization of x.

Let M = {2, 3, 5, … } ∪ {∞} be all the places of ℚ. For a nonzero rational number x and a place μ in M, we let the normalized μ-valuation of x, denoted by |x|_{μ}, be |x|_{μ} = |x| if μ = ∞, and |x|_{μ} = p^{−ord_{p}(x)} if μ = p is finite.

These valuations satisfy the product formula

∏_{μ∈M} |x|_{μ} = 1, for all x ∈ ℚ^{∗}.

Our basic tool is the following simplified version of a result of Schlickewei (see [18], [19]), which is commonly known as the Schmidt Subspace Theorem.

Lemma 3.1. Let P′ be a finite set of places of ℚ containing the infinite place. For any μ ∈ P′, let {L_{1,μ}, …, L_{N,μ}} be a set of linearly independent linear forms in N variables with coefficients in ℚ. Then, for every fixed 0 < ε < 1, the set of solutions x = (x_{1}, …, x_{N}) ∈ ℤ^{N}\{0} to the inequality

∏_{μ∈P′} ∏_{i=1}^{N} |L_{i,μ}(x)|_{μ} < max{|x_{i}| : i = 1, …, N}^{−ε} (6)

is contained in finitely many proper linear subspaces of ℚ^{N}.

Let P and S be as in Section 2. An S-unit x is a nonzero rational number such that |x|_{w} = 1 for every finite valuation w stemming for a prime outside P. We shall need the following version of a theorem of Evertse [8] on S-unit equations.

Lemma 3.2. Let a_{1}, …, a_{N} be nonzero rational numbers. Then the equation

∑_{i=1}^{N} a_{i}u_{i} = 1

in S-unit unknowns u_{i} for i = 1, …, N, and such that ∑_{i∈I} a_{i}u_{i} ≠ 0 for each nonempty proper subset I ⊂ {1, …, N}, has only finitely many solutions.

Finally, we will need lower bounds for linear forms in p-adic logarithms, due to Yu [24], and for linear forms in complex logarithms, due to Matveev [12].

Lemma 3.3. Let p be a fixed prime and a_{1}, …, a_{N} be fixed rational numbers. Let x_{1}, …, x_{N} be integers such that a_{1}^{x_{1}} … a_{N}^{x_{N}} ≠ 1. Let X ≥ max{|x_{i}| : i = 1, …, N}, and assume that X ≥ 3. Then,

ord_{p} (a_{1}^{x_{1}} … a_{N}^{x_{N}} − 1) ≪ log X,

where the constant implied by ≪ depends only on p, N, a_{1}, …, a_{N}.

Lemma 3.4. Let a_{1}, …, a_{N} be fixed rational numbers and, for 1 ≤ i ≤ N, let A_{i} ≥ 3 be an upper bound for the numerator and for the denominator of a_{i}, written in its lowest form. Let x_{1}, …, x_{N} be integers such that a_{1}^{x_{1}} … a_{N}^{x_{N}} ≠ 1. Let

X ≥ max \{\frac{|x_{N}|}{log A_{i}} + \frac{|x_{i}|}{log A_{N}} : i = 1, …, N − 1\} ,

and assume that X ≥ 3. Then,

log |a_{1}^{x_{1}} … a_{N}^{x_{N}} − 1| ≫ −(log A_{1}) … (log A_{n})(log X),

where the constant implied by ≫ depends only on N.

4. Preliminary results

Let P and S be as in Section 2. We start with the following result regarding the size of the coefficient A in Equation (5).

Lemma 4.1. Assume that the Diophantine equation

A(a^{x_{1}} − a^{x_{2}}) = B(q^{y_{1}} − q^{y_{2}}) (7)

admits infinitely many positive integer solutions (A, B, a, q, x_{1}, x_{2}, y_{1}, y_{2}) such that A, B, q in S, x_{1} > x_{2}, y_{1} > y_{2}, a > 1, and gcd(Aa, Bq) = 1. Let M be the common value of the number appearing in either side of Equation (7). We then have A = M^{o(1)} as max{A, B, q, x_{1}, x_{2}, y_{1}, y_{2}} tends to infinity.

Proof. Let q = ∏_{p∈P} p^{z_{p}} and let Z = max{3, z_{p} : p ∈ P}. Assume that p^{a_{p}}||A. Since Aa and Bq are coprime, it follows that p^{a_{p}}|(q^{y_{1}−y_{2}} − 1). By Lemma 3.3, we have that

a_{p} ≪ log(Zy_{1}).

Since this is true for all p ∈ P, it follows that

(8)  log A = ∑_{p∈P} a_{p} log p ≪ log(Zy_{1}) ≪ log(q^{y_{1}}) (\frac{log(Zy_{1})}{Zy_{1}})

≪ (log M) (\frac{log(Zy_{1})}{Zy_{1}}) .

Thus, it suffices to show that Zy_{1} → ∞ when M → ∞. Suppose, on the contrary, that Zy_{1} remains bounded for infinitely many solutions. Then, we may assume that q and y_{1} are fixed, and, since y_{1} > y_{2}, we may assume that y_{2} is fixed as well. Since Aa^{x_{2}}|q^{y_{1}−y_{2}} − 1, it follows that we may further assume that a and A are fixed. It then follows that the largest prime factor of a^{x_{1}−x_{2}} − 1 remains bounded. However, (a^{n} −1)_{n≥1} is a nondegenerate binary recurrent sequence, and it is known that P (a^{n} − 1) tends to infinity with n (in fact, by the well-known properties of primitive divisors to Lucas sequences, see e.g., [6] and [3], P (a^{n} − 1) ≥ n + 1 holds for all a > 1 and n ≥ 7). Hence, x_{1} − x_{2} is bounded as well, contradicting the fact that M tends to infinity. □

We can now present the following theorem.

Theorem 4.2. Let m > n > 0 be fixed positive integers. Then, the Diophantine equation

A(z^{m} − z^{n}) = B(q^{y_{1}} − q^{y_{2}}) (9)

has only finitely many positive integer solutions (A, B, z, q, y_{1}, y_{2}) with z > 1 and A, B, q in S such that gcd(Az, Bq) = 1.

Proof. We assume that the given equation has infinitely many solutions. We write again M for the common value of the two sides in Equation (9). Thus, we assume that M tends to infinity. By Lemma 4.1, it follows that we may assume that A = M^{o(1)}. In particular, A = z^{o(1)} because M ≍ Az^{m}, m is fixed and z tends to infinity. From Equation (9), we now conclude that z^{m(1+o(1))} ≍ Bq^{y_{1}}. This observation will be used several times in the course of the present proof.

We now prove a lemma about solutions of Equation (9) of a certain type.

Lemma 4.3. Let c_{0} ≠ 1 be a fixed rational number. Then there exist only finitely many solutions of Equation (9) with z = s + c_{0} > 1 and s a rational number which is a S-unit.

Proof. We assume again, for a contradiction, that we have infinitely many such solutions. Since z is an integer, it follows that the denominator of s is ≪ 1. If c_{0} = 0, it follows that z ∈ S. In this case, Equation (9) is the S-unit equation

X_{1} + X_{2} + X_{3} + X_{4} = 0,

where X_{1} = Az^{m}, X_{2} = −Az^{n}, X_{3} = −Bq^{y_{1}} and X_{4} = −By^{y_{2}}. Since z > 1 and gcd(Az, Bq) = 1, it follows that it is nondegenerate. In particular, it can have only finitely many solutions (A, B, z, q, y_{1}, y_{2}). Assume now that c_{0} ≠ 0. Equation (9) can be rewritten as

Q(s) = q^{y_{1}}B/A − q^{y_{2}}B/A,

where Q(s) is a polynomial in s whose constant term is d_{0} = c_{0}^{n}(c_{0}^{m−n} − 1) ≠ 0. Dividing both sides of the above equation by d_{0} and rearranging some terms, it follows that the above equation can be rewritten as

∑_{i=1}^{m+2} a_{i}X_{i} = 1, (10)

where a_{1} = 1/d_{0} ≠ 0, a_{2} = −1/d_{0} ≠ 0, a_{i} are fixed rational numbers for i = 3, …, m + 2, X_{1} = q^{y_{1}}B/A, X_{2} = −q^{y_{2}}B/A, and X_{i} = s^{i−2} for i ∈ {3, …, m + 2}. Let I ⊂ {1, 2, …, m+2} be the subset of those indices i such that a_{i} ≠ 0. Equation (10) is an S-unit equation in the variables X_{i} for i ∈ I. Let J be the subset of I (which can be the full set I) such that

∑_{j∈J} a_{j}X_{j} = 1 (11)

is nondegenerate; i.e., has the property that if K is any nonempty proper subset of J, then ∑_{k∈K} a_{k}X_{k} ≠ 0. It is clear that for each solution of Equation (10) such a subset J exists. Since we have infinitely many solutions, we may assume that J is fixed. By Lemma 3.2, it follows that Equation (11) admits only finitely many solutions (X_{j})_{j∈J}. If 1 ∈ J, then q^{y_{1}}B/A takes only finitely many values, and since gcd(Bq, A) = 1, it follows that A, B, q, y_{1} are all bounded. Since y_{1} > y_{2}, we get that y_{2} is bounded as well. Hence, M is bounded in this case. If i ∈ J for some i ≥ 3, it follows that s^{i−2} is bounded. Hence, z is bounded, which is a contradiction. Finally, if J = {2}, then −q^{y_{2}}B/A is fixed. Hence, we may assume that A, B, q, y_{2} are all fixed. With C = q^{y_{2}}B/A, we get z^{m} − z^{n} + C = q^{y_{1}}B/A. One verifies immediately that if m ≥ 3 or if (m, n) = (2, 1) and C ≠ 1/4, then the polynomial R(z) = z^{m} − z^{n} + C has at least two distinct roots. It is known that if Q(X) ∈ ℚ[X] is a polynomial which has at least two distinct roots and if x is a positive rational number with bounded denominator, then Q(x) is a rational number whose numerator has the property that its largest prime factor tends to infinity with x (see, e.g., [23]). This shows that the equation R(z) = q^{y_{1}}B/A can have only finitely many solutions (z, y_{1}) in this case as well (note that the denominator of z divides A which is fixed). Hence, it remains to look at the case (m, n) = (2, 1) and C = 1/4. But since gcd(Bq, A) = 1, this leads to A = 4, B = 1, q = 1, which is impossible because in this case M = 0; hence, z = 1, which is not allowed. □

We now resume the proof of Theorem 4.2. We rewrite Equation (7) as

Az^{n}(z^{m−n} − 1) = Bq^{y_{2}}(q^{y_{1}−y_{2}} − 1). (12)

Since z and q are coprime, it follows that Bq^{y_{2}} divides z^{m−n} − 1.

We first assume that m ≥ 3. If n = m − 1, then Bq^{y_{2}}|(z − 1), which implies that Bq^{y_{2}} ≪ z. Equation (9), after multiplying both sides of it by m^{m}, can be rewritten as

|A(mz − 1)^{m} − Bm^{m}q^{y_{1}}| = |Af(z) − Bm^{m}q^{y_{2}}|, (13)

where f(z) is a polynomial in z with integer coefficients and of degree m − 2. We now write q = dq_{1}^{m}, A = A_{1}A_{0}^{m}, B = B_{1}B_{0}^{m}, where d, A_{1}, B_{1} are mth power free. Clearly, since A, B, q ∈ S and m is fixed, d, A_{1}, B_{1} can take only finitely many values. In what follows, we assume that d, A_{1}, B_{1} are fixed. Equation (13) implies easily that

|\frac{A_{0}(mz − 1)}{B_{0}q_{1}^{y_{1}}} − m(dB_{1}/A_{1})^{1/m}| ≪ \frac{Az^{m−2}}{Bq^{y_{1}}} ≪ \frac{1}{z^{2}} , (14)

when M is sufficiently large. Since B_{0}q_{1} is in S, Ridout’s Theorem [17] tells us that the above inequality (14) can have only finitely many solutions (A_{0}, B_{0}, z, q_{1}, y_{1}) if (dB_{1}/A_{1})^{1/m} is not rational. Indeed, recall that (a particular version of) Ridout’s Theorem says that if α is algebraic and irrational, then for every ε > 0, the Diophantine inequality

|α − \frac{p}{q}| < \frac{1}{q^{1+ε}}

has only finitely many integer solutions (p, q) with q ∈ S. However, for us, if dB_{1}/A_{1} = c_{1}^{m} for some rational number m, then for large z the above inequality (14) leads to the conclusion that A_{0}(mz−1)−mB_{0}c_{1}q_{1}^{y_{1}} = 0, which gives z = s+c_{0}, where s = c_{1}B_{0}q_{1}^{y_{1}}/A_{0}, and c_{0} = 1/m ≠ 1. However, by Lemma 4.3, Equation (9) can have only finitely many solutions of this type also.

We now assume that m − n ≥ 2. If n ≥ 2, then Bq^{y_{2}}|z^{m−n} − 1, therefore Bq^{y_{2}} ≤ z^{m−2}. Hence,

|Az^{m} − Bq^{y_{1}}| = |Az^{n} − Bq^{y_{2}}| ≪ z^{(m−2)+o(1)}.

With the notation q = dq_{1}^{m}, A = A_{1}A_{0}^{m}, B = B_{1}B_{0}^{m}, we get

|\frac{A_{0}z}{B_{0}q_{1}^{y_{1}}} − (dB_{1}/A_{1})^{1/m}| ≪ \frac{Az^{m−2}}{Bq^{y_{1}}} ≪ \frac{1}{z^{2}} ,

and Ridout’s Theorem implies once again that the above inequality can have only finitely many positive integer solutions (A_{0}, B_{0}, z, q_{1}, y_{1}) with A_{0}, B_{0}, q_{1} ∈ S unless dB_{1}/A_{1} = c_{1}^{m} for a rational number c_{1}. If dB_{1}/A_{1} = c_{1}^{m}, we then get for large z that z = c_{1}q_{1}^{y_{1}}B_{0}/A_{0} = s ∈ S, and Equation (9) has only finitely many solutions of this type by Lemma 4.3.

We now assume that n = 1. We then write

z^{m−1} − 1 = (z − 1) (\frac{z^{m−1} − 1}{z − 1}) ,

and note that

gcd (z − 1, \frac{z^{m−1} − 1}{z − 1}) | m − 1.

From Equation (12), it follows that we may write B = B_{2}B_{3}, q = q_{2}q_{3},

z − 1 = B_{2}q_{2}^{y_{2}}u  and  \frac{z^{m−1} − 1}{z − 1} = B_{3}q_{3}^{y_{2}}v,

where B_{2}, B_{3}, q_{2}, q_{3} are positive integers and u, v are positive rational numbers with bounded denominators. Let δ > 0 be some small number to be fixed later. If either

u > z^{δ}  or  v > z^{δ},

then either

B_{2}q_{2}^{y_{2}} < z^{1−δ}  or  B_{3}q_{3}^{y_{2}} ≪ z^{m−2−δ},

and in both cases we have that Bq^{y_{2}} = B_{2}B_{3}(q_{2}q_{3})^{y_{2}} ≪ z^{m−1−δ}. We now get that

|Az^{m} − Bq^{y_{1}}| = |Az − Bq^{y_{2}}| ≪ z^{m−1−δ},

and again with the notations q = dq_{1}^{m}, A = A_{1}A_{0}^{m}, B = B_{1}B_{0}^{m} we arrive at

|\frac{A_{0}z}{B_{0}q_{1}^{y_{1}}} − (dB_{1}/A_{1})^{1/m}| ≪ \frac{z^{m−1−δ}}{Bq^{y_{1}}} ≪ \frac{1}{z^{1+δ+o(1)}} ≪ \frac{1}{z^{1+δ/2}} .

Here, we used the fact that δ is fixed and that A = z^{o(1)}. Since δ > 0 is fixed, Ridout’s Theorem implies once again that the above inequality can have only finitely many positive integer solutions (A_{0}, B_{0}, z, q_{1}, y_{1}) with B_{0}, q_{1} ∈ S unless dB_{1}/A_{1} = c_{1}^{m} for some rational number c_{1}, and as we have already seen, when this last condition holds, then for large z, we get that z = q_{1}^{y_{1}}B/A = s ∈ S, and there can be only finitely many solutions of this type by Lemma 4.3.

From now on, we consider only those solutions for which both inequalities

u < z^{δ}  and  v < z^{δ}

hold. Write D ≪ 1 for the least common multiple of the denominators of u and v. Note that the greatest prime divisor of D is at most m. We now get

B_{3}q_{3}^{y_{2}}v = \frac{z^{m−1} − 1}{z − 1} = \frac{(B_{2}q_{2}^{y_{2}}u + 1)^{m−1} − 1}{B_{2}q_{2}^{y_{2}}u} = ∑_{k=1}^{m−1} \binom{m − 1}{k} (B_{2}q_{2}^{y_{2}}u)^{k−1} ,

which can be rewritten as

−(m − 1)D^{m−2} = −B_{3}q_{3}^{y_{2}}vD^{m−2} + ∑_{k=2}^{m−1} \binom{m − 1}{k}B_{2}^{(k−1)}q_{2}^{(k−1)y_{2}}u^{k−1}D^{m−2}. (15)

We now apply Lemma 3.1 to (15). Put N = m − 1, P′ = P ∪ {∞}. Let x = (x_{1}, …, x_{N}) ∈ ℚ^{N}. For all μ ∈ P′ and all i = 1, …, N, we set L_{i}(x) = x_{i} except for (i, μ) = (1, ∞), for which we put

L_{1,∞} = −x_{1} + ∑_{k=2}^{m−1} \binom{m − 1}{k}x_{k}.

We evaluate the double product appearing at inequality (6) for our system of forms and points x = (x_{1}, …, x_{N}) given by

x_{1} = B_{3}q_{3}^{y_{2}}vD^{m−2}, and
x_{k} = B_{2}^{(k−1)}q_{2}^{(k−1)y_{2}}u^{k−1}D^{m−2}, k = 2, …, m − 1.

It is clear that x_{i} ∈ ℤ for i = 1, …, N. We may also enlarge P in such a way as to contain all the primes p ≤ m. Clearly,

∏_{μ∈P′} |L_{k}(x)|_{μ} ≤ u^{k−1}  for k ≥ 2,

∏_{μ∈P} |L_{1}(x)|_{μ} ≤ \frac{1}{B_{3}q_{3}^{y_{2}}} ,  and
|L_{1}(x)|_{∞} = (m − 1)D^{m−2}.

Thus,

∏_{μ∈P′} ∏_{i=1}^{N} |L_{i}(x)|_{μ} ≤ \frac{(m − 1)D^{m−2}u^{N^{2}}}{B_{3}q_{3}^{y_{2}}} ≪ \frac{(z^{δ})^{m^{2}}}{z^{m−1−δ}} = \frac{1}{z^{m−1−δ(m^{2}+1)}} . (16)

We now observe that

max{|x_{i}| : i = 1, …, N} = B_{3}q_{3}^{y_{2}}vD^{m−2} ≪ z^{m−2},

therefore inequality (16) implies that

∏_{μ∈P′} ∏_{i=1}^{N} |L_{i}(x)|_{μ} ≪ (max\{|x_{i}| : i = 1, …, N\})^{− \frac{m−1−δ(m^{2}+1)}{m−2}} .

Choosing δ = \frac{m − 1}{2(m^{2} + 1)}, we get that the inequality

∏_{μ∈P′} ∏_{i=1}^{N} |L_{i}(x)|_{μ} ≪ (max{|x_{i}| : i = 1, …, N})^{−ε}

holds with ε = \frac{m − 1}{2(m − 2)}. Lemma 3.1 now immediately implies that there exist

only finitely many proper subspaces of ℚ^{N} such that each one of our points x lies on one of those subspaces. This leads to an equation of the form

∑_{i=1}^{N} C_{i}x_{i} = 0,

with some integer coefficients C_{i} for i = 1, …, N not all zero, which is equivalent to

C_{1}B_{3}q_{3}^{y_{2}}vD^{m−2} + ∑_{k=2}^{m−1} C_{k}B_{2}^{k−1}q_{2}^{(k−1)y_{2}}u^{k−1}D^{m−2} = 0.

If C_{1} = 0, then we divide by D^{m−2} and the above relation becomes g(w) = 0, where w = B_{2}q_{2}^{y_{2}}u, and g(X) is the nonzero polynomial

∑_{k=2}^{m−1} C_{k}X^{k−1}.

Hence, w can take only finitely many values, and, since w = z − 1, it follows that z can take only finitely many values. If C_{1} ≠ 0, then w|C_{1}B_{3}q_{3}^{y_{2}}uD^{m−2}. Further, the greatest common divisor of w = z − 1 and B_{3}q_{3}^{y_{2}}vD^{m−2} = D^{m−2}(z^{m−1} − 1)/(z − 1) divides D^{m−2}(m − 1). Hence, this greatest common divisor is O(1). It then follows that w ≪ C_{1}. In particular, w = z − 1 can take only finitely many values in this case as well.

This completes the discussion for the case when m ≥ 3. We now deal with the case (m, n) = (2, 1). In this last case, we have

Az(z − 1) = Bq^{y_{2}}(q^{y_{1}−y_{2}} − 1).

Since Bq and Az are coprime, we get z − 1 = Bq^{y_{2}}λ for some positive integer λ. Hence,

\frac{q^{y_{1}−y_{2}} − 1}{Aλ} = z = Bq^{y_{2}}λ + 1,

therefore q^{y_{1}−y_{2}} − ABλ^{2}q^{y_{2}} = Aλ + 1. We let δ be some small positive number, and we show that the above equation has only finitely many solutions with Aλ < (Bq^{y_{2}})^{1−δ}. Indeed, assume that this is not the case. We then take N = 2, P′ = P∪{∞}, and L_{i,μ}(X_{1}, X_{2}) = X_{i} for all (i, μ) ∈ {1, 2}×P′, except for (i, μ) = (2, ∞), case in which we put L_{2,∞}(X_{1}, X_{2}) = X_{1} − X_{2}. It is easy to see that L_{1,μ} and L_{2,μ} are linearly independent for all μ ∈ P′. Taking x_{1} = q^{y_{1}−y_{2}} and x_{2} = ABλ^{2}, we get easily that

∏_{i=1}^{2} ∏_{μ∈P′} |L_{i,μ}(x_{1}, x_{2})|_{μ} = \frac{Aλ + 1}{ABq^{y_{2}}} ≪ \frac{1}{(Bq^{y_{2}})^{δ}} .

Furthermore, since Aλ < (Bq^{y_{2}})^{1−δ}, it follows that

Aλ^{2}Bq^{y_{2}} ≤ (Aλ)^{2}(Bq^{y_{2}}) ≤ (Bq^{y_{2}})^{2(1−δ)+1},

and

q^{y_{1}−y_{2}} = ABλ^{2}q^{y_{2}} + Aλ + 1 ≤ 2ABλ^{2}q^{y_{2}} ≪ (Bq^{y_{2}})^{3−2δ}.

Hence,

∏_{i=1}^{2} ∏_{μ∈P′} |L_{i,μ}(x_{1}, x_{2})|_{μ} ≪ (max\{x_{1}, x_{2}\})^{− \frac{δ}{3−2δ}} .

Applying Lemma 3.1, it follows that once δ is fixed there are only finitely many choices for the ratio x_{1}/x_{2}. In particular, q^{y_{1}−2y_{2}}/(ABλ^{2}) can take only finitely many values. Enlarging P, if needed, it follows that we may assume that λ is also an S-unit. In this case, the equation q^{y_{1}−y_{2}} − ABλ^{2}q^{y_{2}} = Aλ + 1 becomes a S-unit equation which is obviously nondegenerate, therefore it has only finitely many solutions (A, B, q, λ, y_{1}, y_{2}). Hence, there are only finitely many solutions of Equation (9) which satisfy the above property. From now on, we assume that Aλ > (Bq^{y_{2}})^{1−δ} with some small δ. We now set δ = 1/2 and get that

z = Bq^{y_{2}}λ + 1 ≫ (Bq^{y_{2}})^{2−δ}A^{−1} ≥ (Bq^{y_{2}})^{3/2}z^{o(1)}.

Thus, Bq^{y_{2}} ≪ z^{2/3+o(1)} < z^{3/4}. We now write again q = dq_{1}^{2}, A = A_{1}A_{0}^{2}, B = B_{1}B_{0}^{2} and rewrite Equation (9) as

|A_{1}(A_{0}(2z − 1))^{2} − 4dB_{1}B_{0}^{2}q_{1}^{2y_{1}}| = |4Bq^{y_{2}} − A| ≪ z^{3/4},

which gives

|\frac{A_{0}(2z − 1)}{B_{0}q_{1}^{y_{1}}} − 2(dB_{1}/A_{1})^{1/2}| ≪ \frac{z^{3/4}}{Bq^{y_{1}}} ≪ \frac{1}{z^{5/4}} .

Ridout’s Theorem implies once again that the above inequality can have only finitely many positive integer solutions (A_{0}, B_{0}, z, q_{1}, y_{1}) with q_{1} ∈ S unless dB_{1}/A_{1} = c_{1}^{m} with some rational number m. In this last case, for large z we get that z = s + c_{0}, where s = c_{1}q_{1}^{y_{1}}B_{0}/A_{0} ∈ S and c_{0} = 1/2 ≠ 1, and there are only finitely many solutions of this kind by Lemma 4.3. □

5. Proof of Theorem 2.1

We follow the method of proof of Theorem 1 from [11].

We assume that b is not a perfect power of some integer and that x_{1} > x_{2}. Thus, y_{1} > y_{2}. We also assume that Equation (5) has infinitely many positive integer solutions (A, B, a, x_{1}, x_{2}, y_{1}, y_{2}) with a prime, A, B in S, gcd(Aa, Bb) = 1 and x_{1} > x_{2}. We shall eventually reach a contradiction.

Note that, if x_{1} ≪ 1 holds for all such solutions, then the contradiction will follow from Theorem 4.2. Hence, it suffices to show that x_{1} ≪ 1. In Steps 1 to 3, we will establish that, if x_{1} and y_{1} are sufficiently large, then there exists δ > 0, depending only on b, such that all the solutions of Equation (5) have

max{x_{2}/x_{1}, y_{2}/y_{1}} < 1 − δ. (17)

Then, in Step 4, we adapt the argument used at Step 4 of the proof of Theorem 1 from [11], based on a result of Shorey and Stewart from [22], to get that x_{1} ≪ 1.

We already know that A = M^{o(1)}. We shall show that B = M^{o(1)} as well. Let p^{b_{p}}||B. Then p^{b_{p}}|a^{x_{1}−x_{2}} − 1. It is known that

b_{p} ≤ log(a^{2} − 1) + O(log(x_{1} − x_{2})) ≪ log a + log x_{1} ≪ (log a)(log x_{1}).

Hence,

log B = ∑_{p∈P} b_{p} log p ≪ (log a)(log x_{1}) = log(a^{x_{1}}) (\frac{log x_{1}}{x_{1}} ) ,

therefore B = M^{o(1)} because A = M^{o(1)} and x_{1} tends to infinity.

We now proceed in several steps.

Step 1. The case a is fixed.

In this case, Equation (5) is a particular case of an S-unit equation in four terms, which is obviously nondegenerate. In particular, there are only finitely many such solutions. These solutions are even effectively computable by using the theory of lower bounds for linear forms in logarithms, as in [14].

From now on, by Step 1, we may assume that a > b^{2}. Since

b^{2x_{1}} ≪ \frac{1}{2}a^{x_{1}} < a^{x_{1}−1}(a − 1) ≤ a^{x_{1}} − a^{x_{2}} = (b^{y_{1}} − b^{y_{2}})B/A < b^{y_{1}(1+o(1))}, (18)

we get that x_{1} < y_{1}.

Moreover, inequality (18) shows that there are only finitely many solutions (A, B, a, x_{1}, y_{1}, x_{2}, y_{2}) of Equation (5) with bounded y_{1}, and so, from now on, we shall assume that y_{1} is as large as we wish.

Step 2. There exists a constant δ_{1} > 0 depending only b such that the inequality y_{2} < y_{1}(1 − δ_{1}) holds for large values of y_{1}.

For positive integers m and r, with r a prime number, we write ord_{r}(m) for the exact order at which the prime r divides m. We write b = ∏_{i=1}^{t} r_{i}^{β_{i}}, where r_{1} < r_{2} ⋯ < r_{t} are distinct primes and β_{i} are positive integers for i = 1, …, t. Rewriting Equation (5) as

a^{x_{2}}(a^{x_{1}−x_{2}} − 1) = b^{y_{2}}(b^{y_{1}−y_{2}} − 1)B/A, (19)

we recognize that β_{i}y_{2} ≤ ord_{r_{i}}(a^{x_{1}−x_{2}} −1). Let f_{i} be the following positive integer: If r_{i} is odd, we then let f_{i} be the multiplicative order of a modulo r_{i}. If r_{i} = 2, and x_{1} − x_{2} is odd, we then let f_{i} = 1, and if x_{1} − x_{2} is even, we then let f_{i} = 2. Since y_{2} > 0, it is clear that f_{i}|x_{1} − x_{2}. We write u_{i} = ord_{r_{i}}(a^{f_{i}} − 1). We then have

(20)  β_{i}y_{2} ≤ ord_{r_{i}}(a^{x_{1}−x_{2}} − 1) ≤ u_{i} + ord_{r_{i}}(\frac{x_{1} − x_{2}}{f_{i}} )

≤ u_{i} + \frac{log(x_{1} − x_{2})}{log r_{i}} < u_{i} + \frac{log y_{1}}{log r_{i}} .

For a positive integer m, we write F_{m}(X) = Φ_{m}(X) ∈ ℤ[X] for the mth cyclotomic polynomial if m ≥ 3 and F_{m}(X) = X^{m} − 1 for m = 1, 2. From the definition of f_{i} and u_{i}, we have that

r_{i}^{u_{i}}|F_{f_{i}}(a).

Let F = {f_{i} : i = 1, …, t}, and let ℓ = #F. Observe that

(21)  M^{o(1)}b^{y_{1}} = (b^{y_{1}} − b^{y_{2}})B/A = a^{x_{2}}(a^{x_{1}−x_{2}} − 1)

≥ a ∏_{f∈F} F_{f}(a) = ∏_{f∈F} (a^{1/ℓ}F_{f}(a)) .

For f ∈ F, we put d_{f} = deg(F_{f}). Hence, d_{f} = f if f ≤ 2, and d_{f} = ϕ(f) otherwise, where ϕ is the Euler function. We now remark that

a^{1/ℓ}F_{f}(a) ≫ F_{f}(a)^{\frac{ℓd_{f} +1}{ℓd_{f}}} . (22)

Indeed, since ℓ ≤ t = ω(b) is bounded, the above inequality is equivalent to

a^{d_{f}} ≫ F_{f}(a).

Since all the roots of F_{f}(X) are roots of unity, the above inequality is implied by

a^{d_{f}} ≫ (a + 1)^{d_{f}} ,

which is equivalent to

(1 + \frac{1}{a})^{d_{f}} ≪ 1.

In turn, this last inequality follows from the fact that d_{f} ≤ f together with the fact that f_{i}|r_{i} − 1 whenever r_{i} > 2 by Fermat’s Little Theorem. Let d = max{d_{f} : f ∈ F}. Inequalities (21), (22) and (20) show that

b^{y_{1}(1+o(1))} ≫ ( ∏_{f∈F} F_{f}(a))^{\frac{ℓd+1}{ℓd}} ≫ (∏_{i=1}^{t} r_{i}^{u_{i}})^{\frac{ℓd+1}{ℓd}}

≫ (∏_{i=1}^{t} \frac{r_{i}^{β_{i}y_{2}}}{y_{1}} )^{\frac{ℓd+1}{ℓd}} ≫ \frac{b^{(^{\frac{ℓd+1}{ℓd}})y_{2}}}{y_{1}^{2t}} .

Therefore

y_{1}(1 + o(1)) > (\frac{ℓd + 1}{ℓd}) y_{2} − 2t log y_{1} + O(1),

and so

y_{2} < (\frac{ℓd}{ℓd + 1}) y_{1}(1 + o(1)) + O(log y_{1}) = (\frac{ℓd}{ℓd + 1}) y_{1}(1 + o(1)),

which implies the assertion of Step 2 with δ_{1} = 1/(2ℓd) once y_{1} is sufficiently large.

Step 3. There exists a constant δ_{2} > 0 depending only on b such that the inequality x_{2} < (1 − δ_{2})x_{1} holds for large values of y_{1}.

We look again at Equation (19). We put z = y_{1} − y_{2}, and we notice that, by Step 2, the inequality z/y_{1} ≫ 1 holds for all positive integer solutions of Equation (5), with a a prime not dividing b, and x_{1} > x_{2}. From Equation (19), we learn that x_{2} = ord_{a}(b^{z} − 1). We let g be the multiplicative order of b modulo a. It then follows that a|Φ_{g}(b). Furthermore, if we put v = ord_{a}(Φ_{g}(b)), we then have that

x_{2} = ord_{a} (a^{x_{2}}(a^{x_{1}−x_{2}} − 1)) = ord_{a} (b^{y_{2}}(b^{y_{1}−y_{2}} − 1)) − ord_{a}(A)
≤ ord_{a}(b^{z} − 1) = v + O (\frac{log z}{log a}) .

Consequently,

\frac{b^{z} − 1}{a^{x_{2}}} ≥ \frac{b^{z} − 1}{z^{O(1)}Φ_{g}(b)}.

Since g|z, and since

Φ_{z}(m) = m^{ϕ(z)+O(τ(z))}

holds for all positive integers m, where τ(z) is the number of divisors of z (see [6]), we get that

\frac{b^{z} − 1}{a^{x_{2}}} ≥ b^{z−ϕ(g)+O(τ(z)+log z)} = b^{z−ϕ(g)+O(z^{1/2})},

where we used the well-known fact that τ(z) ≪ z^{1/2}. Note that since z/y_{1} ≫ 1 and since y_{1} is as large as we wish, it follows that z_{1} is as large as we wish. Since

b^{z−ϕ(g)+O(τ(z))} ≤ \frac{b^{z} − 1}{a^{x_{2}}} = M^{o(1)} (\frac{a^{x_{1}−x_{2}} − 1}{b^{y_{2}}}) = b^{o(z)} (\frac{a^{x_{1}−x_{2}} − 1}{b^{y_{2}}}) ,

it suffices to show that z − ϕ(g) ≫ z. If g < z, then z − ϕ(g) ≥ z − g ≥ z/2. Thus, we may assume that g = z. Since the order of b modulo a is g, we get that a ≡ 1 (mod g), therefore z|a − 1. In particular, z|b^{x_{2}}(b^{z} − 1). The argument from the end of Step 3 of the proof of Theorem 1 in [11] shows that if we write p(m) for the smallest prime factor of m, then p(z)|b(b − 1). Hence, p(z) ≪ 1, so

z − ϕ(g) = z − ϕ(z) ≥ z/p(z) ≫ z,

which completes the proof of the assertion of Step 3.

The combination of Steps 2 and 3 shows that Equation (17) holds with δ = min{δ_{1}, δ_{2}}.

Step 4. The exponent x_{1} is bounded.

Recall that A = M^{o(1)} and B = M^{o(1)}. It then follows from (17) that there exists a positive real number η such that

|AB^{−1}a^{x_{1}}b^{−y_{1}} − 1| < a^{−ηx_{1}}. (23)

Write AB^{−1} = p_{1}^{u_{1}} … p_{t}^{u_{t}}. If x_{1} is sufficiently large, then, for 1 ≤ j ≤ t, we have p_{j}^{|u_{j}|} ≤ a^{x_{1}}; hence, |u_{j}|/ log a ≤ 2x_{1}. Furthermore, we have y_{1}/ log a ≪_{b} x_{1}. Applying Lemma 3.4, we get

log |AB^{−1}a^{x_{1}}b^{−y_{1}} − 1| ≫ −(log a)(log x_{1}),

where the constant implied in ≫ depends only on b and P. Combined with (23), this gives an upper bound for x_{1}, in terms of b and P. According to the observation made at the beginning of Section 5, this finishes the proof of our theorem.

6. The ABC conjecture and the equation a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}}

In this section, we discuss conditional results. Bennett [1] conjectured that there exist only finitely many triples of positive integers (a, b, c) with a and b coprime such that the Diophantine equation a^{x} − b^{y} = c has two positive integer solutions (x, y). Note that if (a, b, c) is such a triple, then there exists (x, y) ≠ (x_{1}, y_{1}) such that

a^{x} − b^{y} = c = a^{x_{1}} − b^{y_{1}}.

Thus, we are led to a nontrivial solution (x, y) ≠ (x_{1}, y_{1}) of the equation

a^{x} − a^{x_{1}} = b^{y} − b^{y_{1}}. (24)

In Theorem 2 in [11], it is shown that the ABC-conjecture implies that there are only finitely many positive integer solutions (a, b, x, x_{1}, y, y_{1}) with a and b coprime and (x, y) ≠ (x_{1}, y_{1}), subject to the additional restriction that both a and b are primes. We point out that an equation related to (24), namely x^{p} − x = y^{q} − y, was treated by Mignotte and Pethő in [13]. For example, it is shown there that if 2 ≤ p < q and q ≥ 4 are fixed, then the above equation has only finitely many rational solutions (x, y). Further, the ABC conjecture is used there to suggest that perhaps the above equation has only finitely many integer solutions in all four unknowns (x, y, p, q) with 2 ≤ p < q.

Here, we remove both the restrictions that a and b are coprime, as well as the arithmetic restrictions that a and b are prime in Equation (24), and we prove the following result.

Theorem 6.1. The ABC conjecture implies that the diophantine equation

a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}} (25)

has only finitely many positive integer solutions (a, b, x_{1}, x_{2}, y_{1}, y_{2}) with a > 1, b > 1, x_{1} ≠ x_{2} and a^{x_{1}} ≠ b^{y_{1}}.

Recall that the ABC conjecture asserts that for every ε > 0 there exists a constant C_{ε} depending on ε such that whenever

A + B = C

with nonzero integers A, B and C such that A and B are coprime, then

max{|A|, |B|, |C|} ≤ C_{ε} ( ∏_{p|ABC} p)^{1+ε} .

Proof. We may assume that x_{1} > x_{2} and that a > b (in particular, a ≥ 3). We then also get that y_{1} > y_{2}. We may further assume that gcd(x_{1}, x_{2}) = 1 and that gcd(y_{1}, y_{2}) = 1. Indeed, for if say gcd(x_{1}, x_{2}) = d, we may then write z_{1} = x_{1}/d, z_{2} = x_{2}/d and a_{1} = a^{d}, and note that (a_{1}, b, z_{1}, z_{2}, y_{1}, y_{2}) is also a solution in positive integers of the given equation, and it satisfies as well the conditions a_{1} > 1, z_{1} ≠ z_{2} and a_{1}^{z_{1}} ≠ b^{y_{1}}.

Note also that x_{1} ≠ y_{1}, for if x_{1} = y_{1}, then

max{a^{x_{2}}, b^{y_{2}}} > |a^{x_{2}} − b^{y_{2}}| = |a^{x_{1}} − b^{x_{1}}| = |a − b|(a^{x_{1}−1} + ⋯ + b^{x_{1}−1})
≥ a^{x_{2}} + b^{y_{2}},

which is a contradiction.

Further, since

b^{y_{1}} > b^{y_{1}} − b^{y_{2}} = a^{x_{1}} − a^{x_{2}} ≥ a^{x_{1}}(1 − \frac{1}{a}) > \frac{a^{x_{1}}}{2} > a^{x_{1}−1} > b^{x_{1}−1},

we get that y_{1} > x_{1} − 1, and since y_{1} ≠ x_{1}, we have y_{1} > x_{1}. Thus, y_{1} is the largest one of all four exponents.

We shall first show that x_{1} = O(1), later that y_{1} = O(1) and finally that a is O(1).

Step 1. x_{1} = O(1).

Let p be a prime dividing both a and b, and put p^{r_{p}} ∥ a and p^{s_{p}} ∥ b. It is then clear that the order at which p divides the left-hand side of Equation (25) is p^{r_{p}x_{2}}, while the order at which p divides the right-hand side of the same equation is p^{s_{p}y_{2}}. Identifying those orders we get

r_{p}x_{2} = s_{p}y_{2},

showing that there exists a positive integer t_{p} such that r_{p} = t_{p}y_{2}/ gcd(x_{2}, y_{2}) and s_{p} = t_{p}x_{2}/ gcd(x_{2}, y_{2}). Setting

c = ∏_{p | gcd(a,b)} p^{t_{p}},

we get a = a_{0}c^{y_{2}/ gcd(x_{2},y_{2})} and b = b_{0}c^{x_{2}/ gcd(x_{2},y_{2})}, where now a_{0} and b_{0} are coprime and free of primes p dividing c. Inserting the above expressions for a and b in Equation (25) and cancelling in both sides a factor of c^{x_{2}y_{2}/ gcd(x_{2},y_{2})}, we get

a_{0}^{x_{1}}c^{y_{2}(x_{1}−x_{2})/ gcd(x_{2},y_{2})} − a_{0}^{x_{2}} = b_{0}^{y_{1}}c^{x_{2}(y_{1}−y_{2})/ gcd(x_{2},y_{2})} − b_{0}^{y_{2}}, (26)

which we rewrite as

a_{0}^{x_{2}} (a_{0}^{x_{1}−x_{2}}c^{y_{2}(x_{1}−x_{2})/ gcd(x_{2},y_{2})} − 1) = b_{0}^{y_{2}} (b_{0}^{y_{1}−y_{2}}c^{x_{2}(y_{1}−y_{2})/ gcd(x_{2},y_{2})} − 1) . (27)

Since a_{0} and b_{0} are coprime, we deduce the existence of a positive integer D (note that one of a_{0}, b_{0} or c is larger than 1 so the two sides of the above equation are nonzero) such that both equations

\begin{aligned}a_{0}^{x_{1}−x_{2}}c^{y_{2}(x_{1}−x_{2})/ gcd(x_{2},y_{2})} − 1 & = Db_{0}^{y_{2}} \\ b_{0}^{y_{1}−y_{2}}c^{x_{2}(y_{1}−y_{2})/ gcd(x_{2},y_{2})} − 1 & = Da_{0}^{x_{2}} \end{aligned} (28)

are satisfied. In the sequel, ε denotes a very small positive real number. We now apply the ABC conjecture to each of the above equations (28) (note that the coprimality conditions are satisfied as in both cases one of the terms is −1), getting

Db_{0}^{y_{2}} ≪ (a_{0}b_{0}cD)^{1+ε}  and  Da_{0}^{x_{2}} ≪ (a_{0}b_{0}cD)^{1+ε},

which, after multiplying them side by side and cancelling a factor of D^{2}, lead to the inequality

a_{0}^{x_{2}}b_{0}^{y_{2}} ≪ D^{2ε}(a_{0}b_{0}c)^{3}. (29)

We now rewrite Equation (26) as

a_{0}^{x_{1}}c^{y_{2}(x_{1}−x_{2})/ gcd(x_{2},y_{2})} − b_{0}^{y_{1}}c^{x_{2}(y_{1}−y_{2})/ gcd(x_{2},y_{2})} = a_{0}^{x_{2}} − b_{0}^{y_{2}}, (30)

and put {M_{−}, M_{+}} = {y_{2}(x_{1} − x_{2})/ gcd(x_{2}, y_{2}), x_{2}(y_{1} − y_{2})/ gcd(x_{2}, y_{2})}, where M_{−} ≤ M_{+}. Note first that 1 ≤ M_{−} < M_{+}. Indeed, if not then x_{2}y_{1} = x_{1}y_{2}; hence, x_{2}/x_{1} = y_{2}/y_{1}, and since gcd(x_{2}, x_{1}) = gcd(y_{2}, y_{1}) = 1, we get that x_{2} = y_{2} and x_{1} = y_{1}, which we have already seen to be impossible. Note also that both sides of Equation (30) are nonzero, for if a_{0}^{x_{2}} = b_{0}^{y_{2}}, we then have a^{x_{2}} = b^{y_{2}}; hence, a^{x_{1}} = b^{y_{1}}, which is impossible. Equation (30) immediately implies that c^{M_{−}} | (a_{0}^{x_{2}} − b_{0}^{y_{2}}), so c^{M_{−}} ≤ |a_{0}^{x_{2}} − b_{0}^{y_{2}}|, and further, after cancelling a factor of c^{M_{−}} from both sides of Equation (30), we get a relation of the type

A − B = C, (31)

where C = |a_{0}^{x_{2}} − b_{0}^{y_{2}}|/c^{M_{−}}, and

(A, B) = (a_{0}^{x_{1}}c^{M_{+}−M_{−}}, b_{0}^{y_{1}}), or (a_{0}^{x_{1}}, b_{0}^{y_{1}}c^{M_{+}−M_{−}}),

according to whether y_{2}(x_{1} − x_{2})/gcd(x_{2}, y_{2}) = M_{+}, or M_{−}, respectively. We can now apply the ABC conjecture to Equation (31) (note that the above A and B are coprime), getting that

max\{a_{0}^{x_{1}}, c^{M_{+}−M_{−}}, b_{0}^{y_{1}}\} ≪ (a_{0}b_{0}c\frac{|a_{0}^{x_{2}} − b_{0}^{y_{2}}|}{c^{M_{−}}})^{1+ε} ≪ (a_{0}b_{0}(a_{0}^{x_{2}}b_{0}^{y_{2}}))^{1+ε}, (32)

where we used the fact that M_{−} ≥ 1 and |a_{0}^{x_{2}} − b_{0}^{y_{2}}| ≤ a_{0}^{x_{2}}b_{0}^{y_{2}}. Inserting estimate (29) into (32), we get

max{a_{0}^{x_{1}}, b_{0}^{y_{1}}, c^{M_{+}−M_{−}}} ≪ (a_{0}b_{0}c)^{5}D^{3ε}. (33)

Using Equations (28), and inequalities (29) and (33), we get

D^{2} ≪ (a_{0}^{x_{1}−x_{2}}b_{0}^{y_{1}−y_{2}}c^{M_{+}+M_{−}})

= (a_{0}^{x_{1}}b_{0}^{y_{1}}c^{M_{+}−M_{−}}) (\frac{c^{2M_{−}}}{a_{0}^{x_{2}}b_{0}^{y_{2}}} )

≤ (a_{0}^{x_{1}}b_{0}^{y_{1}}c^{M_{+}−M_{−}})a_{0}^{x_{2}}b_{0}^{y_{2}}

≤ (a_{0}b_{0}c)^{18}D^{11ε},

leading to

D ≪ (a_{0}b_{0}c)^{10}. (34)

Let L = max{a_{0}, b_{0}, c}.

Assume that L = b_{0}. Then, D ≪ b_{0}^{30}, and now inequality (33) shows that

b_{0}^{y_{1}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ b_{0}^{15+90ε} ≪ b_{0}^{16},

and since b_{0} > 1 (because L = b_{0} and a ≥ 3), we get that y_{1} is bounded. Hence, all four exponents x_{1}, x_{2}, y_{1} and y_{2} are bounded in this case.

Assume now that L = c. Then, D ≪ c^{30}. Further, inequality (33) shows that

c^{M_{+}−M_{−}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ c^{15+90ε} ≪ c^{16}.

Since also c^{M_{−}} divides |a_{0}^{x_{2}} − b_{0}^{y_{2}}|, we have that, by inequality (29),

c^{M_{−}} ≤ |a_{0}^{x_{2}} − b_{0}^{y_{2}}| ≤ a_{0}^{x_{2}}b_{0}^{y_{2}} ≪ (a_{0}b_{0}c)^{3}D^{2ε} ≪ c^{9+60ε} ≪ c^{10},

therefore

c^{M_{+}} ≪ c^{26}. (35)

We may also assume that c is as large as we wish, otherwise Equation (25) becomes just a nondegenerate S-unit equation (recall that we have assumed that c = max{a_{0}, b_{0}, c}), and therefore it has only finitely many solutions. So, if c is larger than the constant implied in inequality (35), then M_{+} ≤ 27. This shows now that each one of the four numbers x_{1}−x_{2}, y_{1}−y_{2}, x_{2}/ gcd(x_{2}, y_{2}) and y_{2}/ gcd(x_{2}, y_{2}) is at most 27. From Equations (28), we get

c^{(x_{1}−x_{2})(y_{1}−y_{2})x_{2}y_{2}/ gcd(x_{2},y_{2})^{2}} = (\frac{Db_{0}^{y_{2}} + 1}{a_{0}^{x_{1}−x_{2}}})^{x_{2}(y_{1}−y_{2})/ gcd(x_{2},y_{2})}

= (\frac{Da_{0}^{x_{2}} + 1}{b_{0}^{y_{1}−y_{2}}})^{y_{2}(x_{1}−x_{2})/ gcd(x_{2},y_{2})} .

The last equality above leads to the conclusion that

D | a_{0}^{(x_{1}−x_{2})(y_{1}−y_{2})x_{2}/ gcd(x_{2},y_{2})} − b_{0}^{(x_{1}−x_{2})(y_{1}−y_{2})y_{2}/ gcd(x_{2},y_{2})}.

If the right-hand side of the above divisibility relation is zero, we then get a_{0} = b_{0} = 1, therefore a^{x_{2}} = b^{y_{2}}; hence, a^{x_{1}} = b^{y_{1}}, which is impossible. Hence, the right-hand side of the above relation is nonzero, and since M_{+} ≤ 27, we get that

D ≪ max{a_{0}, b_{0}}^{729}.

Further, the ABC conjecture applied to the appropriate one of Equations (28) gives

c^{2} ≤ c^{M_{+}} ≪ (Da_{0}b_{0}c)^{1+ε} ≪ (Da_{0}b_{0}c)^{3/2}

giving

c ≪ (Da_{0}b_{0})^{2} ≪ (max{a_{0}, b_{0}})^{1462}. (36)

If b_{0} > a_{0}, then inequality (33) gives

b_{0}^{y_{1}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ b_{0}^{7320+2187ε} ≪ b_{0}^{7321},

which gives again the conclusion that all four exponents are bounded. Thus, we may assume that a_{0} > b_{0}, and we get, from inequalities (33) and (36), that

a_{0}^{x_{1}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ a_{0}^{7320+2187ε} ≪ a_{0}^{7321}, (37)

giving that x_{1} = O(1). Note further that from estimate (36), we also deduce that

c^{y_{2}/ gcd(x_{2},y_{2})} ≤ c^{27} ≪ a_{0}^{39,474} (38)

in this case, since we have established that y_{2}/(x_{2}, y_{2}) ≤ 27.

Finally, let us assume that L = a_{0}. We then get that D ≪ a_{0}^{30}, and now estimate (33) leads to

a_{0}^{x_{1}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ a_{0}^{15+90ε} ≤ a_{0}^{16}, (39)

which gives x_{1} ≪ 1. Furthermore, from estimate (33), we also have that

c^{M_{+}−M_{−}} ≪ (a_{0}b_{0}c)^{5}D^{3ε} ≪ a_{0}^{15+90ε} ≤ a_{0}^{16},

while from relation (30) and estimate (29) we get

c^{M_{−}} ≪ |a_{0}^{x_{2}} − b_{0}^{y_{2}}| ≤ a_{0}^{x_{2}}b_{0}^{y_{2}} ≪ (a_{0}b_{0}c)^{3}D^{2ε} ≪ a_{0}^{3+60ε} ≪ a_{0}^{4},

which leads to the conclusion that c^{M_{+}} ≪ a_{0}^{20}. Since y_{2}/ gcd(x_{2}, y_{2}) ≤ M_{+}, we deduce, in particular, that in both cases when L = c and when L = a_{0}, inequality (38) holds.

Step 2. y_{1} = O(1).

From now on, we may assume that both x_{1} and x_{2} are fixed. We show that y_{1} = O(1). Note that in either case when L = a_{0} or L = c, we have, by estimates (37), (39), and (38), that

a_{0}^{3·10^{8}} > (a_{0}^{7321})^{40,000} ≫ (a_{0}^{x_{1}})^{40,000} = (a_{0} · a_{0}^{39,474})^{x_{1}}
≥ (a_{0} · c^{y_{2}/ gcd(x_{2},y_{2})})^{x_{1}} = a^{x_{1}} > a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}} ≫ b^{y_{1}};

hence, b = a_{0}^{O(1/y_{1})}. Since c | b, we also have that c = a_{0}^{O(1/y_{1})}. Hence, b_{0}c = a_{0}^{O(1/y_{1})}. From now on, we will assume that y_{1} is so large so that b_{0}c < a_{0}. In particular, we are in the case L = a_{0}.

Case 1. x_{2} ≥ 2.

Applying the ABC conjecture to the second equation (28), we get

Da_{0}^{x_{2}} ≪ (Da_{0}b_{0}c)^{1+ε},

so

a_{0}^{x_{2}−1−ε} ≪ (b_{0}c)^{1+ε}D^{ε} = a_{0}^{O(1/y_{1})}D^{ε}. (40)

Since clearly D < a^{x_{1}} = a^{O(1)}, we get

D ≪ a^{O(1)} ≤ (a_{0}c^{y_{2}})^{O(1)} = a_{0}^{O(1+y_{2}/y_{1})} = a_{0}^{O(1)},

and now from inequality (40), we get that

a_{0}^{x_{2}−1−ε} ≪ a_{0}^{O(1/y_{1}+ε)}.

The last inequality above leads to the conclusion that y_{1} ≪ 1, because x_{2} ≥ 2 and ε > 0 is arbitrarily small.

Case 2. x_{2} = 1 but x_{1} ≥ 3.

Applying the ABC conjecture to the first of Equations (28), we get

a^{x_{1}−1} = (a_{0}c^{y_{2}/gcd(x_{2},y_{2})})^{x_{1}−1} ≪ (a_{0}b_{0}cD)^{1+ε} = (a_{0}b_{0}c\frac{a^{x_{1}−1}}{b_{0}^{y_{2}}} )^{1+ε} ,

giving

b_{0}^{y_{2}} ≪ (a_{0}b_{0}c)^{1+ε}a^{(x_{1}−1)ε} = a_{0}^{1+O(1/y_{1}+ε)}(a_{0}c^{y_{2}})^{O(ε)} = a_{0}^{1+O(1/y_{1}+ε)}. (41)

On Pillai’s Diophantine equation 211

Now inequality (32) becomes

a_{0}^{x_{1}} ≪ (a_{0}b_{0}|a_{0} − b_{0}^{y_{2}}|)^{1+ε} .

If a_{0} > b_{0}^{y_{2}}, we then get

a_{0}^{x_{1}−2(1+ε)} ≪ b_{0}^{2} = a_{0}^{O(1/y_{1})},

which shows that y_{1} ≪ 1 if ε > 0 is small (because x_{1} ≥ 3), while if b_{0}^{y_{2}} > a_{0}, then, by estimate (41), we get

a_{0}^{x_{1}} ≪ (a_{0}b_{0}b_{0}^{y_{2}})^{1+ε} ≪ a_{0}^{2+O(1/y_{1}+ε)},

which again leads to the conclusion that y_{1} ≪ 1 if ε > 0 is sufficiently small, since x_{1} ≥ 3. Case 3. x_{1} = 2 and x_{2} = 1.

Here, Equations (28) become

\begin{aligned}a_{0}c^{y_{2}} − 1 & = Db_{0}^{y_{2}} \\ b_{0}^{y_{1}−y_{2}}c^{y_{1}−y_{2}} − 1 & = Da_{0}. \end{aligned} (42)

The ABC conjecture applied to the first of Equations (42) above gives

a_{0}c^{y_{2}} ≪ (a_{0}b_{0}cD)^{1+ε} < (a_{0}b_{0}c\frac{a_{0}c^{y_{2}}}{b_{0}^{y_{2}}} )^{1+ε} ,

giving b_{0}^{y_{2}} < b_{0}^{y_{2}(1+ε)} ≪ (a_{0}b_{0}c)^{1+2ε}c^{εy_{2}} = a_{0}^{1+O(1/y_{1}+ε)}. (43) On the other hand, rewriting Equation (25) as

(2a − 1)^{2} = 4b^{y_{1}} − (4b^{y_{2}} − 1),

and applying the ABC conjecture to the above equation we get

(a_{0}c^{y_{2}})^{2} = a^{2} ≪ (ab(b^{y_{2}}))^{1+ε} = (a_{0}b_{0}cb_{0}^{y_{2}}c^{2y_{2}})^{1+ε},

giving a_{0} ≪ (b_{0}c)^{1+2ε}b_{0}^{y_{2}(1+ε)}c_{0}^{2εy_{2}} = b_{0}^{y_{2}}a_{0}^{O(1/y_{1}+ε)}, (44) and comparing estimates (43) and (44), we get

b_{0}^{y_{2}} = a_{0}^{1+O(1/y_{1}+ε)}.

The above estimate and the first equation (42) gives

a_{0}c^{y_{2}} − 1 = Db_{0}^{y_{2}} = Da_{0}^{1+O(1/y_{1}+ε)},

so D = c^{y_{2}}a_{0}^{O(1/y_{1}+ε)}. (45) Further,

b^{y_{2}} = b_{0}^{y_{2}}c^{y_{2}} = a_{0}^{1+O(1/y_{1}+ε)}c^{y_{2}} = (a_{0}c^{y_{2}})^{1+O(1/y_{1}+ε)} = a^{1+O(1/y_{1}+ε)},

and since

a^{2} ≍ a^{2} − a = b^{y_{1}} − b^{y_{2}} ≍ b^{y_{1}},

we get that b^{y_{1}/2} = a^{1+O(ε)} provided that a is sufficiently large with respect to ε. (Of course, if a is bounded, we then get only finitely many solutions). Hence, y_{2} = (y_{1}/2)(1 + O(1/y_{1} + ε)). We thus get that

M_{−} = min{y_{1} − y_{2}, y_{2}} = y_{2}(1 + O(1/y_{1} + ε)).

Inequality (32) now gives

a_{0}^{2} ≪ (a_{0}b_{0}c\frac{|a_{0} − b_{0}^{y_{2}}|}{c^{M_{−}}})^{1+ε} ,

which together with the fact that b_{0}^{y_{2}} = a_{0}^{1+O(1/y_{1}+ε)} gives

c^{y_{2}(1+O(1/y_{1}+ε))} = c^{M_{−}} < c^{M_{−}(1+ε)} ≪ (b_{0}c)^{1+ε}a_{0}^{−1+ε}|a_{0} − b_{0}^{y_{2}}|^{1+ε}
= a_{0}^{O(1/y_{1}+ε)},

giving

c^{y_{2}} = a_{0}^{O(1/y_{1}+ε)}.

Thus, by (45), we also have

D = c^{y_{2}}a_{0}^{O(1/y_{1}+ε)} = a_{0}^{O(1/y_{1}+ε)}.

Subtracting the first equation (42) from the second one, and reducing the resulting equation modulo b^{M_{−}}, we get

a_{0}(c^{y_{2}} + D) ≡ 0 (mod b_{0}^{M_{−}}).

Since a_{0} and b_{0} are coprime,

0 < c^{y_{2}} + D = a_{0}^{O(1/y_{1}+ε)},

while

b_{0}^{M_{−}} = (b_{0}^{y_{2}})^{1+O(1/y_{1}+ε)} = a_{0}^{1+O(1/y_{1}+ε)},

we get a contradiction if ε is sufficiently small and y_{1} is sufficiently large. Thus, y_{1} ≪ 1.

Step 3. a = O(1).

This part of the proof is unconditional and somewhat independent of the previous parts of the proof, which is why we record it as follows:

Theorem 6.2. Assume that x_{1} > x_{2} > 0, y_{1} > y_{2} > 0 are fixed integers with gcd(x_{1}, x_{2}) = gcd(y_{1}, y_{2}) = 1 and y_{1} > x_{1}. Then the diophantine equation

a^{x_{1}} − a^{x_{2}} = b^{y_{1}} − b^{y_{2}} (46)

has only finitely many positive integer solutions (a, b).

The case when x_{2} = y_{2} = 1 of Theorem 6.2 appears in [13]. Further, a partial result along the lines of Theorem 6.2 above appears as Proposition 3 on page 211 in [11]. However, the proof of that result in [11] uses the condition (imposed in that statement) that a and b are coprime, which is why we choose to give a complete and self-contained proof of Theorem 6.2 here.

Proof. It is easy to check that both curves X^{2}−X = Y^{3}−Y and X^{2}−X = Y^{3}−Y^{2} are elliptic, therefore there are only finitely many integer solutions (a, b) of Equation (46) when (x_{1}, x_{2}, y_{1}, y_{2}) = (2, 1, 3, 1) or (2, 1, 3, 2). From now on, we assume that y_{1} ≥ 4. Further, when x_{1} = 2, then x_{2} = 1, and the given equation becomes

(2a − 1)^{2} = 4b^{y_{1}} − 4b^{y_{2}} + 1.

It is easy to check that the polynomial g(Y) = 4Y^{y_{1}} − 4Y^{y_{2}} + 1 has only simple roots. Indeed, if z is a double root of g, then

z^{y_{2}−1}(y_{1}z^{y_{1}−y_{2}} − y_{2}) = \frac{g′(z)}{4} = 0,

and since 0 is not a root of g, we get z^{y_{1}−y_{2}} = y_{2}/y_{1}. Furthermore, the relation g(z) = 0 becomes 4z^{y_{2}}(z^{y_{1}−y_{2}} − 1) = −1, which implies that 4z^{y_{2}}(y_{2}/y_{1} − 1) = −1, therefore 4z^{y_{2}} = y_{1}/(y_{1} − y_{2}). Thus,

(\frac{y_{1}}{y_{1} − y_{2}} )^{y_{1}−y_{2}} = 4^{y_{1}−y_{2}}(z^{y_{1}−y_{2}})^{y_{2}} = 4^{y_{1}−y_{2}} (\frac{y_{2}}{y_{1}} )^{y_{2}} .

Since y_{1} and y_{2} are coprime, the above relation gives, by looking at the denominators, that y_{1} is a power of 2 and that y_{1}−y_{2} = 1. Now by looking at the numerators, we get that y_{2} = y_{1} − 1 is an odd number, and that every prime factor of it divides y_{1}, which is obviously impossible.

From now on, we assume that x_{1} ≥ 3.

Write f(X) = X^{x_{1}} −X^{x_{2}} and let g(Y) = Y^{y_{1}} −Y^{y_{2}}. Recall the following theorem due to Davenport, Lewis and Schinzel (see [7]).

Theorem 6.3. Let f(X) and g(Y) be polynomials with integer coefficients of degrees > 1. Let D_{f}(λ) = disc(f(X) − λ) and D_{g}(λ) = disc(g(Y) − λ). Assume further that there are at least ⌈deg(f)/2⌉ distinct roots of D_{f}(λ) = 0 for which D_{g}(λ) ≠ 0. Then, the polynomial f(X)−g(Y) is irreducible over the complex numbers. Further, the genus of the curve given by the equation f(X) = g(Y) is strictly positive except possibly when deg(f) = 2 or deg(f) = deg(g) = 3. Apart from these possible exceptions, the equation f(x) = g(y) has at most a finite number of integral solutions (x, y).

In our case, one checks easily that all the roots λ of D_{f} are:

(i) λ = 0 with multiplicity x_{2} − 1,

(ii) λ = −ζ^{x_{2}} · (\frac{x_{1} − x_{2}}{x_{1}} ) · (\frac{x_{2}}{x_{1}} )^{\frac{x_{2}}{x_{1}−x_{2}}} , with ζ = e^{\frac{2πik}{x_{1}−x_{2}}} , k = 0, …, x_{1} −x_{2} −1.

The roots λ of D_{g} are obtained in the same way as at (i) and (ii) above by replacing the pair (x_{1}, x_{2}) with the pair (y_{1}, y_{2}). Since gcd(x_{1}, x_{2}) = gcd(y_{1}, y_{2}) = 1, we see that all the nonzero roots of both D_{f} and D_{g} are simple roots. Hence, assuming that either x_{2} < x_{1}/2 or y_{2} < y_{1}/2 holds, we get that either x_{1} − x_{2} ≥ ⌈x_{1}/2⌉ or y_{1} − y_{2} ≥ ⌈y_{1}/2⌉ holds. Since we also have that deg(g) ≥ 4, we conclude that in this instance we may apply Theorem 6.3 above to our pair of polynomials f(X) and g(Y) and conclude that Equation (46) has only finitely many integer solutions (a, b) provided that we can show that the only common root of both D_{f} and D_{g} is λ = 0. Let us assume that D_{f} and D_{g} have a nonzero common root, say z. Using (ii) above, and identifying the absolute value of this nonzero common root, we get the diophantine equation

(1 − \frac{x_{2}}{x_{1}} ) · (\frac{x_{2}}{x_{1}} )^{\frac{x_{2}/x_{1}}{1−x_{2}/x_{1}}} = (1 − \frac{y_{2}}{y_{1}} ) · (\frac{y_{2}}{y_{1}} )^{\frac{y_{2}/y_{1}}{1−y_{2}/y_{1}}} . (47)

The above equation can be rewritten as h(u) = h(v), where u = x_{2}/x_{1}, v = y_{2}/y_{1}, and h(z) is the function defined on (0, 1) and given by

h(z) = (1 − z)z^{\frac{z}{1−z}} .

We shall first use the above equation to rule out some possible instances.

Case 1. u < 1/2 or v < 1/2.

In this case, it suffices to prove that the function h(z) shown above is one-to-one in the interval (0, 1). Indeed, once we have proved this fact, then the equation h(u) = h(v) will force u = v, and since both u and v are rational numbers in reduced form, we read that x_{1} = y_{1} and x_{2} = y_{2}, which is a contradiction.

To prove the injectivity of h(z) in (0, 1), note that with

k(z) = log h(z) = log(1 − z) + \frac{z log z}{1 − z} ,

we have

\frac{dk}{dz} = \frac{log z}{(1 − z)^{2}} < 0, when z ∈ (0, 1),

which proves that h(z) is one-to-one in (0, 1).

From now on, we assume that both u ≥ 1/2 and v ≥ 1/2 hold. Hence, x_{1} ≤ 2x_{2} and y_{1} ≤ 2y_{2}. Since x_{1} ≥ 3 and y_{1} ≥ 4, we conclude that both x_{2} ≥ 2 and y_{2} ≥ 2. Furthermore, let us show that y_{2} ≥ 4. Indeed, assume that y_{2} ≤ 3. Since y_{1} ≥ 4 and gcd(y_{1}, y_{2}) = 1, it follows that y_{2}/y_{1} < 1/2 except when y_{1} = 5. In this case, that is, in the case (y_{1}, y_{2}) = (5, 3), we can still have (x_{1}, x_{2}) = (4, 3), (3, 2), and the Theorem 6.3 does not apply to those cases. However, one checks by hand that the two curves obtained in this way are irreducible and of genus > 1, so we can have only finitely many solutions integer solutions (a, b) to the diophantine equation f(a) = g(b) in these cases.

Case 2. The final contradiction.

From now on, we assume that both u > 1/2 and v > 1/2, that both x_{2} and y_{2} are at least 2, and that one of them is at least 4. To deal with these remaining cases, we use a finiteness criterion of Bilu and Tichy from [5]. We follow the presentation from [4]. To use the criterion, we need to define five kinds of standard pairs of polynomials. In what follows, α and β are nonzero rational numbers, μ, ν and q are positive integers, ρ is a nonnegative integer and ν(X) ∈ ℚ[X] is a nonzero polynomial, which may be constant.

A standard pair of the first kind is a pair of polynomials of the form

(X^{q}, αX^{ρ}ν(X)^{q}),

or switched, where 0 ≤ ρ < q, gcd(ρ, q) = 1 and ρ + deg(ν) > 0.

A standard pair of the second kind is (X^{2}, (αX^{2} + β)ν(X)^{2}), or switched.

Denote by D_{μ}(X, δ) the μth Dickson polynomial, defined by the functional equation D_{μ}(z + δ/z) = z^{μ} + (δ/z)^{μ}, or by the explicit formula

D_{μ}(X, δ) = ∑_{i=0}^{⌊μ/2⌋} d_{μ,i}X^{μ−2i}  with  d_{μ,i} = \frac{μ}{μ − i}\binom{μ − i}{i}(−δ)^{i}.

A standard pair of the third kind is a pair of polynomials of the form

(D_{μ}(X, α^{ν}), D_{ν}(X, α^{μ})),

where gcd(μ, ν) = 1.

A standard pair of the fourth kind is (α^{−μ/2}D_{μ}(X, α), −β^{−ν/2}D_{ν}(X, β)), where gcd(μ, ν) = 2.

A standard pair of the fifth kind is ((αX^{2} − 1)^{3}, 3X^{4} − 4X^{3}) (or switched). The following theorem is the main result of [5]. It extends and completes Theo-

rem 6.3.

Theorem 6.4. Let f(X), g(X) ∈ ℚ[X] be nonconstant polynomials such that the equation f(a) = g(b) has infinitely many integer solutions (a, b). Then f = ϕ∘f_{1} ∘κ and g = ϕ ∘ g_{1} ∘ λ, where κ(X) ∈ ℚ[X] and λ(X) ∈ ℚ[X] are linear polynomials, ϕ(X) ∈ ℚ[X] and (f_{1}(X), g_{1}(X)) is a standard pair.

We assume that the equation f(a) = f(b) has infinitely many positive integer solutions (a, b), and we use the above Theorem 6.4 to reach a contradiction. Let us assume that f = ϕ∘f_{1} ∘κ and g = ϕ∘g_{1} ∘λ. Since, f(X) = X^{x_{1}} −X^{x_{2}} has z = 1 as a simple root, we get that ϕ cannot be a perfect power of exponent > 1 of another polynomial. Writing ϕ(X) = a_{0}(X − α_{1})^{λ_{1}} … (X − α_{t})^{λ_{t}} with t ≥ 1 distinct roots α_{1}, …, α_{t}, where α_{1} = f_{1}(κ(0)), and positive integers λ_{1}, …, λ_{t}, we get that

f(X) = X^{x_{2}}(X^{x_{1}−x_{2}} − 1) = a_{0}(f_{1}(κ(X)) − α_{1})^{λ_{1}} … (f_{1}(κ(X)) − α_{t})^{λ_{t}}.

From the above remark, we have λ_{1} = 1 if t = 1. We now show that this must always be the case. Note that the roots of f(X) are among the roots of f_{1}(κ(X))−α_{i} for i = 1, …, t, and that if t ≥ 2 and i ≠ j, then f_{1}(κ(X)) − α_{i} and f_{1}(κ(X)) − α_{j} are coprime as polynomials. Hence, they do not share any root. Thus, since α_{1} = f_{1}(κ(0)), we get that z = 0 is a root of multiplicity x_{2} of (f_{1}(κ(X)) − α_{1})^{λ_{1}}, and that all other roots have multiplicity one, and are roots of unity of order x_{1}−x_{2}. Thus, when t ≥ 2, we get at once that λ_{2} = ⋯ = λ_{t} = 1, and that if λ_{1} > 1, then f_{1}(κ(X)) − α_{1} = γX^{d} for some nonzero number γ, where d = deg(f_{1}). Identifying the multiplicity of the root z = 0, we get x_{2} = dλ_{1} and x_{1} − x_{2} = (t − 1)d, giving that d | gcd(x_{1}, x_{2}). Thus, d = 1, which is not allowed. In conclusion, even if t > 1, we must then have λ_{1} = ⋯ = λ_{t} = 1, therefore all the roots of ϕ are simple.

Assume now that (f_{1}, g_{1}) is a pair of the first or second kind. Then either f_{1}(X) = X^{q} or g_{1}(X) = X^{q}. Assume say that f_{1}(X) = X^{q}. Then f_{1}(κ(X)) − α_{1} = (γX + δ)^{q} − δ^{q} and this polynomial has z = 0 as a simple root if δ ≠ 0, and as root of multiplicity exactly q = d = deg(f) if δ = 0. The case of the simple root at z = 0 is not convenient (because x_{2} ≥ 2), and the remaining case leads to d = x_{2}, and dt = x_{1}, which is again impossible because it gives x_{2} | x_{1}. The case when g_{1}(X) = X^{q} can be dealt with analogously.

Assume now that (f_{1}, g_{1}) is a pair of the third, fourth or fifth kind. We know that min{x_{2}, y_{2}} ≥ 4. Then either f_{1}(κ(X)) − α_{1} or g_{1}(λ(X)) − α_{s} has a root of multiplicity at least four, where s ∈ {1, …, t} is such that α_{s} = g_{1}(λ(0)). But this implies that either (f_{1}(κ(X)) − α_{1})′ = f′_{1}(κ(X))κ′(X) or (g_{1}(λ(X)) − α_{s})′ = g′_{1}(λ(X))λ′(X) has a triple root. Since κ and λ are nonconstant linear polynomials, we conclude immediately that this last fact is equivalent to the fact that either f′_{1}(X) has a triple root or g′_{1}(X) has a triple root. It is clear that this is not so if (f_{1}, g_{1}) is a pair of the fifth kind. We now verify that the derivative of any Dickson polynomial does not have double roots, therefore it does not have triple roots either. Setting x_{1} = x/\sqrt{δ}, we easily get that D_{μ}(x, δ) = δ^{μ}D_{μ}(x_{1}, 1), therefore it is enough to verify our claim when δ = 1. But since the roots D_{μ}(X, 1) are precisely 2 cos πk/m for k = 0, …, m − 1, which are all real and simple, it follows, by Rolle’s Theorem, that the roots of its derivative D′_{μ}(X) are also real and simple.

This completes the proof of Theorem 6.2 and of Theorem 6.1 as well. □

□

7. Comments and remarks

It would certainly be of interest to extend the results of this paper in order to cover a wider class of equations of the same type as (5). For example, it would be interesting to relax the condition ‘a is a prime’, to, say, ‘a is an integer’ (or, even, to ‘a has a bounded number of prime factors’), or to replace the condition ‘b is fixed’ by the condition ‘b is an S-unit’. We have not succeeded in proving any of such results. The most difficult point seems to lie in Step 3 of Section 5.

Furthermore, we stress that, as in [11], our results are ineffective, since they ultimately depend on the Schmidt Subspace Theorem. It would be very interesting to provide an effective version of even a weaker form of our main theorem.

Moreover, it would be nice to relax the coprimality condition occurring in Theorem 2.1. This, however, seems to be quite difficult.

References

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Université Louis Pasteur, UFR de mathématiques, 7 rue René Descartes, 67084 Strasbourg, France bugeaud@math.u-strasbg.fr

Instituto de Matemáticas, Universidad Nacional Autónoma de México, C.P. 58089, Morelia, Michoacán, México fluca@matmor.unam.mx

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