The Prouhet–Tarry–Escott problem, indecomposability of polynomials and Diophantine equations

In this paper, we show how the subjects mentioned in the title are related. First we study the structure of partitions of A⊆{1,⋯,n}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A \subseteq \{1, \dots , n\}$$\end{document} in k-sets such that the first k-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$k-1$$\end{document} symmetric polynomials of the elements of the k-sets coincide. Then we apply this result to derive a decomposability result for the polynomial fA(x):=∏x∈A(x-a)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f_A(x) := \prod _{x \in A} (x-a)$$\end{document}. Finally we prove two theorems on the structure of the solutions (x, y) of the Diophantine equation fA(x)=P(y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f_A(x)=P(y)$$\end{document} where P(y)∈Q[y]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P(y)\in \mathbb {Q}[y]$$\end{document} and on shifted power values of fA(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f_A(x)$$\end{document}.


Introduction
The Prouhet-Tarry-Escott problem, shortly PTE, asks to describe disjoint pairs A and B of sets of integers such that their first k power sum symmetric polynomials are equal (cf. [18]). For example, if A = {2, 3, 7} and B = {1, 5, 6} then we can take k = 2, since we have 2 + 3 + 7 = 1 + 5 + 6 and 2 2 + 3 2 + 7 2 = 1 2 + 5 2 + 6 2 . (1) In this paper, we connect the PTE problem, and the question for which polynomials f (x), g(x) ∈ Z[x] the equation f (x) = g(y) has infinitely many solutions (x, y) ∈ Z 2 if the zeros of f are simple and form (almost) an arithmetical progression. Both problems have attracted a lot of attention. Already at this point we mention that the latter question (through a deep result of Bilu and Tichy [4]) is closely related to decomposability of polynomials.
Problem 1 was solved independently by Aart Blokhuis (private communication) and by the third author of the present paper (see Corollary 3.1).
In this paper, we study the following more general problem.
The problem asks: is it possible to omit a 'few' elements from the set {1, . . . , n} such that the remaining set can be split into t subsets which have pairwise the PTE-property? Observe that Problem 1 is the special case r = 1, t = 2. In Theorem 2.1, we show that if r is small enough with respect to n, then only k = 2 is possible and A 1 , A 2 , . . . , A t are symmetric. We call a set A = {a 1 , . . . , a k } ⊂ R with a 1 < · · · < a k symmetric if the sums a i + a k+1−i (i = 1, . . . , k) are all equal. It is obvious that such a symmetry implies a PTE-structure.
Next we establish a new link between PTE problems and the indecomposability of certain polynomials. We recall some standard notions. Let K be a field and is another decomposition of f (x). In the sequel we do not distinguish between such equivalent decompositions. Further, we consider the polynomials f (x), f (λ(x)), as well as the polynomials f (x), λ( f (x)) to be equivalent. There is a vast literature on (in)decomposability of polynomials (see, e.g., [2,4,5,7,11,12,19] and the references there). In Theorem 2.2, we show that the studied variant of the PTE problem is equivalent to asking for the indecomposability of certain polynomials.
Using this connection, we show in Corollary 2.1 for given integers n > r ≥ 0 with r small enough with respect to n that if for A ⊆ {1, . . . , n} with |A| = n − r the polynomial is decomposable over Q as h 1 (h 2 (x)), then h 1 and h 2 can be given explicitly. Note that the polynomial f A,c,d (x) represents the product with terms of an arithmetic progression of length n with r terms missing. For example, if is decomposable as h 1 (h 2 (x)), then, apart from equivalence, Next, using the above results, we establish a finiteness theorem for the number of times that a polynomial f A,c,d of the form (3) assumes a value which is also assumed by a given polynomial P with rational coefficients. Related problems are investigated in the papers [2,3,16,26] for consecutive integers and in [15] for arithmetic progressions with at most one term missing. Generalizing and extending many of the above mentioned results, in Theorem 2.3, we provide a finiteness result for the number of values of f A,c,d also taken by another polynomial P(x) ∈ Q[x]. This result, similarly to the above mentioned ones, is ineffective.
Finally, we consider shifted power values (i.e., values of the shape ay + b) of f A,c,d . Related problems have been investigated by many authors. We recall some important results. (For a more detailed survey see, e.g., the introduction of [15].) A celebrated result of Erdős and Selfridge [10] says that the product of two or more consecutive positive integers is never a perfect power. Papers of Erdős [9] and Győry [14] give similar results for binomial coefficients. A recent result of Bennett and Siksek [1] states that if the number k of terms of the arithmetic progression is fixed and large enough, then there are only finitely many instances that the product yields a perfect power. For results with r = 1 (just one term missing), see, e.g., [20,21] (for consecutive integers) and [22] (for general arithmetic progressions). In the equation f A,c,d (x) = ay + b we give an effective upper bound for the exponent and for the integer values x, y for which this equation holds, in Theorem 2.4. This result implies for example that for every integer n ≥ 24 and rational numbers a, b with a = 0 there exists an effectively computable number C such that the equation f A (x) = ay + b with A ⊂ {1, 2, . . . , n}, |A| = n − 2 implies max(|x|, |y|, ) < C.
Our results make a step forward toward the solution of the problem how much one can 'mutilate' an arithmetic progression such that the corresponding product of terms still can take only finitely many values of a given polynomial, or shifted power values.

Results
In connection with Problem 2, we prove the following result.
Remark 1 Theorem 2.1 yields a complete answer to Problem 2 for every n > 2r 3/2 + 5r + 8. On the other hand, for any r and n with n − r even, if A = {1, . . . , n} \ A 0 is symmetric with respect toā (i.e., a ∈ A implies that 2ā − a ∈ A), then we have a partition as in Problem 2 with k = 2.
Indeed, writing b s = c + da s with a s ∈ A \ A 0 and c, d ∈ Q, d = 0, it can be easily seen by induction on n that c can be taken to be zero. Then clearly, we may take d = 1, and the claim follows.
The next result establishes a link between partitions as in Problem 2 and decomposability of certain polynomials. and

Remark 3
From the proof of the theorem, it will be clear that in fact h 2 is independent of which A i we use in its definition.
As a simple consequence of Theorems 2.1 and 2.2 we obtain the following statement.
. . , n} with |A| = n − r where n and r are integers with r ≥ 0 and n > 2r 3/2 + 5r is decomposable over Q if and only if n − r is even and A is symmetric with respect toā

when (up to equivalence) the only decomposition of f A,c,d (x) is given by f
Here, h 1 is the polynomial defined in Theorem 2.2 corresponding to the partition Next we apply our results to the equation f A,c,d (x) = P(y) where P is a given polynomial. The first theorem of this type is general, but ineffective: it only guarantees the finiteness of the number of integral solutions.
. . , n} with |A| = n − r for integers r ≥ 0 and n > 2r 3/2 + 5r + 8 and let c, d ∈ Q with d = 0. Let f A,c,d (x) be as in (7) and let P(y) ∈ Q[y] with deg P ≥ 2. Then the equation has only finitely many integer solutions x, y, unless we are in one of the following cases: where T is an arbitrary non-constant polynomial with rational coefficients, (ii) P(y) = ϕ * (Q(y)), where ϕ * is given by (8) and Q is a non-constant polynomial with rational coefficients having at most two roots of odd multiplicities.

Remark 4
In cases (i) and (ii) one can easily give examples where equation (9) has infinitely many integer solutions x, y.
If the right hand side of (9) is of the shape ay + b where is also unknown, then we can give an effective result.
. . , n} with |A| = n − r with integers r ≥ 0 and n > 2r 3/2 + 5r + 8 and let c, d ∈ Q with d = 0. Let f A,c,d (x) be given by (7) and let a, b ∈ Q with a = 0. Then all solutions of the equation in integers x, y, with ≥ 2 satisfy max(|x|, |y|, ) < C for some effectively computable constant C depending only on a, b, c, d, n. Here we use the convention that for |y| ≤ 1 we have ≤ 3.
In particular, we have n − r = tk. Observe that (4) implies that n − 1 > 2(r − 1) 3/2 + 5(r − 1) + 8 for r > 0. Therefore, by induction on r , we may assume without loss of generality that n ∈ A. We shall frequently use the identity valid for all j ≥ 0. Also, we shall make use of the fact that it follows from the conditions of the theorem by induction on h that As we shall see, our choice of will depend on the parity of k.
Suppose first k is odd. Then k ≥ 3. We choose = −r − 2 in (11) and let Since deg f = k − 1, by our assumptions we have Recall that A = A 1 ∪ · · · ∪ A t and n ∈ A. Observe that Thus we have Hence for any j with 1 ≤ j ≤ t we get On the other hand, assuming without loss of generality that n ∈ A 1 , we also have Combining (12) and (13), we obtain Since k ≥ 3, we can rewrite this inequality as We show that it is impossible. On the one hand, in view of k ≤ (n − r )/2 and (4), we have On the other hand, we get from (4) that This yields a contradiction, which proves our claim for k odd. Suppose k is even and k ≥ 4. Here we choose = −2r − 2 in (11) and let Since deg f = k − 1, by our assumptions we have Observe that the negative values of f are Thus we have Furthermore, Hence for any j with 1 ≤ j ≤ t we get On the other hand, assuming without loss of generality that n ∈ A 1 , we also have, by , Combining (16) and (17), we obtain Since k ≥ 4, we can rewrite this inequality as We show that it is impossible. In view of k ≤ (n − r )/2 and (4), we have On using k ≥ 4 and writing m = n − 2r , it follows that Since (4) implies m > 2r 3/2 + 3r + 8, this yields a contradiction. Finally, let k = 2. Then we have t = (n − r )/2; in particular, n − r is even. That is, we have pairs of elements of A having the same sum. Obviously, this is possible only if we take the largest number with the smallest one, and so on, so the pairs are symmetric with respect toā.

Corollary 3.1 The only solution of Problem 1 with k > 2 is for n = 7.
Proof We apply the proof of Theorem 2.1 with r = 1 and t = 2. It follows that k = 2 if n > 15. On the other hand, n has to be odd and if k > 2, then n ≥ 7. Hence it remains to check the odd values of n between 7 and 15.
If n = 15, 13, 11 or 9, then k = (n−1)/2 and we apply (11) with h = k−1, = −3. If n = 9, then the largest coefficient 2k−2 k−1 is larger than the sum of the absolute values of the other 2k binomial coefficients. Hence the sums in (11) cannot be equal. If n = 9, the largest binomial coefficient, 6 3 = 20, is equal to the sum of the absolute values of the other terms. It follows that 9, 1, 2 ∈ A 1 and 8, 7, 6 ∈ A 2 . However, when applying (11) with h = 1, = −3 we see that the sum of the elements in A 2 exceeds that of A 1 for all possible choices of the remaining elements 3, 4 and 5.

Remark 5
Remark 2 implies that the symmetric polynomials σ 1 , σ 2 of 1, 2, 6 and of 0, 4, 5, and also of 3, 5, 13 and of 1, 9, 11, coincide too. A 0 , A 1 , . . . , A t be a partition as stated in Problem 2. Put A = {1, . . . , n} \ A 0 and let f A , h 1 , h 2 be as in the theorem. We want to show that f A (x) = h 1 (h 2 (x)). If two polynomials of degree n − r have the same values at n − r + 1 points, then they coincide. It is clear that

Proof of Theorem 2.2 Let
and that both f A (x) and h 1 (h 2 (x)) have all a ∈ A 1 as roots. In view of for 2 ≤ i ≤ t, we see that every a ∈ A is both a root of f A (x) and of h 1 (h 2 (x)). Thus f A (x) and h 1 (h 2 (x)) assume the same value at n − r + 1 points, hence f A = h 1 (h 2 ). This proves the "only if" statement and the second statement of the theorem.
To prove the "if" statement, let A ⊆ {1, . . . , n} with |A| = n − r , and suppose that . Clearly, we may assume that both h 1 and h 2 are monic polynomials. Set h 1 Observe that these roots are pairwise distinct. Then Let A i consist of the roots of the polynomial h 2 (x) − α i (i = 1, . . . , t). Then all the symmetric polynomials of the elements of A i for i = 1, . . . , t coincide. So putting A 0 = {1, . . . , n} \ A, the sets A 0 , A 1 , . . . , A t form a partition as in Problem 2. To get the specific decomposition observe that

Proof of Corollary 2.1 Clearly, by
we obtain the decomposition as given in the theorem.
To prove the uniqueness let f A (x) = P(Bx 2 + C x + D) any decomposition of ] and B, C, D ∈ Q, B = 0. Without loss of generality we may assume B = 1. Then f A (x) = P (x + C/2) 2 + D − C 2 /4 . Hence the roots of f A form a symmetric set with respect to −C/2, but they also form a symmetric set with respect to a. Thus C = −2a. This proves the uniqueness.

Proofs of results on Diophantine equations
We start with the proof of Theorem 2.4. For this, we introduce some notation and state three lemmas.
Let f (x) ∈ Z[x] of degree d and height (i.e, the maximum of the absolute values of the coefficients) H , and let a be a non-zero integer. Consider the equation (19) in x, y, ∈ Z with ≥ 2. The next lemma is due to Schinzel and Tijdeman [23]. Actually already Tijdeman [25] suffices.

Lemma 5.1 Suppose that f (x) has at least two different roots. Then for all solutions
x, y, of (19) with |y| > 1 we have

, H ) is an effectively computable constant depending only on a, d and H .
The second lemma is a result of Brindza [6]. Let S be a finite set of primes, and write Z S for the set of those rational numbers whose denominators have no prime divisors outside S. For a rational number q (given in its minimal form), by its height h(q) we mean the maximum of the absolute values of its denominator and numerator.
Then for any finite set S of primes, for the solutions x, y ∈ Z S of (19) we have max(h(x), h(y)) < C 2 .
Here C 2 = C 2 (a, , d, H , S) is an effectively computable constant depending only on a, , d, H , S.
Finally, we formulate a statement taking care of the cases r ≤ 1.

Lemma 5.3
Let k, j be integers with k ≥ 8 and 1 ≤ j ≤ k, and put Further, let a, b ∈ Q with a = 0. Then for all solutions of the equation in integers x, y, with ≥ 2 we have max(|x|, |y|, ) < C 3 , where C 3 is an effectively computable constant depending only on k, a, b. Here we use the convention that for |y| ≤ 1 we have ≤ 3.
Proof In case of j = 1 or j = k, the statement follows from the main result of [26], while in the other cases it is a consequence of Theorem 2.2 of [15]. Now we are ready to give the proof of our effective result. A, c, d, a, b in integers x, y, with ≥ 2. Our proof relies on Lemmas 5.1 and 5.2, hence ultimately on the multiplicities of the roots of f A,c,d (x) and its shifts f A,c,d (x) − b. Thus as by a simple rational substitution and multiplication by appropriate rationals we can transform f A,c,d (x) into f A (x), we may consider f A (x) in place of f A,c,d (x). In view of Lemma 5.3 and n ≥ 9, we may assume r = n − |A| ≥ 2 as well.

Proof of Theorem 2.4 Consider (10) with fixed
As all the roots of f A (x) are simple and real, the same is valid for the polynomial f A (x), and consequently for ( f A (x) − b) . Thus the polynomial f A (x) − b can have at most double roots. Since its degree is n − r ≥ 22, the statement immediately follows from Lemmas 5.1 and 5.2, unless = 2 and f A (x) is of the form with some p, q ∈ Q[x] with deg p ≤ 2. In particular, has the same parity as deg p has. Write a 1 < · · · < a N for the elements of A. Taking derivatives, (20) gives Let α 1 , . . . , α N −1 be the roots of f A (x). Then by Rolle's theorem these are distinct real numbers with We only consider the case deg p = 2. In fact, it is the most complicated possibility, the other cases are simpler and can be handled similarly. Then clearly, deg q = N /2−1, and (21) shows that the roots of q(x) are among the α i -s. Further, (20) implies that for these α i -s we have f A (α i ) = b. Observe that, by (6), f A (α i ) < 0 for i odd, while f A (α i ) > 0 for i even. Altogether, we have two options: (a) either the roots of q(x) are given by α 2 , α 4 , . . . , α N −2 (i.e., all the roots with even indices are involved), (b) or the N /2 − 1 roots of q(x) are among α 1 , α 3 , . . . , α N −1 (that is, all the roots with odd indices with one exception are involved). Put and set Observe that |A * | ≥ N − r − 2 and with deg H ≤ r + 2. Further, among the (not disjoint) quadruples {2i − 2, 2i − 1, 2i, 2i + 1} (i = 2, . . . , (n − 1)/2 ) at least n/2 − 2 − 2r are subsets of A. So by (4), there is a quadruple 2i − 2, 2i − 1, 2i, 2i + 1 contained in A, such that H (x), and thus G(x) has no root in some interval (2i, 2i + 1). However, then the sign of G(x) does not change in this interval.
Here we use that α 2i−2 is the maximum of f A on (2i − 2, 2i − 1). If G(x) < 0 for x ∈ (2i, 2i + 1) then f A (x − 2) < f A (x) and choosing x = α 2i−2 + 2 the same reasoning gives Hence in both cases which shows that the option (a) above concerning the roots of q(x) is not possible. On the other hand, among the quadruples at least n/2 − 2r − 2 are subsets of A. So by (4), there are three quadruples as above contained in A, such that H (x), and thus G(x) has no root in three distinct intervals (2i j + 1, 2i j + 2) ( j = 1, 2, 3). Similarly to (22) we obtain which shows that the option (b) above concerning the roots of q(x) is also impossible.
Now we give the proof of Theorem 2.3. For this we need some more results and notation.
Let δ be a non-zero rational number and μ be a positive integer. Then is the μ-th Dickson polynomial. For properties of these polynomials see, e.g., [17]. We shall use a deep result of Bilu and Tichy [4] concerning equations of the type in integers x, y, where f , g are polynomials with rational coefficients. To describe this result, we introduce some notation. We say that F, G ∈ Q[x] form a standard pair over Q if either (F(x), G(x)) or (G(x), F(x)) appears in Table 1. Now we recall the main result of [4], which will play a key role in the proof of Theorem 2.3. Table 1 Standard pairs. Here α, β are non-zero rational numbers, μ, ν, q are positive integers, p is a non-negative integer, v(x) ∈ Q[x] is a non-zero, but possibly constant polynomial

Kind
Standard pair Parameter restrictions If deg ϕ = n − r then deg F = 1, and we easily get that we are in case (i) of Theorem 2.3.
Suppose deg ϕ = (n − r )/2. Then we have deg F = 2. By Corollary 2.1 the decomposition is given up to a linear transformation: f A (x) = ϕ * ((x − a) 2 ). If we have infinitely many solutions then by Lemma 5.4 we have ϕ * ((x − a) 2 ) = P(y) = ϕ * (G(y)) for some G(y) ∈ Q[y] such that (x − a) 2 = G(y) has infinitely many solutions. Lemma 5.2 implies that we must be in case (ii).
Finally, consider the case deg ϕ = 1. Then deg F = n − r , and we have where F is a member of a standard pair. We check the possible cases. As deg f A ≥ 24, F cannot come from a standard pair of the fifth kind. Since we assumed that deg P ≥ 3, the polynomial F cannot belong to a standard pair of the second type, either.
Assume that F belongs to a standard pair of the first kind. Since all the zeros of f A (x) are real and simple, hence by Rolle's theorem all the roots of f A (x) are real and simple, F(x) = x q is not possible. On the other hand, if with some s 1 , s 2 ∈ Q, s 1 = 0. Using again that the roots of f A (x) are simple, we get q ≤ 2. However, then in view of that the other term in the standard pair in question is x q , we see that deg P ≤ 2, which is excluded.
Finally, assume that F belongs to a standard pair of the third or fourth kind. Then f A (x) should be a linear transform of a Dickson polynomial. More precisely, with some rationals s 1 , s 2 , t 1 , t 2 (s 1 t 1 = 0) and non-negative integer N we can write where D N (x, δ) is the N -th Dickson polynomial, with non-zero parameter δ ∈ Q. (Here we apply the inside and outside linear transformations to f A rather than to D N . In fact, writing f A = ϕ • D N • λ, t 1 x + t 2 and s 1 x + s 2 are the inverses of the linear polynomials ϕ(x) and λ(x), respectively.) Observe that here N = deg f A (x) = |A| must hold. Then, by the well-known identity (see. e.g., formula (2.2) on p. 9 of [17]) Hence as |A| = N , a∈A y 2 + s 2 − a s 1 y + δ = y 2N − t 2 y N + δ N follows. Here we used by comparing the leading coefficients, that t 1 s N 1 = 1 must hold. Write ζ, ξ for the roots of the polynomial y 2 − t 2 y + δ N . Clearly, ζ, ξ are algebraic numbers of degree at most two. Further, we have a∈A y 2 + s 2 − a s 1 y + δ = (y N − ζ )(y N − ξ).
If ζ 0 , ξ 0 are roots of y N − ζ and y N − ξ , respectively, then all the roots of these polynomials are given by ζ 0 ε i and ξ 0 ε i (i = 0, 1, . . . , N − 1), respectively, where ε is a primitive N -th root of unity. By (25) we see that all these roots are algebraic numbers of degrees at most two. This immediately gives that the degree of ε is at most four, hence ϕ(N ) ≤ 4. We conclude that N ≤ 12. This contradicts (24).
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