Diophantine triples in linear recurrences of Pisot type

The study of Diophantine triples taking values in linear recurrence sequences is a variant of a problem going back to Diophantus of Alexandria which has been studied quite a lot in the past. The main questions are, as usual, about existence or finiteness of Diophantine triples in such sequences. Whilst the case of binary recurrence sequences is almost completely solved, not much was known about recurrence sequences of larger order, except for very specialised generalisations of the Fibonacci sequence. Now, we will prove that any linear recurrence sequence with the Pisot property contains only finitely many Diophantine triples, whenever the order is large and a few more not very restrictive conditions are met.

This result is even effective, since an upper bound of the form log 10 (log 10 (max{a i })) < 26 was given on the members of such a quintuple. It is conjectured, that there are no quintuples at all and, even stronger, that if {a, b, c, d} is a Diophantine quadruple and d > max{a, b, c}, then d = d + . The "weaker" conjecture has recently been settled by He, Togbé and Ziegler (cf. [19]), whereas the stronger conjecture still remains open. Now it is an interesting variation of the original problem of Diophantus to consider a linear recurrence sequence instead of the sequence of squares. So we ask for bounds m on the size of tuples of integers {a 1 , a 2 , a 3 , . . . , a m } with a i a j + 1 being members of a given linear recurrence for 1 ≤ i < j ≤ m. We shall call this set a Diophantine m-tuple with values in the linear recurrence (or a Diophantine m-tuple in the recurrences, for short). Here, the first result was due to Fuchs, Luca and Szalay, who proved in [12] that for a binary linear recurrence sequence (u n ) n≥0 , there are only finitely many Diophantine triples, if certain conditions are met. The Fibonacci sequence and the Lucas sequence both satisfy these conditions and all Diophantine triples with values in these sequences were computed in [22] and [23]. Further results in this direction can be found in [2,20] and [21]. Moreover, in [1] it is shown that there are no balancing Diophantine triples; see also [3] for a related result. In [4] it is shown that there are no Diophantine triples taking values in Pellans sequence.
The first result on linear recurrence sequences of higher order than 2 came up in 2015, when the authors jointly with Irmak and Szalay proved (see [13]) that there are only finitely many Diophantine triples with values in the Tribonacci sequence (T n ) n≥0 given by T 0 = T 1 = 0, T 2 = 1, T n+3 = T n+2 + T n+1 + T n for n ≥ 0.
In [17] it was shown that a Tribonacci Diophantine quadruple does not exist. A related result can be found in [18]. One year later in [14], this result was generalized to kgeneralized Fibonacci sequences: For any integer k ≥ 3, define (F The result in this paper deals with a significantly larger class of linear recurrence sequences: Let (F n ) n≥0 be a sequence of integers satisfying a linear recurring relation. Assume that the recurrence is of Pisot type, i.e., that its characteristic polynomial is the minimal polynomial (over Q) of a Pisot number. We denote the power sum representation (Binet formula) by F n = f 1 α n 1 + · · · + f k α n k . Assume w.l.o.g. that α = α 1 is the Pisot number; i.e., α is a real algebraic integer of degree k satisfying α > 1 and if α 2 , . . . , α k denote the conjugates of α over Q then max{|α 2 |, . . . , |α k |} < 1. We remark that by a result of Mignotte (cf. [24]) it immediately follows that the sequence is non-degenerate, and that the characteristic roots are all simple and irrational.
We show that there are only finitely many triples of integers 1 ≤ a < b < c such that 1 + ab = F x , 1 + ac = F y , 1 + bc = F z if at least one of the following conditions holds: • Neither the leading coefficient f 1 nor f 1 α is a square in K = Q(α 1 , . . . , α k ).
• k ≥ 2 and α is not a unit.
The previously treated k-generalized Fibonacci sequences satisfy this Pisot property and neither their leading coefficient f 1 nor f 1 α 1 is a square. However, the new result in this paper helps us to obtain finiteness for many more linear recurrence sequences.
For example, let us consider the irreducible polynomial X 3 − X − 1, which has the Pisot property. Its Pisot root θ := 1.3247179572 . . . is the smallest existing Pisot number by [6]. This number is also known as the plastic constant. Its corresponding linear recurrence sequence (F n ) n≥0 , given by F n+3 = F n+1 + F n , is of Pisot type. If the initial values are not F 0 = 6, F 1 = −9, F 2 = 2, then neither the leading coefficient nor the leading coefficient times θ are squares in the splitting field of X 3 −X −1 over Q. So the theorem can be applied and we obtain, that there are only finitely many Diophantine triples with values in this sequence. However it is yet not clear, what happens in the case F 0 = 6, F 1 = −9, F 2 = 2.
Another example for which the theorem can be applied is the polynomial This polynomial defines a Pisot number of degree 2k + 1 by a result of Siegel (see [25]) and its corresponding linear recurrence sequence is of Pisot type. Independently of its initial values, the result applies to all k ≥ 2 since the degree is sufficiently large. The same applies to Furthermore, all polynomials of the form are known to define Pisot numbers. So, again for k ≥ 2 the theorem applies. We quickly discuss the main shape of the recurrences we study in this paper. Let (F n ) n≥0 be a recurrence of Pisot type as described above. Let us denote K = Q(α 1 , . . . , α k ). Since F n ∈ Z it follows that each element of the Galois group of K over Q permutes the summands in the power sum representation of F n . Moreover, each summand is a conjugate of the leading term f 1 α n 1 over Q and each conjugate of it appears exactly once in the Binet formula. Therefore F n is just the trace Tr K /Q (f 1 α n 1 ). Since f 1 might not be integral, we write f 1 = f /d with d ∈ Z and f being an integral element in K . Thus, conversely starting with a Pisot number α, an integer d ∈ Z and an integral element f in the Galois closure K of α over Q such that dF n = Tr K /Q (f α n ) for every n ∈ N, we can easily construct further examples for which our result applies.
The proof will be given in several steps: First, a more abstract theorem is going to be proved, which guarantees the existence of an algebraic equality, that needs to be satisfied, if there were infinitely many Diophantine triples. This works on utilizing the Subspace theorem (cf. [10]) and a parametrization strategy in a similar manner to that of [14]. If the leading coefficient is not a square in Q(α 1 , . . . , α k ), we obtain the contradiction quite immediately from this equality. In a second step, we will use divisibility arguments and algebraic parity considerations in order to show that this equality can also not be satisfied if the order k is large enough. Let us now state the results.

The results
We start with a general and more abstract statement which gives necessary conditions in case infinitely many Diophantine triples exist. It is derived by using the Subspace theorem (cf. [10]). Theorem 1 Let (F n ) n≥0 be a sequence of integers satisfying a linear recurrence relation of Pisot type of order k ≥ 2. Denote its power sum representation as If there are infinitely many positive integers 1 < a < b < c, such that hold for integers x, y, z, then one can find fixed integers (r 1 , r 2 , r 3 , s 1 , s 2 , s 3 ) with r 1 , r 2 , r 3 positive, gcd(r 1 , r 2 , r 3 ) = 1 such that infinitely many of the solutions (a, b, c, x, y, z) can be parametrized as Furthermore, following the parametrization of x, y, z in , there must exist a power sum c( ) of the form with η ∈ Z ∪ (Z + 1/2), J c an index set, e j being coefficients in Q(α 1 , . . . , α k ) and integers v ij with the property that v ij ≥ 0 if i ∈ {2, . . . , n} and v ij < 0 if i = 1, all independent of , such that Similarly there are a( ) and b( ) of the same shape with The proof is given in Sect. 4. This theorem looks quite abstract. However, it can be applied to a huge family of linear recurrences. Firstly, it can be applied to all linear recurrences, in which the leading coefficient is not a square: Theorem 2 Let (F n ) n≥0 be a sequence of integers satisfying a linear recurring relation F n+k = A 1 F n+k−1 + A 2 F n+k−2 + · · · + A k F n of Pisot type of order k ≥ 2, that is, the characteristic polynomial is an irreducible polynomial of degree k, has integer coefficients A i , and has roots satisfying α 1 > 1 and max{|α 2 |, . . . , |α k |} < 1. If furthermore neither f 1 nor f 1 α 1 are squares in Q(α 1 , . . . , α k ), then there are only finitely many Diophantine triples with values in {F n ; n ≥ 0}.
The proof of this theorem is given in Sect. 5.
Another consequence of Theorem 1 applies to linear recurrences of sufficiently large order. Namely if k ≥ 4, the existence of such a c( ) leads to a contradiction. The same holds already for k = 2, 3, if we assume that the Pisot element α 1 is not a unit in the ring of integers of Q(α 1 , . . . , α k ). Thus, we obtain the following result.
Theorem 3 Let (F n ) n≥0 be a sequence of integers satisfying a linear recurring relation F n+k = A 1 F n+k−1 + A 2 F n+k−2 + · · · + A k F n of Pisot type of order k ≥ 2, that is, the characteristic polynomial is an irreducible polynomial of degree k, has integer coefficients A i , and has roots satisfying α 1 > 1 and max{|α 2 |, . . . , |α k |} < 1. Then there are only finitely many Diophantine triples with values in {F n ; n ≥ 0} if one of the following conditions holds: This theorem is proved in Sect. 6. Before we give the proofs we first start with several useful lemmas that will be used in the sections afterwards.

Some useful lemmas
Assume that we have infinitely many solutions (x, y, z) ∈ N 3 to (1) with 1 < a < b < c. Obviously, we have x < y < z. First, one notices that not only for z, but for all three components, we necessarily have arbitrarily "large" solutions.
Lemma 1 Let us assume, we have infinitely many solutions (x, y, z) ∈ N 3 to (1). Then for each N , there are still infinitely many solutions (x, y, z) ∈ N 3 with x > N .
Proof It is obvious that we must have arbitrarily large solutions for y and for z, since otherwise, a, b, c would all be bounded as well, which is an immediate contradiction to our assumption.
If we had infinitely many solutions (x, y, z) with x < N , then there is at least one fixed x which forms a solution with infinitely many pairs (y, z). Since F x = ab + 1, we have a bound on these two variables as well and can use the same pigeon hole argument again to find fixed a and b, forming a Diophantine triple with infinitely many c ∈ N.
Using these fixed a, b, we obtain from the other two equations in (1), that bF y − aF z = b − a and therefore, the expressions bf 1 α y 1 and af 1 α z 1 (having the largest growth rate) must be equal. So which is a constant. Hence, z − y must be some constant ρ > 0 as well and we can write z = y + ρ for our infinitely many solutions in y and z.
Using the power sum representations in bF y − aF y+ρ = b − a, we get So the terms with the largest growth rate, which are bf 1 α y 1 and af 1 α y+ρ 1 , must be equal and this gives us b = aα ρ 1 . Inserting this into (2) and cancelling on both sides gives us Now for y → ∞, the left hand side converges to 0. The right hand side is a constant larger than 0. So this equality can not be true when y is large enough. This contradiction completes the proof.
Next, we prove the following result, which generalizes Proposition 1 in [13]. Observe that the upper bound depends now on k.

Lemma 2 Let y < z be sufficiently large. Then there is a constant C
(3) . Observe here and below that the numbers F x − 1, F y − 1, F z − 1 are positive integers. Furthermore, let us assume that y (and hence z) is large enough such that Let κ be a constant to be determined later. If y ≤ κz, then Now let us assume that y > κz. We denote λ : Thus, we can write where π is some algebraic integer in Q(α 1 ). Note that the right-hand side above is not zero, for if it were, we would get α λ 1 = (F z − 1)/(F y − 1) ∈ Q, which is false for λ > 0. We compute norms from Q(α 1 ) to Q. Observe that Further, let σ i be any Galois automorphism that maps α 1 to α i . Then for i ≥ 2, we have with C 2 being a suitable constant (e.g. ). Altogether, we obtain In order to balance between (4) and (5), we choose κ such that κ = 1−κ/k, giving κ = k/(k + 1) and The next lemma states the irreducibility (over C) of a certain polynomial. This lemma will be used in the proof of Theorem 3.

Lemma 3
Assume that k ≥ 1. For n ≥ 3, and non-zero complex numbers c 1 , . . . , c n the polynomial . . , d k are all the roots of z k + c 2 /c 1 = 0. This polynomial is square-free, that is it does not have multiple factors of degree ≥ 1. In particular, for n = 3, is such that P(X 2 , X 3 ) = c 2 X k 2 + c 3 X k 3 is square-free. Let p be some irreducible factor of P(X 2 , X 3 ). Then the polynomial above is Eisenstein with respect to p (since p 2 does not divide P(X 2 , X 3 )), so the polynomial is irreducible. Now for n ≥ 4 we apply induction on n noting that where P(X 2 , . . . , X n ) = c 2 X k 2 + · · · + c n X k n is irreducible for n ≥ 4 (by the induction hypothesis), so our polynomial is Eisenstein with respect to the prime p := P(X 2 , . . . , X n ). This proves the lemma.
Proof Indeed, for if not, the homogenized polynomial Now we need to deal with the case when we have a Laurent-polynomial which looks as follows Clearing up the powers of X i from the denominators and calculating P − 1, it will be necessary for the proof of Theorem 3 to look at Lemma 4 Assume that k ≥ 1. Let n ≥ 3 and c 1 , . . . , c n be non-zero complex numbers. Then Proof We rewrite the polynomial as By Capelli's theorem, the given polynomial is irreducible if we succeed to show that: , then α is not of the form β q for some element β ∈ C(X 2 , . . . , X n )(α) and any q | k.
We consider it easier to work with the reciprocal polynomial where c 1 = −c 1 c n+1 , we can work with g(X) instead of f * (X). Note that (i) and (ii) hold for f (X) if and only if they hold for g(X). So, let us check parts (i) and (ii). Part (i) is easy. We just compute the discriminant of g(X): We show that the polynomial in parenthesis is square-free. Assume p 2 is a divisor of it for some irreducible polynomial p of positive degree. Putting and taking derivatives with respect to X 2 , we get that p divides Clearly, since p is irreducible, it is coprime to X 2 , . . . , X n and H, so p must divide H + 2c 2 X k 2 = (3c 2 )X k 2 + c 3 X k 3 + · · · + c n X k n − 1 and by Corollary 1, it must be associated to this last polynomial since this is irreducible. Since n ≥ 3, the same argument using the partial derivative with respect to X 3 instead gives that p is associated to c 2 X k 2 + (3c 3 )X k 3 + · · · + X k n − 1 as well, a contradiction. This proves (i). For part (ii), note that Further, from what we proved above, is square-free as a polynomial in C[X 2 , . . . , X n ]. Let L := C(X 2 , . . . , X n ) and x = (X 2 , . . . , X n ). We need to show that α is not of the form β q for some prime q | k and β ∈ L(α). Assume there is such β and let it be . , X n ], and since A(x) ∈ L, the fraction field of this last ring, it follows that A(x) ∈ C[X 2 , . . . , X n ]. Now the element βγ = A(x) 2 − B(x) 2 is also integral over C[X 2 , . . . , X n ], therefore so is B(x) 2 . Thus, B(x) 2 is a polynomial and since is square-free, it follows that B(x) is itself a polynomial. Now assume q = 2. We then have The right equation above shows that both A(x) and B(x) are non-zero monomials of degree ≤ k/2 in each variable. Thus, deg so the left equation in (6) is impossible since the polynomial on the left-hand side has X 2 -degree deg X 2 ((X 2 · · · X n ) k H) = 2k < 3k. Assume next that q ≥ 3. Taking the trace from L( √ ) to L in the relation α = β q , we get The right-hand side factors into (q + 1)/2 polynomials in C[X 2 , . . . , X n ] as follows. For q . These are all the roots of ζ q = 1. Further, ζ q = 1, and If deg X 2 (A(x) 2 ) = deg X 2 (B(x) 2 ), then each of the polynomials in the above product on the right has X 2 -degree exactly and such a polynomial cannot divide (X 2 · · · X n ) k H, a polynomial of X 2 -degree 2k. For the above deduction we used the fact that B(x) = 0, which is clear. Assume next that deg X 2 (A(x) 2 ) = deg X 2 (B(x) 2 ) and let a 0 , b 0 be the leading X 2 -coefficients (as polynomials in C[X 3 , . . . , X n ]) of A(x) 2 and B(x) 2 . Then the polynomial If that happens then a 0 /b 0 must be constant and determines uniquely the amount ζ k + ζ −1 k = 2 cos(2kπ/q), and since k ∈ {1, . . . , (q − 1)/2}, this in turn determines k uniquely as well. So, this shows that in this case there is at most one k in {1, . . . , (q − 1)/2} for which the polynomials from the product appearing in the right-most side of (7) can have X 2 -degree less than deg X 2 ( ), while all the other (q − 3)/2 factors have degree at least deg X 2 ( ) ≥ 3k but such polynomials cannot be divisors of the polynomial (X 2 · · · X n ) k H of X 2 -degree 2k. This shows that our equation is impossible for q > 3. Thus, q = 3 and we get Recall that H is irreducible by Corollary 1.
≥ 3k, and we get the same contradiction as before. Thus, showing that where a is some non-zero complex number and M = X a 2 2 · · · X a n n is some monomial. We also have the relation where c 1 = 3 c 1 (some cubic root of c 1 ) and M 1 is also a monomial. Further, since we see that B(x) is a divisor of (X 2 · · · X n ) k/2 , so B(x) = M 2 is also a monomial. Thus, we get The right-hand side above is a polynomial and since M, M 1 , M 2 are monomials, it follows that M 2 2 | M and M 2 2 | M 1 . Thus, = cM 3 + dM 4 is a sum of two monomials with some non-zero coefficients. However, this is impossible since a quick look at shows that as a polynomial in X 2 it has non-zero coefficients for X 3k+r 2 , X 2k+r 2 , X k+r 2 and X r 2 , where r = k − 2 k/2 ∈ {0, 1}. This contradiction finishes (ii); hence, the proof.

Proof of Theorem 1
The aim of this section is to prove Theorem 1.
Proof We first show that if there are infinitely many solutions to (1), then all of them can be parametrized by finitely many expressions as given in (14) for c below. The arguments in this section follow the arguments from [13] and [14].
From now on, we assume w.l.o.g. that α 1 = |α 1 | > |α 2 | ≥ · · · ≥ |α k |. We assume that there are infinitely many solutions to (1). Then, for each integer solution (a, b, c), we have Our first aim is to prove, that the growth-rates of these infinitely many x, y and z have to be the same, except for a multiplicative constant. Let us recall that we trivially have x < y < z and that, by Lemma 1, the solutions of x need to diverge to infinity as well. We now want to prove that there exists a constant C 5 > 0 such that C 5 z < x for infinitely many triples (x, y, z).
In order to prove this, we choose x (and hence y, z) large enough. We denote by g := gcd(F y − 1, F z − 1). Then we use Lemma 2 to obtain and hence which implies x > C 7 z for a suitable new constant C 7 (depending only on k) and x, z being sufficiently large. Next, we do a Taylor series expansion for c which was given by Using the power sum representations of F x , F y , F z , we get We then use the binomial expansion to obtain where O has the usual meaning, using estimates from [15] and where T is some index, which we will specify later. Let us write x := (x, y, z). Since x < z and z < x/C 7 , the remainder term can also be written as O(α −T x /C 7 1 ), where x = max{x, y, z} = z. Doing the same for y and z likewise and multiplying those expressions gives T r 1 =0 p 0 +p 2 +···+p k =p 1 q 0 +q 2 +···+q k =q 1 r 0 +r 2 +···+r k =r 1 d p,q,r M p,q,r ⎞ ⎠ where p = (p 0 , p 1 , . . . , p k ), q = (q 0 , q 1 , . . . , q k ), and r = (r 0 , r 1 , . . . , r k ) are vectors of non-negative integers satisfying p 0 + p 2 + · · · + p k = p 1 , q 0 + q 2 + · · · + q k = q 1 , r 0 + r 2 + · · · + r k = r 1 , where the integer n depends only on T , d j are non-zero coefficients in the field K = Q(α 1 , . . . , α k ), and M j is a monomial of the form in which L i,j (x) are linear forms in x ∈ R 3 with integer coefficients which are all nonnegative if i = 2, . . . , k and negative if i = 1. Set J = {1, . . . , n − 1}. Note that each monomial M j is "small", that is there exists a constant κ > 0 (which we can even choose independently of k), such that This follows easily from the following fact: By the Pisot property of F n , we can write α 1 = |α 1 | > 1 + ζ for a suitable ζ > 0 (a conjecture of Lehmer asserts that ζ can be chosen to be an absolute constant). Using this notation and a suitable κ, we have Our next aim is to apply a version of the Subspace theorem given in [10] to show that there is a finite expansion of c involving terms as in (9); the version we are going to use can also be found in Section 3 of [16], whose notation -in particular the notion of heights -we follow.
We work with the field K = Q(α 1 , . . . , α k ) and let S be the finite set of places (which are normalized so that the Product Formula holds, cf. [10]), that are either infinite or in the set {v ∈ M K : |α 1 | v = 1 ∨ · · · ∨ |α k | v = 1}. Observe that we may choose α 1 , . . . , α k in C and therefore view K as a subfield of C. We denote by | · | ∞ the unique place such that |β| ∞ = |β| = (β) 2 + (β) 2 for all β ∈ C. According to whether −x + y + z is even or odd, we set = 0 or = 1 respectively, such that α (−x+y+z− )/2 1 ∈ K . By going to a still infinite subset of the solutions, we may assume that is always either 0 or 1.
Using the fixed integer n (depending on T ) from above, we now define n + 1 linearly independent linear forms in indeterminants (C, Y 0 , . . . , Y n−1 ). For the place ∞ introduced above, we set where ∈ {0, 1} is as explained above, and For all other places v in S, we define We will show, that there is some δ > 0, such that the inequality is satisfied for all vectors We shall use the notation y = (c, y 0 , . . . , y n−1 ) below. The use of the correct ∈ {0, 1} guarantees that these vectors are indeed in K n+1 . First notice, that the determinant in (12) is given by Thus |det(l 0,v , . . . , l n,v )| v = |1| v = 1 for all places v. Notice further, that which is a consequence of c ∈ Z and the remaining components y 0 , . . . , y n−1 of y being S-units (hence satisfy Thus, for (12) it suffices to consider v∈S n i=0 and the double product on the left-hand side can be split up into Now notice that the last double product equals 1 due to the Product Formula and that An upper bound on the number of infinite places in K is k! and hence, for y large enough such that |f 2 α y 2 + · · · + f k α y k | < 1/2. And finally the first expression is just which, by (9), is smaller than some expression of the form C 8 α Now we choose T (and the corresponding n) large enough such that Then we can write v∈S n i=0 For the height of our vector y, we have the estimate with suitable constants C 9 , C 10 , C 11 . For the second estimate, we used that and bounded it by the maximum of those expressions. Furthermore we have , which just changes our constant C 11 . Now finally, the estimate is satisfied provided that we pick δ small enough. So all the conditions for the Subspace theorem are met. Since we assumed that there are infinitely many solutions (x, y, z) of (12), we now can conclude that all of them lie in finitely many proper linear subspaces. Therefore, there must be at least one proper linear subspace, which contains infinitely many solutions and we see that there exists a finite set J c and (new) coefficients e j (for j ∈ J c ) in K such that we have with monomials M j as before.
Likewise, we can find finite expressions of this form for a and b.
Next we use the following parametrization lemma: Lemma 5 Suppose, we have infinitely many solutions for (1). Then there exists a line in R 3 given by with rationals r 1 , r 2 , r 3 , s 1 , s 2 , s 3 , such that infinitely many of the solutions (x, y, z) are of the form (x(n), y(n), z(n)) for some integer n.
Proof Assume that (1) has infinitely many solutions. We already deduced in Section 4 that c can be written in the form  y, z). In the same manner, we can write Since 1 + bc = F z = f 1 α z 1 + · · · + f k α z k , we get We now pick β 1 , . . . , β as a basis for the multiplicative group generated by {α 1 , . . . , α k , −1}. We remark that each element in this group is an S-unit with the set S defined in Sect. 4. We express each α 1 , . . . , α k as a product of β 1 , . . . , β and insert them into (15). We obtain a new equation of the form j∈J e j β where again J is some finite set, e j are new coefficients in K and L i,j are linear forms in x with integer coefficients. Note that the sum on the left hand side is not zero, since it contains the summand −1. This is an S-unit equation.
We may assume that infinitely many of the solutions x are non-degenerate solutions of (16) by replacing the equation by a new equation given by a suitable vanishing subsum if necessary.
Therefore for i = j, the theorem on non-degenerate solutions to S-unit equations (see [11]) yields that the set of is contained in a finite set of numbers. Now since β 1 , . . . , β are multiplicatively independent, the exponents (L 1,i − L 1,j )(x), . . . , (L ,i − L ,j )(x) take the same value for infinitely many x. Since we assumed, that these linear forms are not all identically zero, this implies, that there is some non-trivial linear form L defined over Q and some c ∈ Q with L(x) = c for infinitely many x. So there exist rationals r i , s i , t i for i = 1, 2, 3 such that we can parametrize with infinitely many pairs (p, q) ∈ Z 2 .
Since r i , s i and x, y, z are all integers, r i p 0 + s i q 0 + t i are integers as well. Replacing r i by r i , s i by s i and t i by r i p 0 + s i q 0 + t i , we can indeed assume, that all coefficients r i , s i , t i in our parametrization are integers.
Using a similar argument as in the beginning of the proof, we get that our equation is of the form j∈J e j β where r := (λ, μ), J is a finite set of indices, e j are new non-zero coefficients in K and L i,j (r) are linear forms in r with integer coefficients. Again we may assume that we have (L 1,i (r), . . . , L ,i (r)) = (L 1,j (r), . . . , L ,j (r)) for any i = j.
Applying the theorem of non-degenerate solutions to S-unit equations once more, we obtain a finite set of numbers , such that for some i = j, we have So every r lies on a finite collection of lines and since we had infinitely many r, there must be some line, which contains infinitely many solutions, which proves our lemma.
We apply this lemma and define as the least common multiple of the denominators of r 1 , r 2 , r 3 . Infinitely many of our n will be in the same residue class modulo , which we shall call r. Writing n = m + r, we get (x, y, z) = ((r 1 )m + (rr 1 + s 1 ), (r 2 )m + (rr 2 + s 2 ), (r 3 )m + (rr 3 + s 3 )).
Replacing n by m, r i by r i and s i by rr i + s, we can even assume, that r i , s i are integers. So we have This holds for infinitely many m, so we can choose a still infinite subset such that all of them are in the same residue class χ modulo 2 and we can write m = 2 + χ with fixed χ ∈ {0, 1}. Thus, we have where η ∈ Z or η ∈ Z + 1/2. Using this representation, we can write (14) as for infinitely many , where , and x = x( ) = (x(2 + χ), y(2 + χ), z(2 + χ)). So for infinitely many solutions (x, y, z), we have a parametrization in , such that c is a power sum in this with its roots being products of α 1 , . . . , α k . This, together with (8) gives the functional identity which proves Theorem 1.

Linear recurrences with nonsquare leading coefficient
The aim of this section is to prove Theorem 2.
First we observe, that there are only finitely many solutions of (18) with c( ) = 0. That can be shown by using the fact, that a simple non-degenerate linear recurrence has only finite zero-multiplicity (see [11] for an explicit bound). We will apply this statement here for the linear recurrence in ; it only remains to check, that no quotient of two distinct roots of the form is a root of unity or, in other words, that has no solutions in n ∈ Z/{0}, m 1 < 0 and m i > 0 for i = 2, . . . , k. But this follows at once from Mignotte's result [24]. So, we have confirmed that c( ) = 0 for still infinitely many solutions. We insert the finite expansion (18) in for c into (17). Furthermore, we use the Binet formula and write F x , F y , F z as power sums in x, y and z respectively. We get an equation of the form Using the parametrization (x, y, z) = (r 1 m + s 1 , r 2 m + s 2 , r 3 m + s 3 ) with m = 2 or m = 2 + 1, we have expansions in on both sides of (17). Since there must be infinitely many solutions in , the largest terms on both sides have to grow at the same rate. In order to find the largest terms, let us first note the following: If e 0 = 0 for infinitely many of our solutions, then the largest terms were or some even smaller expression on the left-hand side, if e 1 = 0 as well. Note that there could be more than one term in the expansion of c with the same growth rate, for example if y and z are just translates of x and therefore we have α −y on both sides, we see that the left-hand side converges to 0, when x grows to infinity (which it does by Lemma 1), while the right-hand side is the constant f 1 = 0. This is a contradiction.
So we must have that e 0 = 0 for infinitely many of our solutions. Then e 0 α (−x+y+z− )/2 1 certainly is the largest term in the expansion of c and we have for the largest terms, which implies that e 2 0 = f 1 α 1 . But this is a contradiction, since we assumed that neither f 1 nor f 1 α 1 is a square in K . So, the theorem is proved.

Linear recurrences of large order
We now prove Theorem 3.
Proof We follow the same notation as in the proof of Theorem 1. Supposing that we have infinitely many Diophantine triples with values in {F n ; n ≥ 0}, we get the functional identity where x = r 1 +s 1 , y = r 2 +s 2 , z = r 3 +s 3 , r 1 , r 2 , r 3 positive integers with gcd(r 1 , r 2 , r 3 ) = 1 and s 1 , s 2 , s 3 integers.
We first handle (i) in the theorem. Therefore assume that α is not a unit. Then, by Mignotte's result [24], there is no multiplicative dependence between the roots and thus (e.g. by using Lemma 2.1 in [7]), it follows that if we put X = (X 1 , . . . , X k ) and P i (X) = k j=1 f j α s i j X r i j − 1 ∈ K [X 1 , . . . , X k ] for i = 1, 2, 3, then for each h ∈ {1, 2, 3} putting i, j such that {h, i, j} = {1, 2, 3}, we have that for some Q h (X) ∈ K [X ±1 1 , . . . , X ±1 k ]. For this we have to identify the exponential function → α 1 by X 1 , → α 2 by X 2 and so forth. Actually, Theorem 1 shows that Q h (X) ∈ K [X ±1 1 , X 2 , . . . , X k ]. Since the polynomial on the left-hand side of (21) has no pole at X 1 = 0 it follows that the Laurent-polynomial on the right-hand side is a polynomial in X 1 as well. This imposes some conditions on the degrees: (P) Parity: r 1 + r 2 + r 3 ≡ 0 (mod 2). This is clear from degree considerations since 2deg X 1 (Q h ) = deg X 1 (P i ) + deg X 1 (P j ) − deg X 1 (P h ) = r i + r j − r h .
(T) Triangular inequality: r 1 + r 2 > r 3 . It is clear that r 1 + r 2 ≥ r 3 , otherwise P 1 (X)P 2 (X)/P 3 (X) has negative degree as a polynomial in, say, X 1 , so it cannot be a polynomial in X 1 . To see that the inequality must be in fact strict, assume that equality holds. Then Q 3 (X) = q 3 ∈ K [X 1 ]. Hence, P 1 (X)P 2 (X) = q 2 3 P 3 (X).
In the left, we have the monomial X r 1 1 X r 2 2 with non-zero coefficient f 1 f 2 α s 1 1 α s 2 2 , whenever r 1 < r 2 . However, such monomials do not appear in the right above. Thus, we must have r 1 = r 2 , and since further we also have r 3 = r 1 + r 2 and gcd(r 1 , r 2 , r 3 ) = 1, it follows that (r 1 , r 2 , r 3 ) = (1, 1, 2). In this case, the coefficient of X 1 X 2 in the left is f 1 f 2 α s 1 1 α s 2 2 + α s 1 2 α s 2 1 , and this must be zero since X 1 X 2 does not appear in P 3 (X). This shows that (α 1 /α 2 ) s 1 −s 2 = −1, so s 1 = s 2 . But then x = y, which is not allowed. Observe now that by the corollary to Lemma 3 (proved in Sect. 3) we know that the polynomials P i (X) are irreducible (as a polynomial in C[X]). We have that r 1 ≤ r 2 ≤ r 3 . From (21) for (i, j, h) = (1, 2, 3) it follows that P h (X) divides P i (X) or P j (X). By degree considerations, P h (X) divides P j (X) (otherwise r 1 = r 2 = r 3 and we can take them all to be 1, which is impossible by (P)). So, again by degree considerations, P h (X) and P j (X) are associated and P i (X) is a square which contradicts Lemma 3.
In case (ii) of the theorem, the identification with Laurent-polynomials (e.g. again via Lemma 2.1 in [7]) does not work in the above form. But when α is a unit, then we have the relation α 1 · · · α k = ±1, which allows applying a similar identification as we now explain. We insist again that by Mignotte's result [24] there are no other multiplicative relations between α 1 , . . . , α k . In particular, any k − 1 of these numbers (e.g. α 1 , α 2 , . . . , α k−1 ) are multiplicatively independent. Hence, we may identify → α 1 by X 1 , → α 2 by X 2 and so forth, which implies that → α k must be identified with ±1/(X 1 · · · X k−1 ). Theorem 1 shows that if we put P i (X) = k−1 j=1 f j α s i j X r i j + f k α s i k (X 1 · · · X k−1 ) r i − 1 ∈ K X ±1 1 , . . . , X ±1 k−1 for i = 1, 2, 3, then for each h ∈ {1, 2, 3} putting i, j for the two indices such that {h, i, j} = {1, 2, 3}, we have that for some Q h (X) ∈ K [X ±1 1 , . . . , X ±1 k−1 ]. We clear on each side denominators and put P i (X) = P i (X)(X 1 · · · X k−1 ) r i ∈ K [X 1 , . . . , X k−1 ] for i = 1, 2, 3. Then ((X 1 · · · X k−1 ) (r i +r j −r h )/2 Q h (X)) 2 = P i (X)P j (X) The left-hand side has no non-zero poles at X i while the right-hand side does not have either a zero or a pole at X i = 0 for i = 1, . . . , k − 1, therefore we deduce that Q h (X) = (X 1 · · · X k−1 ) (r i +r j −r h )/2 Q h (X) is a polynomial. By Lemma 4 the polynomials P i (X), P j (X), P h (X) are irreducible (as polynomials in C[X]). Repeating the arguments from above leads again to the sought contradiction.