Abstract. Diophantine triples taking values in recurrence sequences have recently been studied quite a lot. In particular the question was raised whether or not there are finitely many Diophantine triples in the Tribonacci sequence. We answer this question here in the affirmative. We prove that there are only finitely many triples of integers 1 ≤ u < v < w such that uv + 1, uw + 1, vw + 1 are Tribonacci numbers. The proof depends on the Subspace theorem.

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.

We show that if k ≥ 2 is an integer and (F (k) n ) n≥0 is the sequence of k-generalized Fibonacci numbers, then there are only finitely many triples of positive integers 1 < a < b < c such that ab+1, ac+1, bc+1 are all members of {F (k) n : n ≥ 1}. This generalizes a previous result (cf.[8]) where the statement for k = 3 was proved. The result is ineffective since it is based on Schmidt's subspace theorem.

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