“…Therefore, it seems natural to search the binomial coefficients n k in certain recurrences. Special cases of this question were handled by several authors, see, for example, [3].…”
Let Lm denote the mth Lucas number. We show that the solutions to the diophantine equation (2t/k) = Lm, in non-negative integers t, k ≤ 2t−1, and m, are (t, k, m) = (1, 1, 0), (2, 1, 3), and (a, 0, 1) with non-negative integers a.
“…Therefore, it seems natural to search the binomial coefficients n k in certain recurrences. Special cases of this question were handled by several authors, see, for example, [3].…”
Let Lm denote the mth Lucas number. We show that the solutions to the diophantine equation (2t/k) = Lm, in non-negative integers t, k ≤ 2t−1, and m, are (t, k, m) = (1, 1, 0), (2, 1, 3), and (a, 0, 1) with non-negative integers a.
“…Other particular instances of equation ( 4) above have been investigated in [3] (with a Lucas sequence (u n ) n≥0 and k either fixed or a prime number) and in [9] (for k = 3). We set forward the following:…”
mentioning
confidence: 99%
“…Note that the right hand side of equation ( 6) above is a polynomial in x. In this case, one can use the methods of Corvaja and Zannier from [2] to show that equation ( 6) above has only finitely many solutions (n, x) when k ≥ 4 and that the same is true when k ∈ {2, 3} with finitely many exceptional parametric families of quadruples (a, b, c, d), all explicitly computable (see, for example, [3] and [7]).…”
In this paper, we look at a diophantine equation of the form u n = x y , where (u n ) n≥0 is a binary recurrent sequence of integers. We show that if the pair of integers (x, y) belongs to a fixed line of the Pascal triangle, then the above equation has only finitely many positive integer solutions (n, x, y).
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