Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2004. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 31)

LUCA, F., Primitive divisors of Lucas sequences and prime factors of ... and ...

22 F. Luca p is congruent to 1 modulo 8. There are only 9 such primes which are smaller than 233, namely 'PI = {2, 17, 41, 73, 89, 97, 113, 137, 193}. So, with 2 = x

2, we need to find all the solutions of the equation z

2-dy

2 = -1, (6) where d > 1 and y > 1 are integers whose factors belong to V\, and d is squarefree. There are precisely — 1 = 2

9 — 1 = 511 possible values for d. We used Mathematica to find, for every such d, the smallest solution (X\(d), Y\(d)) of the Pell equation (5). Only 255 values of d have the property that equation (5) has a solution with the sign —1 in the right hand side. Out of these values of d, only 13 have the property that all prime factors of Y\(d) are in V\. Now suppose that (z, y) — (X

n(d), Y

n (d)

) is a solution of equation (6) for some odd value of n and one of these 13 values of d. Since P(Y

n(d)) < 197, it follows, by the primitive divisor theorem, that 2n — 1 < 197, i.e. n < 99. Thus, we have computed all the 50 • 13 = 650 values of Y

n(d) (i.e., for each one of the 13 values of d, and for each odd n with n < 99), and we tested each one of these numbers to see if their prime factors are in V\. No new number was found, so n — 1. Thus, z = X\(d) for one of the 13 values of d. Since 2 = x

2

, we tested if X\(d) is a perfect square. Five values of x were found, namely x = 1, 2, 3, 9, 10. So, the largest solution of the inequality P(x

4 + 1) < 233 is 10

4 + I = 73 • 137, and P(x

4 + 1) > 233 holds for all integers x > 11. We conclude this section by remarking that we could have done the final testing for P(x

4 + 1) < 233 by combining the primitive divisor technique with a result of J. H. E. Cohn from [3]. Namely, in [3], the following result is proved: Assume that d > 1 is a squarefree number. Then the equation X

4 — dY

2 = — 1 can have at most one solution in positive integers (A', Y). Moreover, let (X\(d

), Y\(d)) denote the smallest positive solution of A"

2 — dY

2 = — 1, and write X\(d) — AB

2, where A is squarefree. Then the only possible value of the odd integer k for which Xk(d) can be a square is k = A. 3. The running time of the algorithm Given K > 1, an algorithm to compute all positive integer solutions x of the inequality P(a

,

2-|-1) < K was presented in section 1, together with its findings when K = 100. Let f(X) G Z[A

r] be a polynomial having at least two distinct roots. In his PhD thesis, Haristoy (see [4]) improved upon earlier estimates of Shorey and Tijdeman (see chapter 7 of [10]) and showed that the inequality P(f(x)) log

2 x log

3 xj log

4 X holds if X is a sufficiently large positive integer. Here and in what