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 ...

Primitive divisors of Lucas sequences and prime factors of^

2+l and x

4 + l 21 for all n > 1, where a(d) = Xx(d) + Vd\'\(d), ß(d) = X\(d) - VdY\(d). It is wellknown that Y\ (d) | Y

n(d) for all n > 1. Thus, since in equation (4) the number y has P{y) < 101, it follows that P^d)) < 101 must hold. Of the totality of 2672 pairs (X\(d), Y\(d)) with d £ B, only 143 of them satisfy this condition. Testing this took a few minutes with Mathematica. Of course, we did not factor the numbers Yi(d) because some of them are quite large. Instead, we computed, for each given d

, the largest divisor Md of Y\(d) having P(Md) < 101, and we tested if l'i(d) is equal to 4/,/. Let, now C be the set consisting of these 143 elements d £ B for which P(Y\(d)) < 101, and assume that y = Y

n(d) for some odd n > 1 and some d £ C. Since v Í iw~ 11\ QW

n - 3{

dY

' r ii % 1 }

n(d)Yi{d) = —— -77—, for all n > 1, a-(a) — p{d) it follows that the sequence is a Lucas sequence of the first, kind I y 1 (d) J n> 1 with roots o((/) and ß(d). Since a(d) and ß(d) are real, it follows, by a result of Carmichael (see [2]), that the nth term of this sequence has a primitive divisor for all n > 12. We recall that a primitive divisor of the nth term of a Lucas sequence is a prime divisor p of it which, among other properties, it also fulfills the condition that p = ±1 (mod 11). In particular, if n > 12 is odd, then there exists a prime number p | Y

n(d) such that p > 2n — 1. Since we are searching for values of n and d such that P(Y

n(d)) < 07, it follows that n is an odd number such that 2n — 1 < 97, hence, n < 49. Thus, we used Mathematica to compute, for every one of the 143 values of d £ the numbers Y

n(d) for all odd values of n < 49, resulting in a totality of 143 • 25 = 3575 such numbers. For each one of these numbers, we applied the procedure described above to eliminate the ones for which P(Y

n(d)) > 97. The computation took a few minutes, and a totality of 156 numbers Y

n(d) survived (that is, only 13 new numbers Y

n(d) for n > 1 odd and d £ C were found). For each of these numbers we computed x - X

n(d). The conclusion of these computations is that there are precisely 156 positive integer values of x for which P(x

2 + 1) < 101. Of these 156 positive integers, 140 of them are less than 10

5, 10 more of them are between 10

5 and 10

6, and the largest 6 of them are 1984933, 2343692, 3449051, 6225244, 22709274, and 24208144. Thus, the largest positive integer solution x of the inequality P(x

2 + 1) < 101 is 24208

144

2 + 1 = 29

3 • 37

2 -53-6

1

2 • 89. W

7e now turn our attention to P(x

4 +1). Suppose that x is a positive integer such that P(x

4 + 1) < 233. If p is a prime number dividing x

4 + 1, then either p = 2, or