Lucas Numbers of the Form Px 2 , Where P Is Prime

Let L denote the n th Lucas number, where n is a natural number. n 2 Using elementary techniques, we find all solutions of the equation: Ln px where p is prime and p (i000. KEY Wi’RDS AND PHRASES. Lucas number 1985 AMS SUBJECT CLASSIFICATION CODE. IIB39 1. NTRODUCTION Let. n denote a natural number. Let L denote the n th Lucas number, that is, n L1=1, L2=3, Ln Ln_l+Ln_2 for n_ 3. In [1], J.H.E. Cohn found all Lucas nmers which are square or t.wlce a square. As a result of a late paper of Cohn [2], it is known that for each integer c _ 3, there is at. most one Lucas number of the form cx2. Uslnq [3], Definition 2, and (9) below, we see that there are 111 primes, p, such that (i) 2 < p. 1000, and (ii) there exists n such that p[Ln In this paper, we find all solutions of the equation: L px2 (*) n where the prime p satisfies conditions (i) and (ii) above. We find that only 8 such values of p yield solutions of (*). The results aze summarized in Table 3 on the last page. The larger problem of finding all solutions to |*) appears more difficult; its solution would yield all Lucas numbers which ace prime. 2. PRELIMINARIES Let n denote a natural number. Let p denote a p[ime, not necessarily satisfying conditions (i) and (ii) above. Definition i -Let Fn denote the n th Fibonacci number, that is, F 2 I, F + for n_ 3 n Fn-I Fn-2 Definition 2 Let z(n) Mink: k_l and nlFk} Definition 3 Let y(n) 1/2z(n) if 21z(n). For each integer c _3, the equation L cx 2 has at most one solution. n If Ln=q is prin,e, then y(q) =n and y(q2) =qn. (i)