On a Conjecture of Erdos on 3-powerful Numbers

A conjecture of P. Erdos says that the diophantine equation x+y = z has infinitely many solutions with (x,y) — 1 and such that if a prime p divides xyz, then p divides xyz. In this paper, we give a proof of this conjecture. Let k ^ 2 be an integer. A /c-powerful number is a positive integer x with the property that if the prime p divides x, then p divides x (2-powerful numbers were called initially powerful numbers by S. W. Golomb [2]). In [1], P. Erdos studied kpowerful numbers and conjectured for k = 3 (see also [3, B 16]. CONJECTURE 1 (Erdos, 1975). The diophantine equation x + y = z has infinitely many solutions in relatively prime 3-powerful numbers. In this paper, we give a proof of this conjecture. In fact, we construct explicitly three infinite families of solutions. Let a be a positive integer. Let x0, y0 and z0 be relatively prime integers such that = az\. Define the sequence of triples (xn,yn,zn) with the recurrences for n ^ 0. By induction, one may show that xn+y 3 n = az\ for all n ^ 0. If we choose a, x0, y0 and z0 such that ( a is cubic-free, xoyo= I(mod2), ' x0 = 2 (mod 3), y0 = 1 (mod 3), a\zl then we have the following. THEOREM. Let a, JC0, y0 andz0 be as above. For n^ I, let the triples (xn, yn, zn) of integers be defined by recurrences (1). Then for all n^ 1, (xn,yn) = 1, a\z\ and Received 1 February 1994. 1991 Mathematics Subject Classification 11D25. Bull. London Math. Soc. 27 (1995) 317-318 318 ABDERRAHMANE NITAJ Proof. It is obvious that for all n ^ 1, xn and yn satisfy xnyn = 0(mod2), xn = 2 (mod 3) and yn = 1 (mod 3). Suppose that (xn,yn) = 1, and let/? be a prime such that p divides (xn+1,yn+1) = (xl + 2yl2xl+yl). Thenp divides 2 (x 3 +2/ n ) (2x 3 + ; ; 3 ) = 3>> and hence p = 3, which contradicts the congruence xl + 2yn = 1 (mod 3). Then (n+iJn+i) = 1 3 zn # 0. Now, since a\z\, we have a |z '+ 1 . Finally, for all n ^ 1, 2 | zn. Hence l im, ,^ \zn\ = + oo. If we start with integers a, JC0, y0 and z0 satisfying (2), and if we put for all n = mm(\xn\\\yn\\a\zf), W,|jn | 3 ,fl |zn | ), (3) .Yn = Zn-Xn, then Xn, Yn and Zn are relatively prime 3-powerful numbers. Thus each of the triplets (Xn, Yn, Zn) gives a solution of Erdos' conjecture. Numerical examples (i) Let a = 6, x0 = 1,805,723, y0 = -2,237,723 and z0 = -960,540. Then (ii) Let a = 9, JC0 = 2 7 1 , y0 = 919 and z0 = 438. Then x 3 + ^ = 2-3-73. (iii) Let a = 12, xQ = 63,963,143, y0 = -26,919,143 and z0 = 27,226,290. Then In these three examples, the integers a, x0, yQ and z0 satisfy (2), and hence, using the recurrence relations (1), we obtain infinitely many solutions for the conjecture of Erdos. We see that in all these solutions, two terms are perfect cubes. For the remaining cases, we propose the following. CONJECTURE 2. The diophantine equationx+y = 2has infinitely many solutions in relatively prime 3-powerful numbers none of which is aperfect cube or only one of which is a perfect cube. References1. P. ERDOS, 'Problems and results on consecutive integers', Eureka 38 (1975-76) 3-8.2. S. W. GOLOMB, 'Powerful numbers', Amer. Math. Monthly 11 (1970) 848-852.3. R. K. GUY, Unsolved problems in number theory (Springer, New York, 1981). Departement de MathematiquesUniversite de Caen14032, Caen CedexFrance