On a conjecture of Erdös on 3-powerful numbers

Abderrahmane Nitaj

A conjecture of P. Erdös 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^{3}divides xyz. In this paper, we give a proof of this conjecture.

Let k ≥ 2 be an integer. A k-powerful number is a positive integer x with the property that if the prime p divides x, then p^{k} divides x (2-powerful numbers were called initially powerful numbers by S. W. Golomb [2]). In [1], P. Erdös studied k-powerful numbers and stated the following conjecture for k = 3 (see also [3], B 16):

Conjecture 1. (Erdös, 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 give an explicit construction of three infinite families of solutions.

Let a be a positive integer. Let x_{0}, y_{0} and z_{0} be relatively prime integers such that x_{0}^{3} + y_{0}^{3} = az_{0}^{3}. Define the sequence of triples (x_{n}, y_{n}, z_{n}) with the recurrences

(1)  \{ \begin{matrix}x_{n+1} = x_{n} (x_{n}^{3} + 2y_{n}^{3}) , \\ y_{n+1} = −y_{n} (2x_{n}^{3} + y_{n}^{3}) , \\ z_{n+1} = z_{n} (x_{n}^{3} − y_{n}^{3}) , \end{matrix}

for n ≥ 0. By induction one may show that x_{n}^{3} + y_{n}^{3} = az_{n}^{3} for all n ≥ 0. If we choose a, x_{0}, y_{0} and z_{0} such that

(2)  \{ \begin{matrix}a is cube-free, \\ (x_{0}, y_{0}) = 1, \\ x_{0}y_{0} ≡ 1 (mod 2), \\ x_{0} ≡ 2 (mod 3), \\ y_{0} ≡ 1 (mod 3), \\ a|z_{0}^{3}, \end{matrix}

then we have the following.

Theorem. Let a, x_{0}, y_{0} and z_{0} be as obove. For n ≥ 1, let the triples (x_{n}, y_{n}, z_{n}) of integers be defined by recurrences (1). Then for all n ≥ 1, (x_{n}, y_{n}) = 1, a|z_{n}^{3} and

lim_{n→∞} |z_{n}| = +∞.

Proof. It is obvious that for all n ≥ 1, x_{n} and y_{n} satisfy x_{n}y_{n} ≡ 1 (mod 2), x_{n} ≡ 2 (mod 3) and y_{n} ≡ 1 (mod 3). Suppose that (x_{n}, y_{n}) = 1 and let p be a prime such that p divides (x_{n+1}, y_{n+1}) = (x_{n}^{3} + 2y_{n}^{3}, 2x_{n}^{3} + y_{n}^{3}). Then p divides 2 (x_{n}^{3} + 2y_{n}^{3}) − (2x_{n}^{3} + y_{n}^{3}) = 3y_{n}^{3} and hence p = 3, which contradicts the congruence x_{n}^{3} + 2y_{n}^{3} ≡ 1 (mod 3). Then (x_{n+1}, y_{n+1}) = 1 and z_{n} ≠ 0. Now, since a|z_{n}^{3}, then a|z_{n+1}^{3}. Finally, for all n ≥ 1, 2^{n}|z_{n}. Hence lim_{n→∞} |z_{n}| = +∞. ■

If we start with integers a, x_{0}, y_{0} and z_{0} satisfying (2) and if we put for all n ≥ 1,

(3)  \{ \begin{matrix}X_{n} = min(|x_{n}|^{3}, |y_{n}|^{3}, a|z_{n}|^{3}), \\ Z_{n} = max(|x_{n}|^{3}, |y_{n}|^{3}, a|z_{n}|^{3}), \\ Y_{n} = Z_{n} − X_{n}, \end{matrix}

then X_{n}, Y_{n} and Z_{n} are relatively prime 3-powerful numbers. Thus each of the triplets (X_{n}, Y_{n}, Z_{n}) gives a solution of Erdös’ conjecture.

Numerical examples.

(i) Let a = 6, x_{0} = 1, 805, 723, y_{0} = −2, 237, 723 and z_{0} = −960, 540. Then

x_{0}^{3} + y_{0}^{3} = −2^{7} · 3^{4} · 5^{3} · 7^{3} · 2287^{3}.

(ii) Let a = 9, x_{0} = −271, y_{0} = 919 and z_{0} = 438. Then

x_{0}^{3} + y_{0}^{3} = 2^{3} · 3^{5} · 73^{3}.

(iii) Let a = 12, x_{0} = 63, 963, 143, y_{0} = −26, 919, 143 and z_{0} = 27, 226, 290. Then

x_{0}^{3} + y_{0}^{3} = 2^{5} · 3^{4} · 7^{3} · 13^{3} · 9973^{3}.

In these three examples, the integers a, x_{0}, y_{0} and z_{0} satisfy (2), and hence, using the recurrence relations (1), we get infinitely many solutions for the conjecture of Erdös. We see that in all these solutions two terms are perfect cubes. For the remaining cases, we state the following:

Conjecture 2. The diophantine equation

x + y = z,

has infinitely many solutions in relatively prime 3-powerful numbers none of which is a perfect cube or only one of them is a perfect cube.

References

1. P. Erdös, ‘Problems and results on consecutives integers’, Eureka 38 (1975−76), 3−8.

2. S. W. Golomb, ‘Powerful numbers’, Amer. Math. Month. 77 (1970), 848 − 852.

3. R. K. Guy, Unsolved Problems in Number Theory, Springer Verlag, New York, Heidelberg, Berlin, 1981.

Département de Mathématiques Université de Caen 14032, Caen Cedex France (email nitaj@univ-caen.fr)