The proof is elementary but quite complicated and uses combinatorial arguments. In a previous paper [P. Erdös, J. London Math. Soc. 10, 126-128 (1935; Zbl 012.05202)] the following weaker result was proved: Assume that no ai divides any other than (1) holds. In another paper [J. Math. Anal. Appl. 15, 60-64 (1966; Zbl 151.03502)] the authors point out that (1) does not hold if we assume that [ai,aj] =
Classif.: * 11B83 Special sequences of integers and polynomials
11B75 Combinatorial number theory
Index Words: number theory
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