For each integer n > 1, let p(n) denote the largest prime factor of n. Determine all(adsbygoogle = window.adsbygoogle || []).push({});

triples (x; y; z) of distinct positive integers satisfying

x; y; z are in arithmetic progression,

p(xyz) <= 3.

So far I have come up with 2^{2k + 1}, 2^{2k + 1}+ 2^{2k}, and 2^{2k + 2}other than the solutions 1,2,3, or 2,3,4 or 6,9,12. Is this sufficient?

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# Consecutive integers such that the prime divisors of each is less or equal to 3

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