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

[Wiz14]

For each integer n > 1, let p(n) denote the largest prime factor of n. Determine all
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?

 

[Norwegian]

If we restrict the discussion to the primitive solutions, the complete list will be (1,2,3), (2,3,4) and (2,9,16). You can exclude the rest by simple considerations modulo 3 and 8.

 

[Wiz14]

If we restrict the discussion to the primitive solutions, the complete list will be (1,2,3), (2,3,4) and (2,9,16). You can exclude the rest by simple considerations modulo 3 and 8.

Thank you for your answer, but can you be more specific? How can we prove that your solutions and my solutions are the only solutions?

 

[ramsey2879]

Thank you for your answer, but can you be more specific? How can we prove that your solutions and my solutions are the only solutions?

Your solutions are not the only solutions if we allow multiplying each primitive solutions by 3^a * 2^b as you did as you left out solutions such as 18, 27, 36.

 

[Wiz14]

Your solutions are not the only solutions if we allow multiplying each primitive solutions by 3^a * 2^b as you did as you left out solutions such as 18, 27, 36.

but how to exclude the rest by considerations mod 3 and 8? thank you.

 

[Norwegian]

By a primitive solution, I mean a triple (a,b,c), where a, b and c have no common factor. In that sense, the list (1,2,3), (2,3,4), (2,9,16) is complete.

To exclude the rest, you need to do some case by case analysis. One useful initial observation is that exactly one of the numbers is divisible by 3, as alternatives quickly lead to contradictions. Then the case (odd, odd, odd) should be trivial to deal with. Looking at (odd, even, odd), this has to be of the form (1, 2^m, 3ⁿ) giving the equation 2^m+1=3ⁿ+1. Here the RHS\equiv2 or 4 (mod 8), so m<3, giving the solution (1,2,3). The case (even, odd, even) has to be of the form (2^t,3ⁿ,2^m). For m<3, we have (2,3,4). For m≥3, the last number is \equiv 0 mod 8, the middle is \equiv1 or 3 mod 8, so the first must be \equiv2 or 6 mod 8, implying t=1, and n even. The resulting equation is 3ⁿ=2^m-1+1 or (3^c-1)(3^c+1)=2^m-1 where c=n/2 can only take the value 1, and we get (2,9,16), and we are done.

 

[ramsey2879]

If we restrict the discussion to the primitive solutions, the complete list will be (1,2,3), (2,3,4) and (2,9,16). You can exclude the rest by simple considerations modulo 3 and 8.

Other soluion is (1,1,1) ? Is that an arithmetic progression? See http://en.wikipedia.org/wiki/Arithmetic_progression Never mind the problem asked for distinct numbers