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

1. Oct 6, 2012

### 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 22k + 1, 22k + 1 + 22k, and 22k + 2 other than the solutions 1,2,3, or 2,3,4 or 6,9,12. Is this sufficient?

2. Oct 7, 2012

### 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.

3. Oct 7, 2012

### Wiz14

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?

4. Oct 7, 2012

### ramsey2879

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.

5. Oct 7, 2012

### Wiz14

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

6. Oct 8, 2012

### 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, 2m, 3n) giving the equation 2m+1=3n+1. Here the RHS$\equiv$2 or 4 (mod 8), so m<3, giving the solution (1,2,3). The case (even, odd, even) has to be of the form (2t,3n,2m). For m<3, we have (2,3,4). For m≥3, the last number is $\equiv$ 0 mod 8, the middle is $\equiv$1 or 3 mod 8, so the first must be $\equiv$2 or 6 mod 8, implying t=1, and n even. The resulting equation is 3n=2m-1+1 or (3c-1)(3c+1)=2m-1 where c=n/2 can only take the value 1, and we get (2,9,16), and we are done.

7. Oct 9, 2012

### ramsey2879

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

Last edited: Oct 9, 2012