(In reply to
Observations, some values by Jer)
Let's prove that 3 is the only possible prime factor of n.
The initial triplet has only one multiple of n and the goal triplet has three multiples of n. Then there must be some application of the transformation that takes two non-multiples of n and changes them into multiples of n.
Let that set be (x, y, z) prior to the transformation. Wlog assume z is the multiple of n.
Let p be some prime factor of n. After the transformation is applied the result should become (0, 0, 0) mod p.
Jer has shown x+y+z=3n. Then apply mod p to each side to get x+y = 0 mod p. Let x = r mod p, which makes y = p-r mod p. Note that r is strictly in the range [1,p-1] and must be coprime to p.
Now apply the transformation mod p to (x, y, z) to yield (2*r-(p-r), 2*(2*(p-r)-r, 0) mod p. This simplifies to (3r, 3*(p-r), 0) mod p, but this is to be (0, 0, 0) mod p.
(3r, 3*(p-r), 0) = (0, 0, 0) mod p then implies 3r = 0 mod p. r is coprime to p, therefore p must equal 3. This then means 3 is the only prime factor of n which limits n's form to that of n=3^m for nonnegative integer m.
re: Observations, some values
