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dividing the sum of entire functions

Statement

In this issue, it's best to try to try to build you an example. Obviously, if we have to divide many times, it is best to try 1, which is the universal divisor, and the first three candidates that tends to happen is 1, 1, 1, which clearly, each number divides the sum of the others.

already have one, we can now begin to work more. Since there can be other cases relating to same (it would not be a primitive triplet), what happens if there are two equal and the other is no longer the same? That is, if we have a triple (n, n, m) or (m, n, n) with this property, we have that m divides 2n. If different n, or exactly 2n, in which case we reach the triple (1, 1, 2), and they can not have common divisors, or is less than n. In the latter case, once you take away all common divisors with n, can only be 2 or 1. However, if 1, this requires that n must divide n + 1, which is impossible for n greater than 1, or n must divide n + 2, which can not be for n greater than 2.

Well, we have already removed the case question of two or three equal numbers, we will deal with different numbers.

Suppose the lowest is 1. The second, ie, b, c should be divided, but the third, c, must divide ab + 1, so it must be b + 1 (otherwise you would not be greater than that by less than or equal to b + 1) . However, b must divide ab + 2, so it must divide 2, and that means it's 2 (can not be 1, as it would like to). Therefore, we have another short list: (1, 2, 3).

If the child is not 1, there is a prime p which divides it. Working with cousins \u200b\u200bis very comfortable as you can keep track on a par. That cousin can not divide another issue boc, because if so, divides the three. This is proved as follows: as to divide divides ab + cyp a, p divides ab + c, so who can divide only one of them.

Now for the greatest number. Since it is greater than b, and divide a + b, must be less than a + b, and greater than b. As the largest factor of a + b (after himself) is less than b, since a + b is less than 2b, c must match a + b. So our slate is of the form a, b and a + b. However, b must divide 2a + b, so you must divide 2a. That is reduced, since it has no common factors with a, b is 2 to 1, cases have been treated. So

the only valid ordered triples are (1, 1, 1), (1, 1, 2) and (1, 2, 3).