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 Mutual dependence (Posted on 2017-11-01)
Some triplets of distinct positive integers (k,l,m) exist
such that each one of the three numbers divides the sum of the other two.

Define such sets.

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 Analytical solution Comment 3 of 3 |
Let the integers be (c, b, a) where a > b > c

(b+c)/a < 2 , so we know that (b+c)/a = 1.

Solving for c gives c = a-b.
And b must be greater than a/2, because it is greater than c.

So the integers are (a-b, b, a).

Then (a+c)/b = (2a-b)/b is an integer.
Let (2a-b)/b = k.
Solving for b gives  b = 2a/(k+1).
By inspection, the only k for which b is less than a and greater than a/2 is k = 2.
So b = 2a/3 and c = a-b = a/3.

Checking, this works because (a+b)/c = 5.

So any multiple of (1,2,3) works and there are no other solutions (except for permutations).

 Posted by Steve Herman on 2017-11-03 11:38:23

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