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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.

 See The Solution Submitted by Ady TZIDON Rating: 3.0000 (1 votes)

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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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