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**jozou** Find all integers of the form 1 ≤ a < b < c, s.t.:

1* ∃k a+b = kc

2* ∃l a+c = lb

3* ∃m b+c = ma

Attempt: Well, I have answer that solution is of the form a, 2a, 3a. I can prove that if b = 2a, then c = 3a:

From 2* we have c = a(2l - 1)

Then from 1* and previous we have 3a = ka(2l - 1). So k = 1 or k = 3:

If k = 3, then

1 = 2l - 1, l = 1, a +c = 1.b = 2a what contradicts assumption that a < c.

So k = 1. Then a + b = 3a = 1.c

But I can't prove that b=2a. Any hints? Thanks!