1. ## divisibility of numbers

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!

2. ## Re: divisibility of numbers

Originally Posted by 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!
We can find that $b=2a$ and $c=3a$ without using divisibility properties, only with inequalities.

None of $k,l,m$ is $0$, since $a,b,c$ are positive.
Spoiler:
So the first inequality I've found is from 1*: $kc=a+b implying that $k<2$ and therefore $k=1$. Then $a+b=c$.

We use the above equation to rewrite equation 2* in two ways replacing $a$ or $c$: $lb=a+c=2c-b>2b-b=b$, then $l>1$; $lb=a+c=2a+b<2b+b=3b$, then $l<3$. From $1 it follows that $l=2$.

Finallly, if you add equations 1* and 2* we find that $2a+b+c=kc+lb=c+2b$, which gives $b=2a$. Substituting $b$ in $a+b=c$ we find that $c=3a$.

Edit: I didn't even use equation 3*!

3. ## Re: divisibility of numbers

Thanks melese, I like it, that is very elegant proof!