strange number-theory

do exist five different number such as $a_1,a_2,a_3,a_4,a_5\in\mathbb{N}$ which sum of every three of them be divisible by the sum of two other?

I guess I have to prove

$i<j \implies a_i<a_j$

$a_4+a_5\le a_1+a_2+a_3$ is not possible.

but I don't know how to prove it

• A parity argument might be useful here. For example, we cannot have all five numbers odd because the sum of any three would be odd and therefore not divisible by the sum of the other two (which must be even). Also if the numbers all have a common divisor, we could factor that out to get an example in which the five numbers are without common divisor. – hardmath Apr 10 '18 at 16:46
• No, this problem does not require that the sum of the last two be less than or equal to the sum of the first three. In fact, the order in which the numbers are given is not relevant at all.. What you need to prove is that if i, j, k are any three different integers from 1 to 5 and p and q are the other two such integers then $\frac{a_i+ a_j+ a_k}{a_p+ a_q}$. – user247327 Apr 10 '18 at 16:48
• @user247327: Since the problem asks about natural numbers, the fact that $a_4+a_5$ divides $a_1+a_2+a_3$ implies the former is less than or equal to the latter (regardless of how the five numbers are arranged). – hardmath Apr 10 '18 at 16:51

Assume w.n.l.g. that $a_1<a_2<a_3<a_4<a_5$. The two largest account for more than $2/5$ of the sum, so the three smallest account for less than $3/5$, and so $(a_1 + a_2 + a_3) / (a_4 + a_5) < 3/2$ and can't be any integer besides $1$. That is, $a_1+a_2+a_3 = a_4+a_5$. Now consider $(a_1+a_2+a_4)/(a_3+a_5)$. If this is $1$ as well, then $a_3=a_4$, which can't be true. So it's at least $2$: $a_1+a_2+a_4 \ge 2a_3 + 2a_5$. Now we have $$-a_3+a_4 = a_1+a_2-a_5 > a_1+a_2-2a_5 \ge 2a_3 - a_4,$$ or $a_4 \ge (3/2)a_3$. But then $a_4+a_5 > 2a_4 \ge 3a_3 > a_1+a_2+a_3$, which contradicts the equality $a_4+a_5=a_1+a_2+a_3$ that we already proved.