Each of the eight natural numbers is less than 16, and all numbers are different.

Each of the eight natural numbers is less than 16, and all numbers are different. Prove that among their pairwise differences there are at least three identical ones.

1. Given:
a1, a2, a3, a4, a5, a6, a7, a8 ∈ N;
1 ≤ a1 <a2 <a3 <a4 <a5 <a6 <a7 <a8 ≤ 15. (1)
2. From 8 numbers you can make:
N = C (8, 2) = 8! / (2! * 6!) = 28 pairs.
3. The absolute value of the difference of numbers in each pair can take values ​​from 1 to 14 (with a1 = 1 and a8 = 15). Therefore, if there are no three identical among them, then we get two pairs for each value:
28: 14 = 2.
4. Then for the sum of all differences we get:
S = 2 * (1 + 2 + … + 13 + 14);
S = 2 * 14 * (1 + 14) / 2;
S = 14 * 15 = 210. (2)
5.On the other hand, based on inequalities (1), for the same sum we obtain:
S = 7 (a8 – a1) + 5 (a7 – a2) + 3 (a6 – a3) + (a5 – a4);
S ≤ 7 * 14 + 5 * 12 + 3 * 10 + 8;
S ≤ 98 + 60 + 30 + 8;
S ≤ 196. (3)
6. Equalities (2) and (3) contradict each other, so there are three identical pairs.
Q.E.D.