The problem states the following:
We color the first $\frac{n^3 + 5n}{6}$ numbers with red and blue. Prove that we can find $n$ numbers $a_1 < a_2 < ... < a_n$, colored with the same color and with the relation $a_2 - a_1 \leq a_3 - a_2 \leq \dots \leq a_n - a_{n - 1}$.
Firstly, I proved by induction that $n^3 + 5n \equiv 0 \pmod 6$. Base case: $n = 1$; $1 + 5 \equiv 0 \pmod 6$. Induction step: supose that $n^3 + 5n \equiv 0 \pmod 6$. Then, $(n + 1)^3 + 5(n + 1) = n^3 + 3n^2 + 3n + 1 + 5n + 5 \equiv 3n^2 + 3n + 6 \pmod 6 \equiv 3n(n + 1) \pmod 6$. As $n(n + 1) \equiv 0 \pmod 2$, $3n(n + 1) \equiv 0 \pmod 6$, which means $\top$ for induction. (we have finished the proof by induction).
Then, I should prove, again by induction, that there is no coloring of the first $\frac{n^3 + 5n}{6}$ numbers for which I can't find a sequence with the given property. Then I think Wan der Waerden Theorem may help, but I don't know clearly how to apply it. Because Wan der Waerden numbers are huge and barely aproximated, not calculated, and they aren't linked with $\frac{n^3 + 5n}{6}$. So, this aproach looks like an inductive reasoning/proof for a strengthened form of Wan der Waerden Theorem on a particular case ($r = 2$ colors, not an arithmetical progression with invariant ration, but one with monovariant ratio).
Thanks in advance!