Not enough information... JBMO2020 wrote: The integer numbers are colored in 3 colors. Prove that there are two numbers whose difference is a perfect square. integer numbers are colored in 3 colors, does that mean the first 3? or 1 is red, 2 is blue, 3 is green, 4 is red... "prove that there are two numbers whose difference is a perfect square" meaning that two different colored numbers?

I can't understand in totally. Let $A,B,C $ be the set of the colored numbers. Are you trying to say $$A=\{3k+a:0\le k\in\mathbb {Z}:0\le a\le 2\}$$$$B=\{3k+b:0\le k\in\mathbb {Z}:0\le b\le 2\}$$$$C=\{3k+c:0\le k\in\mathbb {Z}:0\le c\le 2\}$$ Where $A\cap B\cap C=\emptyset $.

Assume the contrary. Consider the coloring of $n, n+9, n+16, n+25$. The color of $n$ is different from that of $n+9, n+16, n+25$, so two of $n+9, n+16, n+25$ must be of the same color. However $n+25$ is different from both $n+9$ and $n+16$, so $n+9$ and $n+16$ must be the same color. Since the argument holds for arbitrary $n$, $0, 7, \dots, 49$ have the same color, a contradiction.