Peshnin A.

What is the minimum number of colors needed to color the natural numbers so that any two numbers differing by 2 or by a factor of two are colored differently?

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Example. We will sequentially paint natural numbers in ascending order. 1 and 2 will be painted in two different colors. For the color of each number $k>2$, there are no more than two restrictions: it cannot be the same color as the number $k / 2$ or the number $k-2$. Therefore, for any such $k$, there will definitely be a third color.

Estimate. Consider the numbers 4, 6, and 8. Any two of them must be painted in different colors, so there cannot be fewer than three colors.

Comment. An example of coloring with three colors can be constructed differently. We will paint all numbers in two colors, adhering only to the first condition. For example, all numbers that give a remainder of 0 or 1 when divided by 4 will be painted red, and the rest of the numbers will be painted blue. Now we need to ensure the fulfillment of the first condition. For this, we will repaint in purple all numbers in which the number 2 appears in an odd power in their prime factorization. Then the first condition will obviously not be violated, and the second condition will also be met, since numbers differing by two times have different parity of the exponent of the power of two in their prime factorization.

## Answer

In three colors.