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There is a finite coloring of the positive integers for which no infinite set has all pairwise sums and pairwise products in one color. Refining this coloring by the parity of the 2-adic valuation also prevents a constant infinite sequence from evading the obstruction.

No. It is enough to take the prime number $t=2$. By the monochromatic sums-and-products obstruction, there is a finite coloring $\chi$ of $\mathbb{N}$ for which no infinite set has all its pairwise sums and pairwise products in one color. Refine $\chi$ by also recording the parity of the 2-adic valuation.

If a sequence $(x_i)$ made both requested families monochromatic, put $y_i=x_i^2$. If the set of distinct $y_i$ were infinite, an injective subsequence would make all pairwise sums $y_i+y_j$ and products $y_i{y}_j$ monochromatic under $\chi$, a contradiction. Otherwise some $y=x_i^2$ occurs infinitely often. Two occurrences give the sum $2y$ and the product $y^2$, butbecause $v_2(y)$ is even. The refining colors differ, another contradiction.

Solved by gpt-5.6-sol high.