Lemma. For every positive integer $k$ we can find primes $p_1, \ldots, p_k$ such that for all $S \subseteq [k]$, the color of $\textstyle \prod_{i \in S} p_i$ depends only on $\lvert S \rvert$. Proof. For positive integers $r$ and $k$, let $P(r, k)$ be the statement that there exist primes $p_1, \ldots, p_k$ such that for all $S \subseteq [k]$ of size at most $r$, the color of $\textstyle \prod_{i \in S} p_i$ depends only on $\lvert S \rvert$. We will prove $P(r, k)$ for all $r$ and $k$ via induction on $r$, which will suffice as the statement of the lemma is just $P(k,k)$. As a base case, $P(1, k)$ follows from the pigeonhole principle. For the inductive step, we apply Ramsey's theorem on hypergraphs, which states that given positive integers $r$, $c$, and $k$, there exists a positive integer $R^{(r)}(k;c)$ such that for any coloring of the subsets of $[R^{(r)}(k;c)]$ of size $r$ with $c$ colors, we may find a subset $X \subseteq [R^{(r)}(k;c)]$ of size $k$ such that all subsets of $X$ of size $r$ have the same color. This immediately yields the implication $P(r-1, R^{(r)}(k; 4)) \implies P(r, k)$. $\blacksquare$ Here's a funny way to finish. Let $k$ be some prime $\ell \geq A + 2$ and let $\textstyle n = \prod_{i \in [\ell]} p_i$. Note that if we exclude $1$ and $n$, the number of divisors of each color must be a multiple of $\ell$. If (2) is false, then since $\ell \geq A + 2$ this number of divisors must be the same for each color. But $2^\ell - 2$ is not divisible by $4$, contradiction.

numbersandnumbers wrote: Lemma. For every positive integer $k$ we can find primes $p_1, \ldots, p_k$ such that for all $S \subseteq [k]$, the color of $\textstyle \prod_{i \in S} p_i$ depends only on $\lvert S \rvert$. Proof sketch. Spam hypergraph Ramsey's theorem. $\blacksquare$ Here's a funny way to finish. Let $k$ be some prime $\ell > A + 2$ and let $\textstyle n = \prod_{i \in [\ell]} p_i$. Note that if we exclude $1$ and $n$, the number of divisors of each color must be a multiple of $\ell$. Since $\ell > A + 2$, this number of divisors must be the same for each color. But $2^\ell - 2$ is not divisible by $4$, contradiction. Seems magical, but can you make your proof more clear? (especially what is the Spam hypergraph Ramsey's theorem ) Thanks!

EthanWYX2009 wrote: numbersandnumbers wrote: Lemma. For every positive integer $k$ we can find primes $p_1, \ldots, p_k$ such that for all $S \subseteq [k]$, the color of $\textstyle \prod_{i \in S} p_i$ depends only on $\lvert S \rvert$. Proof sketch. Spam hypergraph Ramsey's theorem. $\blacksquare$ Here's a funny way to finish. Let $k$ be some prime $\ell > A + 2$ and let $\textstyle n = \prod_{i \in [\ell]} p_i$. Note that if we exclude $1$ and $n$, the number of divisors of each color must be a multiple of $\ell$. Since $\ell > A + 2$, this number of divisors must be the same for each color. But $2^\ell - 2$ is not divisible by $4$, contradiction. Seems magical, but can you make your proof more clear? (especially what is the Spam hypergraph Ramsey's theorem ) Thanks! Fix $k$. We prove that for a fixed $N$, if there are $N_i$ primes, then some subset satisfies the property that the product of $1, \dots, i$ elements in the set is all the same color. For the base case of $k = 2$, we can take a graph of $p_1, \dots, p_{R(k, k, k, k)}$ and color in to get a $K_k$ clique to finish, so $N_2 = R(k, k, k, k)$. Now, suppose that we have the result for $i$. Take the $i+1$-hypergraph $G$ on $i+1$ tuples from each primes. Ramsey's theorem again gives some $N_{i+1} = R(N_i)$ such that some arbitrary $N_i$ element subset of $G$ is all the same color. Anyways, this problem should be true regardless of the number of colors.

Frankly , it took me more time to figure out the best restrictions of the problem are actually mod ,after the technical procedure of Ramsey Thm.

By the way ,I think it’s the most illusive one among these 12 problems ,since I’ve made a mediocre mistake by claiming that the identity is still working while the number of colors is just two.