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Show that there is an infinite set of natural numbers such that the sum of any two elements has an even number of prime factors.


My attempt:

Define a coloring on the doubletons of $\mathbb{N}$, that is a coloring $f:\mathbb{N}^{(2)}\to \{0,1\}$ by: $$ f(\{x,y\}) = \left\{ \begin{array}{ll} 0 & \quad \text{if $x+y$ is has an even number of prime factors} \\ 1 & \quad \text{otherwise} \end{array} \right. $$ By the Infinite Ramsey Theorem, there exists an infinite subset $N\subseteq \mathbb{N}$ such that all of the doubletons of $N$ are colored $0$ or all of the doubletons of $N$ are colored $1$. If $N$ is monochromatic in $0$, then we are done, as $N$ witnesses the desired property. Otherwise, assume, $N$ is monochromatic in $1$. Then for every doubleton of $\{x,y\}\subseteq N$, $x+y$ has an odd number of prime factors. If there exists a prime $p$ such that for all $x\in N$, $p\nmid x$. Then define: $$ pN=\{px \mid x\in N\} $$ We have that for arbitrary $\{x,y\} \subseteq N$, $x+y$ has an odd number of prime factors. Therefore, we have that $px + py = p(x+y)$ will have an even number of prime factors and we are done as $pN$ witnesses the desired property. Otherwise, we have that $N$ satisfies: $$\forall p \exists x (p\mid x)$$ If there is a $p$ such that $p$ only divides finitely many $\{x_1,x_2,\ldots, x_n\} \subseteq N$
then $N\setminus \{x_1,x_2,\ldots, x_n\}$ is still an infinite set and we do the same trick as before: $$ p(N\setminus \{x_1,x_2\ldots,x_n\})=\{px: x\in N\setminus \{x_1,x_2\ldots,x_n\}\} $$ witnesses the desired property. If not then for all $p$, $p$ divides infinitely many of the elements of $N$. In fact, for every prime $p$, $p$ divides all but finitely many elements of $N$. Because if there is an infinite subset $A\subseteq N$ and a prime $p$ such that $p$ divides each element of $A$ but $N\setminus A$ is infinite, then we can appeal to the same argument as above to obtain a set that has the desired properties. Now enumerate $N$ as $N=\{n_1,n_2,n_3,\ldots\}$, we know that for each prime $p$, there is an index $i$, such that $p\mid n_j $ for all $j\ge i$. We can deduce that there is a finite index $k$, such that every prime $p$ divides $n_i$ for $i\ge k$. If not, then there is a prime $p$, such that for all $k\in \mathbb{N}$, $p\nmid n_i$ for $i\ge k$ which is not the case. So every prime number divides $\{n_k,n_{k+1},n_{k+2},\ldots \}$ a wild contradiction to the fact that $N$ is a subset of natural numbers.

My Question:

Am I coming to a false conclusion in the last paragraph? I feels ridiculous that I should end up deducing that the set: N has numbers that infinitely many primes divide.

Wrong conclusion. $``$We can deduce that there is a finite index $k$, such that every prime $p$ divides $n_i$ for $i\ge k$." is not true, we cannot deduce that...

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  • $\begingroup$ Everything goes well, until the last paragraph. You have that for each $p$ there is an $i$ such that $p\mid n_j$ for all $j\ge i$. This does not imply that there is a fixed $k$ such that $p\mid n_i$ for all $i\ge k$. For instance, let $n_i$ be the product of the first $i$ primes, for all $i$. $\endgroup$ – Andrés E. Caicedo May 7 '14 at 3:11
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    $\begingroup$ @AndresCaicedo I just realized I could have used the set: $\{pn+1: n \in \omega\}$ for fixed prime $p$, which makes the problem a triviality.... Also, I realized, the false logic there too $\endgroup$ – Rustyn May 7 '14 at 3:12
  • $\begingroup$ Yes. The problem was asked here a few years ago (of course), and that was the approach, but I like the additional analysis that you provided. $\endgroup$ – Andrés E. Caicedo May 7 '14 at 3:14
  • $\begingroup$ I guess what I've proved is sort of interesting. Under this coloring, either there is a monochromatic set such that the sum of every two numbers of that set has an even number of prime factors or there is a monochromatic set such that each prime $p$ divides all but finitely many of its elements. $\endgroup$ – Rustyn May 7 '14 at 3:34
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Ah we just use the set: $\{7n+1: n \in \omega\}$ to begin with.

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