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## Subset of natural numbers

Number theory
Papapetros Vaggelis
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### Subset of natural numbers

Let $\displaystyle{S\in\mathcal{P}(\mathbb{N})\setminus \left\{\varnothing\right\}}$ such that the

set $\displaystyle{\mathbb{N}\setminus S}$ is infinite. Suppose that $\displaystyle{x\,,y\in S\implies x+y\in S}$.

Prove that there exists $\displaystyle{n\in\mathbb{N}\,,n\geq 2}$ which divides every element of $\displaystyle{S}$.
jason1996
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### Re: Subset of natural numbers

Suppose that is no $n\in \mathbb{N}$, n>1 which divides every element of S. Then there exist $a,b\in S$ with (a,b)=1. It is trivial to prove that for every $k,l\in \mathbb{N}$, $ka+lb\in S$. Let q>ab. Then if (k,l) solution of the equation ka+lb=q (1) there is either k>b or l>a.(2) Also we notice that if (k,l) solution of (1) then (k+b,l-a) is also solution of (1) This means that (1) has infinite number of solutions. Suppose that there is no solution in positive integers. Let $(k_{1},l_{1})$ the solution where $k_{1}$ takes the maximum negative value possible. Then $l_{1}>0$ and also $k_{1}+b>0$ and $l_{1}-a<0\Rightarrow l_{1}<a\Rightarrow k_{1}>b$ (from (2)) contradiction because we defined $k_{1}$ as negative. So for every q>ab (1) has a solution in positive integers so $\mathbb{N}/S\subseteq \left \{ 1,2,...ab\left. \right \} \right.$ contradiction. So the proof has finished.
Why there exist $a,b\in S$ with $(a,b)=1$?