The question is to prove
A set of nonnegative integers with greatest common divisor 1 and closed under addition has all but finite many nonnegative integers.
I know there is a number theoretic proof which has presented by my textbook, but I don't have much background in number theory. I would appreciate it if someone can give an understandable elementary proof. Thank you!
The number theoretic proof is attached below.
A try for a simpler solution.
We may work with a subset $B$ of $\mathbb{N}$ (taking $B=-S=\{-a, a\in S\}$) if $S$ is your subset.
1) There exists $b_1,\cdots,b_s\in B$, $u_1,\cdots,u_s\in \mathbb{Z}$ such that $u_1b_1+\cdots+u_sb_s=1$. If all the $u_j$ are $\geq 0$, the result is trivial, as this imply $1\in B$. Suppose not, and that $u_1,\cdots u_l$ are $\geq 0$, $u_{l+1},\cdots u_s$ are $<0$. Let $c_1=u_1b_1+\cdots+u_lb_l \in B$ (as $B$ is stable under addition) and $c_2=-(u_{l+1}b_{l+1}+\cdots+u_sb_s)\in B$, we have $c_1-c_2=1$.
2) If $c_2=0$, the result is again trivial. Suppose $c_2\geq 1$, and take $n\geq c_2^2$. By Euclidean division, we can write $n=kc_2+r$, with $0\leq r<c_2$. We have $c_2^2\leq n=kc_2+r<(k+1)c_2$, hence $k\geq c_2-1$, and $k-r\geq 0$. Now $$n=kc_2+r=kc_2+rc_1-rc_2=(k-r)c_2+rc_1\in B$$ and we are done.