Let $M$ be a set of numbers such that $M\subset\mathbb{N^*}$ and $2018\in M$
And If $m\in M$ then all the postive divisors of $m$ are in $M$
And If $k,m \in M$ and $1 < k < m $ then $km+1 \in M$
prove that $M=\mathbb{N^*}$
I think we should do induction Let's suppose that all numbers $\le n$ are in $M$
let s take 2 divisors of $n$ , $k$ and $m$
so $k.m+1=n+1\in M$
the problem is for prime numbers and perfect squares
From $2018$, we get $1,2,1009,2018$. From these, $2\cdot 1009+1=2019$ (and then $3$ and $673$) and $2\cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers $$1,2,3,4,5,6,7,8,9,11,15,16,18 $$ How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10\pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,\ldots, n-1\in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=\frac {n-1}2$ (so $1<k<m<n-1$) and find $n\in M$. If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=\frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1\in M$ and then $n^2=(n+1)(n-1)+1\in M$ and then also $n$ as a positive divisor thereof.