semisimplicity wrote:
Let $h \ge 3$ be an integer and $X$ the set of all positive integers that are greater than or equal to $2h$. Let $S$ be a nonempty subset of $X$ such that the following two conditions hold:
if $a + b \in S$ with $a \ge h, b \ge h$, then $ab \in S$;
if $ab \in S$ with $a \ge h, b \ge h$, then $a + b \in S$.
Prove that $S = X$.
For $a\in S$, $h(a-h)\in S$ and $h(a-h)>a$. So $S$ is unbounded.
Let $h_0=3*[\frac h3]+3\ge h$. Pick $a\in S$ with $a>9h_0$. Then $b=h_0(a-h_0)\in S$ and $3|b$ and $\frac b3>2h$.
Then $(\frac b3+2)(\frac {2b}3-2)\in S$ and $(\frac {2b}3+4)+(\frac b3-1)=b+3\in S$.
By induction, we have $b+3n\in S$ for any integer $n>0$.
For any $x\ge 2h$, we construct sequence $\{x_0=x,x_{n+1}=h(x_n-h)\}$. Pick large $N$ such that $x_N>b+h_0$.
Since $c=h_0(x_N-h_0)>b$ and $3|c$, $c\in S$.
$c\in S$ implies $x_N\in S$. $x_{n+1}\in S$ implies $x_n\in S$ and $x=x_0\in S$.
Q.E.D.