On writing every integer from $(a-1)(b-1)$ onwards as a sum of two non-zero integers from the semigroup generated by $a,b$

Question

Let $\mathbb N$ be the semigroup (even a monoid) of non-negative integers. Let $a<b$ be relatively prime integers such that $2< a$. Let $S :=\mathbb N a +\mathbb N b$ be the semigroup generated by $a,b$. Put $c:=(a-1)(b-1)$. It is well-known that $c-1 \notin S$, and $\mathbb N +c \subseteq S$ (Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why?). My question is: Under what kind of conditions, can we say that $\mathbb N +c \subseteq (S\setminus \{ 0 \}) + (S\setminus \{0\})$ ?

Here, for subsets $A,B$ of $\mathbb N$, we denote $A+B := \{ a+b : a\in A, b \in B \}$.

Answer 1

The only elements of $S$ that you can't get as a sum of two non-zero elements of $S$ are the elements $a$ and $b$. Every other element of $S$ is of the form $ar+bs$ with at least one of the following being true: $r\ge2$, $s\ge2$, or $r$ and $s$ both positive. In the first case, $ar+bs=(a)+(a(r-1)+bs)$; in the second, $ar+bs=(b)+(ar+b(s-1))$; in the third, $ar+bs=(ar)+(bs)$; in all cases, $ar+bs$ is a sum of two nonzero elements of $S$.

If $a=2$, then $c=b-1$, and you can't get $c+1=b$ as a sum of two nonzero elements of $S$. But if $a>2$, then $c>b-1$, so you can get everything starting from $c$ as a sum of two nonzero elements of $S$.