Additive basis of a set union the square of the set

Question

Recall a set of integers $S$ is said to be an additive basis for the natural numbers if there is a $k$ such that every positive integer is expressible as a sum of at most $k$ elements of $S$. Similarly, a set $S$ is said to be an asymptotic additive basis for the natural numbers if there is a $k$ such that every sufficiently large integer is the sum of at most $k$ (not necessarily distinct) elements of $S$. Lagrange's four-square theorem can be thought of as a statement that the squares are an additive basis with $k=4$.

Given a set $S$, we will write $S^2= \{s^2: s \in S\}$.

Question: Is there an example of a set $S$ which is not an additive basis but where $S \cup S^2$ is an additive basis? (The same question then for asymptotic additive basis but I will not focus on that here.)

Note that any set with positive Schnirelmann density is an additive basis, so one naive way of solving this would be to exhibit a set $S$ which is not an additive basis but where $S \cup S^2$ has positive Schnirelmann density but this does not work; if $S$ has Schnirelmann density density zero then so will $S^2$.

Answer 1

In the paper "On additive bases. II", Deshouillers and Fouvry prove a conjecture (made in part I, by a different set of authors) that for each sequence $K$ of positive integers, there is a set $A$ such that $A^k$ is a basis precisely when $k$ belongs to $K$. See J. London Math. Soc. (2) 14 (1976), no. 3, 413–422.

In particular, one can have $A$ not a basis but $A^2$ is a basis (so that $A\cup A^2$ is also a basis). This case in fact follows from what is done in Part I.

Answer 2

It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Edit: this only works if $S$ is a finite set, otherwise my proof does not hold

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example: $S = -\mathbb{N}$, which is clearly not an additive basis, but $S^2 = \mathbb{N}^2$ obviously is, by Lagrange's 4 squares theorem.

I believe this answers your question in all 4 possible cases!