A set such that $A$, $A+A$ have density zero but $A+A+A$ has positive density.

Question

I have been thinking about Lagrange's theorem in terms of sumsets. Certainly the perfect squares $\square = \{ n^2: n \in \mathbb{Z}\} = \{ 0,1,4,9,\dots\}$ has density $0$. In fact, I think $$ \square + \square = \{ a^2 + b^2 : a,b \in \mathbb{Z}\}$$ also has density zero, but by Lagrange's 3 squares theorem $n = a^2 + b^2 + c^2$ if $n \not \equiv 4^a(8k+7)$. So that density is: $$ d(\square+\square+\square) = 1 - \frac{1}{8}\big[ 1 + \frac{1}{4} + \frac{1}{4^2} + \dots \big] > 0$$ How did this happen? I just added finitely many sets of density $0$ and got a set with positive density. Is this a contradiction?

Answer 1

Density zero has a wide range of refinements. The number of squares up to some large positive $N$ is about $\sqrt N.$ The number of sums of two squares is not much lower than $\sqrt N \cdot \sqrt N = N;$ two effects reduce that, some pairs sum to between $n$ and $2N,$ and there are some repeats such as $64+1 = 49+16.$ The count of distinct sums of two squares up to $N$ is actually about $0.7642 N / \sqrt {\log N}.$ This is asymptotically larger than the count of primes, for example.

By the time you get to four squares, you can't have more than $N$ distinct numbers represented up to $N.$ Some overage is due to represented numbers from $N$ to $4N,$ but most of it is due to considerable repetition, numbers represented many different ways.