Let $\mathscr{A}\subseteq\mathbb{N}$ be an infinite subset of $\mathbb{N}$, and suppose $$ s(n) := |\{(a_1,a_2)\in \mathscr{A}^2 : a_1+a_2 = n\}|$$ is uniformly bounded for every $n$ (that is, there is $M>0$ s.t. $s(n) < M,\forall n\in\mathbb{N}$).
Does this implies that the number of solutions to $$ 2a_1+a_2 = n,\quad a_1,a_2 \in\mathscr{A} $$ is also uniformly bounded for all $n$?
I think it might be false, but I'm having trouble coming up with a counterexample. I thought some IP set could provide such example (e.g. $\{n:n\text{ has only $1$s on its $3$-adic expansion}\}$), but I think IP sets in general does not even satisfy the premise.
I also tried a couple of things to prove it true (like considering $\hat{a}_1+2\hat{a}_2=m$ and adding both equations), but it didn't work. Any hints?
There are sets such that solutions to $a+b = n$ are bounded by $1$ and solutions to $2a+b=n$ are unbounded.
Suppose you have a finite set $\mathcal A$ of size $k$ and a large integer $N$. If you try to add a pair $(a,N-2a)$ without making a redundancy to equations $a+b=n$, you have at most only $2k^3+2k^2$ forbidden values for $a$.
In particular, if $2(k+2m-1)^3 + 2(k+2m-1)^2 < (N-2)/2$ then you have enough room to pick $m$ pairs in a row.
Now to build a set such that forall $m$ there is a $N_m$ such that $2a+b = N_m$ has at least $m$ solutions, you can simply pick $N_m > 2+4[(k+2m-1)^3+(k+2m-1)^2]$ where $k=2+4+\ldots+(2m-2)$ and choose the pairs one at a time.