A subcase of Problem #42

Problem #42 on Bloom's website of Erdős Problems asks about a fascinating property of Sidon sets. While the problem itself is wide open, one can still salvage something when the set is "smaller than dense". This is what we will see here.

Theorem: Fix an integer $M\ge 1$. Then there exists $N_0=N_0(M)$ such that for every $N\ge N_0$ and every Sidon set $A\subset [N]:=\{1,2,\dots, N\}$ with $|A|=o(\sqrt N)$, there exists a Sidon set $B\subset [N]$ with $|B|=M$ such that $(A-A)\cap(B-B)=\{0\}$.

To begin with, for a finite set $S\subset \mathbb Z$, define $S-S := \{s-s': s,s'\in S\}$ and $\Delta(S) := \{s-s': s,s'\in S,\ s>s'\}\subset \mathbb N$. Then, $S-S \,=\, \{0\} \cup \Delta(S) \cup (-\Delta(S))$.

First consider the first two lemmas, which are obvious from definition, and then a third one whose proof is provided.

Lemma 1: If $S\subset\mathbb Z$ is Sidon, then the map
$$\phi:\{(s,s')\in S^2: s>s'\}\to \mathbb N,\quad \phi(s,s')=s-s'$$
is injective. In particular, $|\Delta(S)|=\binom{|S|}{2}$.

Lemma 2: Let $S\subset\mathbb Z$ be finite. If all elements of $\Delta(S)$ are distinct (equivalently, $\phi$ above is injective), then $S$ is Sidon.

Lemma 3: Let $N\ge 1$, $F\subset \{1,2,\dots, N-1\}$, and let $M\ge 1$. Let
$$C_M \,:=\, M \,+\, \sum_{r=1}^{M-1} r\binom{r}{2} \,=\, M+\frac12\sum_{r=1}^{M-1}(r^3-r^2)$$
be a constant depending only on $M$. If $N \ge \binom{M}{2} |F| + C_M$, then there exists a set $B=\{b_1<b_2<\cdots<b_M\}\subset [N]$ such that $\Delta(B)\cap F=\emptyset$, and all elements of $\Delta(B)$ are distinct (implying $B$ is Sidon by Lemma 2).

Proof: We build $b_1<b_2<\dots<b_M$ inductively, always choosing the new element to the right of the previous ones. Let $b_1:=1$. For $t\ge 2$, suppose $b_1<\dots<b_{t-1}$ have been chosen so that $\Delta({b_1,\dots,b_{t-1}})\cap F=\emptyset$ and all positive differences among $\{b_1,\dots,b_{t-1}\}$ are distinct. Let $E_{t-1}:=\Delta({b_1,\dots,b_{t-1}})$. Then, $|E_{t-1}|=\binom{t-1}{2}$.

We seek an integer $b_t>b_{t-1}$ such that for every $i\in\{1,\dots,t-1\}$, $b_t-b_i \notin F$ and $b_t-b_i \notin E_{t-1}$. For fixed $i$, the condition $b_t-b_i\in F$ is equivalent to $b_t\in b_i+F$. Similarly $b_t-b_i\in E_{t-1}$ is equivalent to $b_t\in b_i+E_{t-1}$. Therefore the set of "bad" values of $b_t$ is contained in
$$\mathcal B_t:=\bigcup_{i=1}^{t-1}(b_i+F)\ \cup\ \bigcup_{i=1}^{t-1}(b_i+E_{t-1})$$
implying
$$|\mathcal B_t|\le \sum_{i=1}^{t-1}|b_i+F| + \sum_{i=1}^{t-1}|b_i+E_{t-1}| = (t-1)|F| + (t-1)\binom{t-1}{2}$$
by a union bound.

Now consider the consecutive integers $b_{t-1}+1, b_{t-1}+2,\dots$. Among these, at most $|\mathcal B_t|$ are forbidden. Hence the least integer $>b_{t-1}$ satisfying the two required properties obeys $b_t \,\le\, b_{t-1} \,+\,  |\mathcal B_t| \,+\, 1 \,\le\, b_{t-1} \,+\, (t-1)|F| \,+\, (t-1)\binom{t-1}{2} \,+\, 1$. Define an explicit upper-bound sequence $U_t$ by $U_1:=1$ and
$$U_t:=U_{t-1} + (t-1)|F| + (t-1)\binom{t-1}{2}+1$$
for $t\ge 2$. Then, we can choose $b_t\le U_t$ at each step, so in particular $b_M\le U_M$. This gives
$$U_M = 1 + \sum_{t=2}^M \left((t-1)|F| + (t-1)\binom{t-1}{2}+1\right) = \binom{M}{2}|F| + C_M$$
on summing the recurrence. So, $U_M\le N$ and hence $b_M\le N$, so $B:=\{b_1,\dots,b_M\} \subset[N]$ exists. Finally, by construction, no new positive difference equals an old one, hence completing the proof. $\square$

Equipped with all these, we can finally provide a proof of the advertised theorem.

Proof of Theorem: Let $A\subset [N]$ be Sidon and let $|A|=o(\sqrt{N})$. Let $F:=\Delta(A)\subset{1,\dots,N-1}$. By Lemma 1, $|F| = |\Delta(A)| = \binom{|A|}{2} \le \frac{|A|^2}{2}=o(N)$. For a fixed $M$, we have
$$N \ge \binom{M}{2}|F| + C_M$$
for all sufficiently large $N$. So, there exists $B\subset[N]$ with $|B|=M$ and $B$ Sidon, so that
$$\Delta(B)\cap\Delta(A)=\emptyset$$
using Lemma 3.

It remains to show $(A-A)\cap(B-B)=\{0\}$. Let $x\neq 0$ and $x\in(A-A)\cap(B-B)$. Then $|x|\in\Delta(A)$ and $|x|\in\Delta(B)$, a contradiction. This gives
$$(A-A)\cap(B-B)=\{0\}$$
hence completing the proof. $\square$