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# Recovering Sidon sets from difference sets, part 2.

This is inspired by a recent question. A set $A \subset \mathbb{Z}/n\mathbb{Z}$ with $|A|=m$ is a Sidon set if all the pairwise sums of distinct elements are unequal: $A+A=\{a+a' \mid a,a' \in A, a \ne a'\}$ has $\binom{m}2$ elements. Since $a+b=c+d$ implies $a-c=b-d$ it is equivalent to require that $\{a-a'\mid a,a' \in A\}$ has $m(m-1)+1$ elements (including 0.)

It is clear that when $\lambda$ is coprime to $n$, $B=\lambda A$ is also a Sidon set and $B-B=\lambda(A-A).$ I am going to restrict to the case that $n$ is prime (although more general rings could be of interest.) My question is if the converse is true (I'll explain why I call this the converse):

Suppose that $n$ is prime, $A,B \subset \mathbb{Z}/n\mathbb{Z}$ are Sidon sets and $B-B=A-A.$ Must it be the case that either

1. There are $\lambda=\pm 1,k$ with $B=\lambda A+k$? OR

2. $A-A=B-B=\mathbb{Z}/n\mathbb{Z}$ and there are $\lambda,k$ with $B=\lambda A+k$?

I can report that this is true in all the cases I checked which were most of the possibilities with $n \le 29$ and $m \le 5$ as well as $n=31$ and $m=6.$ However this may simply be the Strong Law of Small Numbers.

In that last case (as well as $n=13$ and $m=4$) we have $A-A= \mathbb{Z}/n\mathbb{Z}$ so it is not a surprise that there are choices of $\lambda \ne \pm 1$ with $\lambda(A-A)=A-A.$ In my limited explorations there were no other examples of $\lambda(A-A)=A-A.$ I think that this is really the only way this could happen (if not, change case 2 to simply have $\lambda(A-A)=A-A$.)

Let $H$ be the multiplicative subgroup generated by $\lambda.$ Then $\lambda(A-A)=A-A$ would require that $A-A$ is a union of translates of $H$ (along with $0$). This might be the start of a proof.

I called the question a converse because we could ask if $B-B=\alpha(A-A)$ requires that $B$ is a translate of a dilation of $A$, but (since I restricted to $n$ prime) there is an inverse with $\alpha \beta=1$ and then $C-C=A-A$ for $C=\beta B.$