Well-spread (weak Sidon) sets

Question

Does anybody have access to the paper A. Kotzig: On well spread sets of integers, 1972, or does anybody know the proof of $\sigma^*(n)\geq 4+\binom{n-1}2$ for $n\geq7$ (as cited in Marr,Wallis: Magic Graphs, Springer 2013, Th. 2.15)

Notions: A set $A=\{a_1,\dots,a_n\}$ of integers is a well-spread (weak Sidon) set of cardinality $n$, if $a_i+a_j \neq a_k+a_\ell$ whenever these four elements of $A$ are distinct. Then $\sigma(A):= \max A - \min A +1$ is the size of $A$, and $\sigma^*(n)$ is the minimal $\sigma(A)$ taken over all well-spread sets $A$ of cardinality $n$.

Answer 1

Suppose that $A\subset[1,l]$ is weak Sidon; we show that $l>\binom{n-1}2$, which is very close to your estimate.

There are $n(n-1)/2$ positive differences of the form $a_1-a_2$ with $a_1,a_2\in A$. All these differences are in $[1,l-1]$, but we cannot conclude that $l-1\ge n(n-1)/2$ because some of the differences can coincide; namely, we have $a_1-a_2=a_3-a_4$ if and only if exactly one of the following holds: (i) $a_1=a_4$ and $a_2,a_1=a_4,a_3$ is a three-term arithmetic progression; (ii) $a_2=a_3$ and $a_1,a_2=a_3,a_4$ is a three-term arithmetic progression. We associate with every equality of the form $a_1-a_2=a_3-a_4$ the middle term of the corresponding progression.

The crucial observation is that for every element $a\in A$, the integer $2a$ is the middle term of at most one progression: for an equality of the form $a_1+a_2=b_1+b_2=2a$ would contradict the weak Sidon property. Therefore, there are at most $n$ equalities of the form $a_1-a_2=a_3-a_4$. Removing for every such equality the corresponding difference from the count, we get $n(n-1)/2-n\le l-1$; that is, $$ l \ge \frac12(n^2-3n+2) = \frac12(n-1)(n-2)=\binom{n-1}2. $$