A subset $A$ of a group $G$ is defined to be *self-linked* if $A\cap gA\ne\emptyset$ for all $g\in G$. This happens if and only if $AA^{-1}=G$.

For a finite group $G$ denote by $sl(G)$ the smallest cardinality of a self-linked set in $G$. It is clear that $sl(G)\ge \sqrt{|G|}$. A more accurate lower bound is $sl(G)\ge \frac{1+\sqrt{4|G|-3}}2$. By a classical result of Singer (1938), for any power $q=p^k$ of a prime number $p$, the cyclic group $C_n$ of cardinality $n=1+q+q^2$ contains a self-linked subset of cardinality $1+q$, which implies that $sl(C_n)=1+q=\frac{1+\sqrt{4n-3}}2$. So, for such numbers $n$ the lower bound $\frac{1+\sqrt{4n-3}}2$ is exact. In this paper we prove the upper bound $sl(C_n)\le \sqrt{2n}$ holding for all $n\ne 4$.

**Problem 1.** Is $sl(C_n)=(1+o(1))\sqrt{n}$?

This problem is equivalent to

**Problem 2.** Does the limit $\lim_{n\to\infty}{sl(C_n)}/{\sqrt{n}}$ exist?

If the answer to Problems 1,2 are negative, then we can also ask

**Problem 3.** Evaluate the constant $\lambda:=\limsup_{n\to\infty}{sl(C_n)}/{\sqrt{n}}$.

At the moment it is known that $1\le\lambda\le\sqrt{2}$.