74th Romanian Mathematical Olympiad

We shall denote by $M$ the set of all positive integers $n$, with $n\ge 2$, such that the equation (1) has a unique solution in the ring $({\mathbb{Z}}_n,+,\cdot )$. We will show that $M=\{11\}$.

In the ring $({\mathbb{Z}}_{11},+,\cdot )$, the equation (1) can be equivalently written as $$x^2-\hat{3}\cdot x+\hat{5}=\hat{0}⟺x^2-\widehat{14}\cdot x+\widehat{49}=\hat{0}⟺(x-\hat{7}{)}^2=\hat{0},$$ with the unique solution $x=\hat{7}$. Hence, $11\in M$.

Since $k^2-3k+5=(k-1)(k-2)+3$ is an odd number for any integer $k\in \mathbb{Z}$, the equation $x^2-\hat{3}\cdot x+\hat{5}=\hat{0}$ has no solutions in the ring $({\mathbb{Z}}_n,+,\cdot )$ for any even integer $n$, so that $M\subseteq 2\cdot \mathbb{N}+1$.

Let $n\in M$ be arbitrary, and $x\in {\mathbb{Z}}_n$ the unique solution of the equation (1). For the element $y=\hat{3}-x$ we have then $$y^2-\hat{3}\cdot y+\hat{5}=\hat{9}-\hat{6}\cdot x+x^2-\hat{3}\cdot \hat{3}+\hat{3}\cdot x+\hat{5}=x^2-\hat{3}\cdot x+\hat{5}=\hat{0},$$ so $y$ is also a solution of the equation (1). Because of the uniqueness condition, we deduce that $x=y=\hat{3}-x$, or, equivalently, $\hat{2}\cdot x=\hat{3}$. Since $n$ is odd, $\hat{2}$ is invertible, and we obtain that $x=\hat{3}\cdot {\hat{2}}^{-1}$.

The fact that $x=\hat{3}\cdot {\hat{2}}^{-1}$ is a solution of the equation can be written equivalently, considering the fact that $n$ is odd, as: $$\begin{gathered} (\hat{3}\cdot {\hat{2}}^{-1}{)}^2-\hat{3}\cdot (\hat{3}\cdot {\hat{2}}^{-1})+\hat{5}=\hat{0}⟺\hat{4}\cdot ((\hat{3}\cdot {\hat{2}}^{-1}{)}^2-\hat{3}\cdot (\hat{3}\cdot {\hat{2}}^{-1})+\hat{5})=\hat{0}⟺ \\ ⟺\hat{9}-\widehat{18}+\widehat{20}=\hat{0}⟺\widehat{11}=\hat{0}⟺n\mid 11. \end{gathered}$$ Since $n\ge 2$, we conclude that $n=11$. Hence, $M\subseteq \{11\}$.

Then $11\in M\subseteq \{11\}$ implies that $M=\{11\}$.