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
$$x^2 - \hat{3} \cdot x + \hat{5} = \hat{0} \iff x^2 - \widehat{14} \cdot x + \widehat{49} = \hat{0} \iff (x - \hat{7})^2 = \hat{0},$$
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:
$$(\hat{3} \cdot \hat{2}^{-1})^2 - \hat{3} \cdot (\hat{3} \cdot \hat{2}^{-1}) + \hat{5} = \hat{0} \iff \hat{4} \cdot ((\hat{3} \cdot \hat{2}^{-1})^2 - \hat{3} \cdot (\hat{3} \cdot \hat{2}^{-1}) + \hat{5}) = \hat{0} \iff \\ \iff \hat{9} - \widehat{18} + \widehat{20} = \hat{0} \iff \widehat{11} = \hat{0} \iff n \mid 11.$$
Since $n \ge 2$, we conclude that $n = 11$. Hence, $M \subseteq \{11\}$.

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