Problem 1. (a) Solve the equation $x^{2}-x+\hat{2}=\hat{0}, x \in \mathbb{Z}_{7}$.

(b) Determine the natural numbers $n \geq 2$, for which the equation $x^{2}-x+\hat{2}=\hat{0}$, $x \in \mathbb{Z}_{n}$, has a unique solution.

Mathematical Gazette
Solution. (a) Since 4 and 7 are coprime, and $\left(\mathbb{Z}_{7},+, \cdot\right)$ is a field, the given equation is equivalent to $\hat{4} x^{2}-\hat{4} x+\hat{1}=\hat{0}$, i.e., $(\hat{2} x-\hat{1})^{2}=\hat{0}$, hence $\hat{2} x=\hat{1}$, from which $x=\hat{4}$.

2 points

(b) Let $n \geq 2$ be a natural number for which the given equation has a unique solution, and let $a \in \mathbb{Z}_{n}$ be the respective solution. Then $(\hat{1}-a)^{2}-(\hat{1}-a)+\hat{2}=a^{2}-a+\hat{2}=\hat{0}$, so $\hat{1}-a$ is a solution to the given equation. ................................... 3 points Therefore, $a=\hat{1}-a$, hence $\hat{2} a=\hat{1}$. In particular, $\hat{2}$ is invertible in the ring $\mathbb{Z}_{n}$ and $a=\hat{2}^{-1}$. It follows that $\hat{2}^{-2}-\hat{2}^{-1}+\hat{2}=\hat{0}$, from which $\hat{1}-\hat{2}+\hat{8}=\hat{0}$, i.e., $\hat{7}=\hat{0}$. Therefore, $n$ is a divisor of 7, so $n=7$. ................. 2 points