I have the next problem: find $n\geq2$ such that the equation $x^2-x+\hat2=\hat0$ has only one solution in $\mathbb Z n$.
Let $a$ be the only solution of the equation.We get that $\hat1-a$ is also a solution so $a=\hat1-a$ , $\hat2a=\hat1$. $a={\hat2}^{-1}$ results that ${\hat2}^{-2}-{\hat2}^{-1}+\hat2=0$ so $\hat7=0$.
I think that $n$ can be any multiple of $7$, but on the solution is written only that $n=7$. Is somewhere a condition that $n$ needs to be prime?
You misinterpreted the relation you obtain: $7\equiv 0\mod n$ means $n$ is a divisor of $7$. As $7$ is prime, this implies $n=1$ (excluded by hypothesis, and uninteresting anyway). So the solution is $n=7$.