modular equation: x(x-1) = 0 mod 6 I have tried solving this equation, but I am not able to find solutions.
x(x-1) = 0 mod 6

Does anyone know how to do it?
Thank you so much in advance! :)
 A: If $$x(x - 1)\equiv 0\pmod 6$$ then for some $y$, $$6y = x(x - 1).$$ It follows, then, that for some $u$ and $v$, we have that $x = 6u$ or $x - 1 = 6v$. Of course, they both cannot be true, so $u\neq v$. Therefore, $$x \equiv 0, 1\pmod 6.$$ Notice that when $y = 1$, we get the equation, $$6 = x(x - 1).$$ Of course, in this case, $x = 3$, because $3\times 2 = 6$. However, $3\not\equiv 0, 1\pmod 6,$ so we get that this is a unique solution. In fact, there is also another solution: $x = -2$. But how do we find these unique solutions if none of them are congruent to $0, 1$ mod $6$? Well, we find these unique values of $x$ in the quadratic equation, $$x^2 - x - 6 = 0.$$ But there are more unique solutions out there! How can we find all of them and not just a certain amount? Well, in that case, I could explain, but @dxiv already has. Go to his answer below. $\ \downarrow\downarrow\downarrow$
A: Since the values of $x(x-1)$ repeat every six numbers, then we have six cases to check - when $x$ equals either $0,1,2,3,4$ or $5$. We can immediately rule out $x=0$ and $x=1$, because one of the factors will be $0$. Modulo $6$, we also have:
$$
\left\{ 
\begin{array}{c}
x=2;\ x(x-1)=2  \\ 
x=3;\ x(x-1)=0  \\ 
x=4;\ x(x-1)=0  \\
x=5;\ x(x-1)=2
\end{array}
\right. 
$$
Summing it all up, $x$ can be $0,1,3$ or $4$.
If the problem was modulo $12$, $24$, or anything too large to check by brute force, we can look at the factors of multiples of that number, and observe which ones are in the form $x(x-1)$.
A: Hint: $\,x(x-1)\,$ is the product of two consecutive numbers. One of them must be even, so the product itself is even i.e. $\,x(x-1) \equiv 0 \pmod{2}\,$. That leaves to solve $\,x(x-1) \equiv 0 \pmod{3}\,$. Since $\,3\,$ is a prime, this means either $\,x \equiv 0 \pmod{3}\,$, or $\,x \equiv 1 \pmod{3}\,$.
