Solve $[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ How can one solve the equation :
$[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ ?
Such that $[x]$ is the integer part of $x$.
By definition : 
$[\frac{x^2-x+1}{2}]<=\frac{x^2-x+1}{2}<[\frac{x^2-x+1}{2}]+1$.
So :
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}<\frac{x-1}{3}+1$.
1)
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}$ <=> $3x^2-5x+5>0$.
2)
$\frac{x^2-x+1}{2}<\frac{x-3}{2}+1$ <=> $3x^2-5x-1<0$.
How can I solve the system ?

Write the original equation as:
$$\left\lfloor\frac{x^2-x+1}{2}\right\rfloor=\frac{x-1}{3} \tag{1}$$
By definition of the floor function, both sides of the equation must be integers, i.e. $x\in\mathbb{Z}$. So then $x^2-x+1$ is an odd integer, whether $x$ is even or odd. Therefore
$$\left\lfloor\frac{x^2-x+1}{2}\right\rfloor=\frac{x^2-x+1}{2}-\frac{1}{2}=\frac{x^2-x}{2}$$
Substituting back into (1):
$$\frac{x^2-x}{2}=\frac{x-1}{3} \implies \frac{x(x-1)}{2}=\frac{x-1}{3} \implies x=\tfrac{2}{3}\text{ or }x=1$$
Of these $x=1$ is the only integer potential solution, and checking against (1) shows that it is a solution to the original equation.
