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 ?
 A: The key is to recognize that the left hand side is, by definition of the floor function, an integer, so the right hand side must also be an integer, and hence we must have $x=3k+1$ for some $k\in\mathbb{Z}$.  This gives (on doing a little algebra)
$${x^2-x+1\over2}={3k(3k+1)\over2}+{1\over2}$$
Note that $2$ divides either $k$ or $3k+1$, so $3k(3k+1)/2$ in an integer for any $k$, hence
$$\big\lfloor{x^2-x+1\over2}\big\rfloor={3k(3k+1)\over2}$$
Thus the equation to solve (for integer $k$) is
$${3k(3k+1)\over2}=k$$
But the only integer solution is $k=0$, since $k\not=0$ implies $3(3k+1)=2$.  So $x=1$ is the only real solution to the original equation.
A: 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.
