Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square? Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square?
Context for this question:  For pedagogical purposes I was trying to create an example of a cubic polynomial with both its critical values and its zeros all at integers.  After fumbling around with it unsuccessfully for a while, I realized that the problem of creating such an example was equivalent to finding a rational number $a$ ($a \ne 0,1$) with the property that $a^2 - a + 1$ is a perfect square.  (Equivalently, find two integers $m,n$ such that $m^2 - mn + n^2$ is a perfect square.)  I haven't been able to find an example and strongly suspect that it's not possible, but can't see a simple proof of why that should be so.
Any proof or counterexample to the conjecture that no such $a$ exists?

EDIT:  I may have found the beginning of an argument, within a few minutes of posting the question.  I'd appreciate feedback.
Suppose $b$ is a positive integer such that $a^2-a+1=b^2$.  Then $a-1 = a^2-b^2$, so $a-1= (a-b)(a+b)$.  Now if $a-b > 1$ then we have
$$a-1 = (a-b)(a+b) > a+b$$ whence $b < -1$, contradicting the assumption that $b$ is a positive integer.
So if such a $b$ exists, it must be that $a-b \le 1$.  Then.... (?)

$$a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}$$
So what you ask are there any rational numbers $u=a-\frac{1}{2}$ and $v$ such that,
$$u^2+\frac{3}{4}=v^2$$
$$v^2-u^2=\frac{3}{4}$$
$$(v-u)(v+u)=\frac{3}{4}$$
Let $x$ and $y$ be two real numbers that multiply to $\frac{3}{4}$, with the condition that $\frac{x+y}{2}$ and $\frac{y-x}{2}$ are both rational. Note that this condition equates to the condition that both $x$ and $y$ are rational. 
Solving the system,
$$v-u=x$$
$$v+u=y$$
Yields the rational solutions $v=\frac{x+y}{2}$ and $u=\frac{y-x}{2}$. Which implies $a=\frac{y-x+1}{2}$. 

If you restrict $v$ to be an integer. Then it must be that,
$$x+y=2v \in \mathbb{Z}$$
$$xy=\frac{3}{4}$$
Eliminating $y$ from this system gives, 
$$x(2v-x)=\frac{3}{4}$$
Or,
$$4x^2-8vx+3=0$$
By the rational root theorem, the only rational solutions to the above equation are,
$$x=\pm \frac{3}{2}, \pm \frac{3}{4}, \pm 3, \pm \frac{1}{2}, \pm \frac{1}{4},  \pm 1$$
The only solutions which correspond to integer values of $v$ are $x=\pm \frac{3}{2}, \pm \frac{1}{2}$. These give $v=\pm 1$. So we have that,
$$x+y=\pm 2$$
$$xy=\frac{3}{4}$$
Solving for $x$ and $y$ leads to the conclusion that the only possible integer solutions for $a$ is $a=0$ or $a=1$. 
