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: For the record, integers $m,n$ such that $m^2-mn+n^2$ is a perfect square (integer) do exist, for instance $m=15$, $n=8$, for which $m^2-mn+n^2=169=13^2$. This is just to say that the answer to at least one unambiguous form of the question is affirmative.
A: $$a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}$$
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 $a=1$. 
A: Here is an alternative solution that I came up with shortly after Ahmed S. Attaalla posted his answer.
If $a^2 - a + 1 = b^2$
for some $b \in \mathbb{Q}$ then $a^2 - a + (1-b^2) = 0$, whence by the quadratic formula we have
$$a = \frac{1 \pm \sqrt{1 - 4\left(1-b^2\right)}}{2} = \frac{1 \pm \sqrt{4b^2-3}}{2}$$
If this is to be rational, then we need $4b^2 - 3$ to be a rational square.  In other words we need to find $c$ such that $4b^2 - 3 = c^2$, or equivalently such that $\left(2b\right)^2 - c^2 = 3$.
Now the problem has been reduced to:  "Find two rational numbers (in this base $2b$ and $c$) whose squares differ by a given number (in this case $3$)."  This is a classical problem whose solution is given in Diophantus's Arithmetica (Book II, Problem 10). The solution given there (generalized and in modern language and notation) runs more or less as follows:
Choose any nonzero rational parameter $k$, and set $2b = c + k$.  Then the condition $\left(2b\right)^2 - c^2 = 3$ reduces to
$$(c+k)^2 - c^2 = 3$$
$$c^2 + 2ck + k^2 - c^2 = 3$$
$$c = \frac{3-k^2}{2k}$$
Now follow the breadcrumbs back to the original problem:  We have
$$2b = \frac{3-k^2}{2k} + k = \frac{3+k^2}{2k}$$
$$b= \frac{3+k^2}{4k}$$
So
$$\begin{align*}a &= \frac{1 \pm \sqrt{4\left(\frac{3+k^2}{4k}\right)^2-3}}{2}\\
&= \frac{1 \pm \sqrt{4\left(\frac{9+6k^2+k^4}{16k^2}\right)-3}}{2}\\
&=\frac{1\pm\sqrt{\frac{9-6k^2+k^4}{4k^2}}}{2}\\
&=\frac{1 \pm \frac{3-k^2}{2k}} {2}\\
&=\frac{2k \pm (3 - k^2)}{4k}\\
\end{align*}$$
To summarize:  for any rational $k\ne 0$, we have $a = \frac{2k \pm (3 - k^2)}{4k}$ are two rational numbers satisfying the conditions of the problem.
For example, with $k$ = 1, we get $a = \frac{2\pm(2)}{4}$, which produces the two trivial solutions $a=0,1$.  (It's interesting to note that $k=3$ also produces the same trivial solution.)
For a nontrivial example, with $k=4$, we get $a = \frac{8\pm 13}{16}$.  We can confirm that
$$\left(\frac{21}{16} \right)^2 - \left(\frac{21}{16} \right) + 1 =\frac{441-336+256}{256} = \frac{361}{256} = \left(\frac{19}{16}\right)^2 $$
and
$$\left(\frac{-5}{16} \right)^2 - \left(\frac{-5}{16} \right) + 1 =\frac{25+80+256}{256} = \frac{361}{256} = \left(\frac{19}{16}\right)^2 $$
