Perfect Square In Polynomial Form

First note that $x = 0$ is a solution. Now let's assume $x \ne 0$ . We have $\begin{aligned} (x^2+\frac12 x + 1)^2 &= x^4 + x^3 + \frac94 x^2 + x + 1 > x^4+x^3+x^2+x+1 \\ (x^2+\frac12 x)^2 &= x^4+x^3+\frac14 x^2 < x^4+x^3+x^2+x+1 \\ \end{aligned}$ where the latter is true because $\frac34 x^2+x+1$ has its minimum at $x = -2/3, y = 2/3$ , so it's always positive.

So the square root of $x^4+x^3+x^2+x+1$ is in the length-1 interval $(x^2+\frac12x,x^2+\frac12x+1)$ . The endpoints of the interval are either integers or halves of odd integers. The only way there can be an integer in this interval is if it is the latter, and the integer is the midpoint $x^2+\frac12x+\frac12$ .

Now we solve $\begin{aligned} (x^2+\frac12x+\frac12)^2 &= x^4+x^3+x^2+x+1\\ x^4+x^3+\frac54x^2+\frac12x+\frac14 &= x^4+x^3+x^2+x+1\\ \frac14x^2-\frac12x-\frac34 &=0\\ \frac14(x-3)(x+1)=0 \end{aligned}$ so the solutions are $x = 0$ (from above), $x = 3$ , $x = -1$ . The sum is $\fbox{2}$ .