Three, given that $a$ is an integer, the equation $x^{2}+(2 a+1) x$ $+a^{2}=0$ has integer roots $x_{1}, x_{2}, x_{1}>x_{2}$. Try to find the value of $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
Three, Solution: From the problem, we know that the discriminant $4a + 1$ is a perfect square, so $a \geqslant 0$. Since $4a + 1$ is odd, we can set $(2k + 1)^2 = 4a + 1$, solving for $a = k(k + 1)$.
$$
x_{1,2} = \frac{1}{2} \left[ \left( -2k^2 - 2k - 1 \right) \pm \sqrt{(2k + 1)^2} \right].
$$

When $k \geqslant 0$, $x_1 = -k^2$, $x_2 = -(k + 1)^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$,
When $k < 0$, $x_1 = -(k + 1)^2$, $x_2 = -k^2$, then $\sqrt[4]{x_1^2} - \sqrt[4]{x_2^2} = -1$.