smc

The solutions to this equation are $z=1$, $z=-1\pm i\sqrt{3}$, and $z=-2\pm i\sqrt{2}$. Consider the five points $(1,0)$, $\left(-1,\pm \sqrt{3}\right)$, and $\left(-2,\pm \sqrt{2}\right)$; these are the five points which lie on $\mathcal{E}$. Note that since these five points are symmetric about the $x$-axis, so must $\mathcal{E}$. Now let $r=b/a$ denote the ratio of the length of the minor axis of $\mathcal{E}$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$-coordinate by a factor of $r$, $\mathcal{E}$ is sent to a circle ${\mathcal{E}}^{\prime }$. Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$, $\left(-r,\pm \sqrt{3}\right)$, and $\left(-2r,\pm \sqrt{2}\right)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1=(r,0)$, $P_2=\left(-r,\sqrt{3}\right)$, and $P_3=\left(-2r,\sqrt{2}\right)$ lies on the $x$-axis. Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1{P}_2}$ and $\overline{P_1{P}_3}$ are$$y=\frac{\sqrt{3}}{2}+\frac{2r}{\sqrt{3}}x\text{and}y=\frac{\sqrt{2}}{2}+\frac{3r}{\sqrt{2}}\left(x+\frac{r}{2}\right)$$respectively. These two lines have different slopes for $r\ne 0$, so indeed they will intersect at some point $(x_0,y_0)$; we want $y_0=0$. Plugging $y=0$ into the first equation yields $x=-\frac{3}{4r}$, and so plugging $y=0$ into the second equation and simplifying yields$$-\frac{1}{3r}=x+\frac{r}{2}=-\frac{3}{4r}+\frac{r}{2}.$$Solving yields $r=\sqrt{\frac{5}{6}}$. Finally, recall that the lengths $a$, $b$, and $c$ (where $c$ is the distance between the foci of $\mathcal{E}$) satisfy $c=\sqrt{a^2-b^2}$. Thus the eccentricity of $\mathcal{E}$ is $\frac{c}{a}=\sqrt{1-{\left(\frac{b}{a}\right)}^2}=\sqrt{\frac{1}{6}}$ and the requested answer is $\boxed{7}$.