Suppose the real numbers $x_1, x_2, \dots, x_5$ satisfy the given set of equations. Let $a = x_1 + x_2 + \dots + x_5$. Then, each of the five numbers $x_j, j = 1, 2, \dots, 5$ satisfies the quadratic equation $x^2 - a x - 1 = 0$. This equation has 2 distinct real roots $\frac{a \pm \sqrt{a^2 + 4}}{2}$. Denote one of the roots equaling $x_1$ by $\alpha$ and the other by $\beta$. Then we have $\alpha \beta = -1$. Furthermore, if we assume that exactly $k$ among the numbers $x_j, j = 1, 2, \dots, 5$ equal $\alpha$, and $5-k$ among them equal $\beta$, we get $a = k\alpha + (5-k)\beta = k\alpha - \frac{5-k}{\alpha}$. Substituting this into the equation $\alpha^2 - a\alpha - 1 = 0$, we obtain $(k-1)\alpha^2 = 4-k$. Since $\alpha^2 > 0$, we must have either $k=2$ or $k=3$. If we have $k=2$, then we get $\alpha = \pm\sqrt{2}$, while if $k=3$, we get $\alpha = \pm\frac{\sqrt{2}}{2}$. It is easy to check that each of these values of $\alpha$ and the corresponding values for $\beta$ satisfy the quadratic equation $x^2 - a x - 1 = 0$, so the desired solution for the problem is given by $\pm\sqrt{2}, \pm\frac{\sqrt{2}}{2}$.