4. For positive integer $a$ and integers $b, c$, in the rectangular coordinate system $O-xyz$, points $O(0,0,0)$, $A(a, b, c)$, and $B\left(x^{2}, x, 1\right)$ satisfy $=\frac{\pi}{2}$. The real number $x$ has exactly two distinct real solutions $x_{1}, x_{2} \in (0,1)$. The minimum value of $a$ is $\qquad$.


4. 5 Detailed Explanation: For the equation $a x^{2}+b x+c=0$ to have two distinct real roots $x_{1}, x_{2}$ in the open interval $(0,1)$, assume $0<c>0, a \geqslant 2$; from $x_{1}+x_{2}=-\frac{b}{a}>0$ we know $b<0$, and $x_{1} x_{2}=\frac{c}{a}>0$, thus $c>0$. Therefore, $4 a c+1 \leqslant(a+c-1)^{2}$, which simplifies to $a^{2}+c^{2}-2 a c-2 a-2 c \geqslant 0$. If $a=2$, then $c^{2}-6 c \geqslant 0$, leading to $c \geqslant 6$, a contradiction! If $a=3$, then $c^{2}-8 c+3 \geqslant 0$, leading to $c \geqslant 8$, a contradiction! If $a=4$, then $c^{2}-10 c+8 \geqslant 0$, leading to $c \geqslant 10$, a contradiction! If $a=5$, then $c^{2}-12 c+15 \geqslant 0$, leading to $c=1$, or $c \geqslant 11$, so we can take $c=1, b=-5$, solving the equation $5 x^{2}-5 x+1=0$, we get $\left(x_{1}, x_{2}\right)=\left(\frac{5+\sqrt{5}}{10}, \frac{5-\sqrt{5}}{10}\right)$. Therefore, $a_{\min }=5$.