The vicious cycle (1)

Nice 3-in-1 problem!!!

Refer to the solution of Problem 3 , we have: $\space P_2 = 1 + P_3$ .

Refer to the solution of Problem 2 , we have: $\space P_1^2 = 25P_2 = 25(1+P_3)$ .

Now: $\space P_1 = min (x^2+P_3x+1)$ .

Let $\space f(x) = x^2+P_3x+1\quad \Rightarrow \dfrac {df(x)}{dx} = 2x + P_3$

$\therefore \dfrac {df(x)}{dx} = 0 \quad \Rightarrow x = \dfrac {-P_3}{2}\text{. }$ Since $\space \dfrac {d^2f(x)}{dx^2} > 0$ , $\Rightarrow P_1 = f_{min} (x) = f \left( \frac {-P_3}{2} \right) = \left( \frac {-P_3}{2} \right)^2 + P_3 \left( \frac {-P_3}{2} \right) +1 = 1 - \dfrac {P_3^2}{4}$

$\Rightarrow 4P_1 = 1-P_3^2\quad \Rightarrow 16P_1^2 = (4 - P_3^2)^2$

$\Rightarrow 16[25(1+P_3)] = 16-8P_3^2+P_3^4$

$\Rightarrow P_3^4 -8P_3^2-400P_3-384= 0$

$\Rightarrow (P_3-8)(P_3^3+8P_3^2+56P_3+48)= 0$

$\Rightarrow P_3 = 8\quad \Rightarrow P_2 = 1 + P_3 = 9$

$\Rightarrow P_1 = 1-\dfrac {P_3^2}{4} = 1 - \dfrac {8^2}{4} = \boxed {-15}$