Given polynomial $P(x) = x^2 + ax + b$ and that there only exists one $c$ such that $P^2(c) = c$. Calculate the minimum value of $a + b + c$. 
Given polynomial $P(x) = x^2 + ax + b$. Knowing that there only exists one value of real $c$ such that $P^2(c) = c$, calculate the minimum value of $a + b + c$.


Notation: $P^2(x) = (P \circ P)(x)$.

We have that $P^2(c) = c \iff (c^2 + ac + b)^2 + a(c^2 + ac + b) + b = c$
$$\iff [c^2 + (a - 1)c + b][c^2 + (a + 1)c + a + b + 1] = 0$$
For the above equation to have only one solution, $$(a - 1)^2 - 4b, (a + 1)^2 - 4(a + b + 1) \le 0 \implies b \ge \frac{(1 - a)^2}{4}$$
$$\implies \left[c^2 + (a - 1)c + \frac{(1 - a)^2}{4}\right] \cdot \left[c^2 + (a + 1)c + a + \frac{(1 - a)^2}{4} + 1\right] \ge 0$$
$$\implies (a + 2c - 1)^2[(a + 2c + 1)^2 + 4] \ge 0$$
It could be implied that 'the only solution' is $c = \dfrac{1 - a}{2}$, where $b = \dfrac{(1 - a)^2}{4}$.
$$a + b + c = a + \frac{1 - a}{2} + \frac{(1 - a)^2}{4} = \frac{a^2 + 3}{4} \ge \frac{3}{4}$$
Is the above solution correct? If not, could you provide with one? Thanks for your attention.

Just a less computational argument for the first part of the solution: if $P(x) = x$ had no real solutions, then $P(x)>x$ for all real $x$ and hence $P(P(x)) > P(x) > x$, contradicting the existence of $c$. Any solution of $P(x) = x$ is also a solution of $P(P(x)) =x$, hence $P(x) = x$ must have exactly one solution, $c$. That implies $P(x) = x+(x-c)^2$.
Therefore $a=1-2c$, $b= c^2$ and $a+b+c = c^2-c+1 = (c-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}$. Equality holds for $c=\frac{1}{2}$, $a=0$, and $b = \frac{1}{4}$.
