We have the equation
$$x_{n+1} = ax_n(1-x_n) - v_n$$
Why are there only fixed points for $(a-1)^2 - 4av_0 \geq 0$?
Show that if $ 1<a<4$, there are 2 fixed points with $0<p_1 < p_2 <1$
For the first problem, I was able to calculate that the fixed points would be given by:
$$ p_{1,2} = \dfrac{a-1}{2a} \pm \sqrt{(\dfrac{a-1}{2a})^2 - \dfrac{v_o}{a}}$$
But after that I'm stuck. The second problem I have no idea how to prove but I think that is because I don't understand the first problem.
Taking from where you stopped, consider the expression under the root and factor out $(2a)^2 = 4a^2$:
$$ \begin{split} p_\pm = \frac{a-1}{2a} \pm \sqrt{\left( \frac{a-1}{2a}\right)^2 - \frac{v_0}{a}} = \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0}, \end{split} $$ which is a real number provided $(a-1)^2 \geq 4av_0$, other wise the $\sqrt{\cdot}$ is complex.
2nd Problem For the last problem it suffices to prove that $$ 0 \leq \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0} \leq 1, $$
or equivalently that $$ 0 \leq a-1 \pm \sqrt{(a-1)^2 - 4av_0} \leq 2a. $$
Right inequality:
(assuming $1\leq a \leq 4$), certainly $a - 1 \leq a$ and also $$ \sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = (a-1) < a, $$ so the sum (and surely the difference) should be no more than $a+a = 2a$.
Left inequality
Since $$ \sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = a-1 $$ and so subtracting both sides from $a-1$ yields the result.