Let $c>0$ be a positive real number. We define the sequence $\left(x_{n}\right)$ by $x_{0}=0$ and

$$
x_{n+1}=x_{n}^{2}+c
$$

for each $n \geq 0$. For which values of $c$ is it true that $\left|x_{n}\right|<2016$ for all $n$ ?
The answer is $c \leq \frac{1}{4}$.

First, we show that $c \leq \frac{1}{4}$ all work. Clearly it suffices to prove the result when $c=\frac{1}{4}$. In that case, the sequence is defined by $x_{n+1}=x_{n}^{2}+\frac{1}{4}$. We claim that $x_{n} \leq \frac{1}{2}$ for all $n$. Indeed, this follows by induction, since it is true for $n=1$ and for the inductive step we have

$$
x_{n+1}=x_{n}^{2}+\frac{1}{4} \leq \frac{1}{4}+\frac{1}{4}=\frac{1}{2} .
$$

Now we show all $c>\frac{1}{4}$ fail. Assume on the contrary that $\left|x_{n}\right|<2016$ for all $n$. Then since the real interval $[0,2016]$ is compact, we must have that $x_{n}$ converges to some limit $x$. This limit must then satisfy

$$
x^{2}=x+c .
$$

Thus the discriminant $1-4 c$ is nonnegative, so $c \leq \frac{1}{4}$ must hold.