Celebrating year 2016!

We can show that $f_{n+1}(x)-2 \; = \; \frac{2-f_n(x)}{\big(\sqrt{7-\sqrt{\,7+f_n(x)}}+2\big)\big(3 + \sqrt{7 + f_n(x)}\big)}$ so that, provided that the formula makes sense (and we can calculate the square roots) $\big|f_{n+1}(x)-2\big| \,\le\, \tfrac16\big|f_n(x) - 2\big|$ While $f_{2016}(2) = f_0(2) = 2$ , $f_{2016}(x)$ is closer to $2$ than is $x$ for any valid $x \neq 2$ , and so $2$ is the only solution.

N.B. We can calculate $f_1(x)$ for any $-7 \le x \le 42$ . From the above observations, $f_1(x)$ will lie in the same interval. By induction, we can calculate $f_n(x)$ for all $n \ge 1$ provided that $-7 \le x \le 42$ .

It is clear that the domain of $f_1(x)$ is the closed interval $[-7, 42],$ and $f_1'(x)= -\frac{1}{4 \sqrt{7+x} \sqrt{7-\sqrt{7+x}}}.$ Since $f_1'(x)<0$ for any $x\in [-7, 42]$ then $f_1(x)$ is decreasing on $[7, 42]$ , and, additionally, its maximum and minimum value, respectively, are $f(-7)=\sqrt 7 \approx 2.6$ and $f(42)=0.$ Thus the range of $f(x)$ is contained in $[0, 3],$ and, therefore, for any natural number $n>1$ and any value of $x$ in $[-7, 42],$ we have that $f_{n}(x)=f(f_{n-1}(x))$ is a number in the interval $[0, 3].$ This implies that for whatever natural number $n>1$ the equation $f_{n}(x)=x$ does not have real solutions outside the interval $[0, 3].$

We claim that for any natural number $n$ the equation $f_{2n}(x)=x,\:\:\:\:\:\:\:\:(1)$ has a unique real solution in the interval $[0, 3],$ and, therefore, according to what we explained above, a unique real solution everywhere.

To prove our claim we have to show first that for any integer $n$ , the following inequality holds $|f_n'(x)|< 1\:\: \text{for any} \:\:x\in [0, 3]. \:\:\:\:(2)$

Let us prove it by induction. Let us start proving that the inequality is true in the case that $n=1.$ It easy to see that $f_1''(x)=\frac{14-3 \sqrt{x+7}}{(16 (x+7)^{3/2} (7-\sqrt{x+7})^{3/2}}$ is always positive for any $x$ in the interval $[0, 3],$ and then $f_1'(x)$ is increasing on that interval and always negative, as we saw before. Therefore $|f_1'(x)|\leq -f_1'(0)=\frac{1}{4 \sqrt{7} \sqrt{7-\sqrt{7}}}<1.$ So this proves the case where $n=1.$ Now let $k$ be any natural number and $x\in[0, 3]$ . Let us assume that the inequality (2) is true for $n=k,$ then since $f_{k+1}'(x)=f_{k}'(f_1(x))f_1'(x)$ and $f_1(x)$ is in the interval $[0, 3]$ when $x$ is in that interval , we obtain that the inequality (2) is also true for $n=k+1.$ This complete the proof of (2).

Since $f_{1}(2)=2$ , then $f_{2n}(2)=2$ for any natural number $n.$ This proves that the equation (1) has at least a real solution in $[0, 3]$ . Now we are going to prove the uniqueness. Let us prove it by contradiction. Assume that $x_1$ and $x_2$ are both solutions of (1) in $[0, 3]$ . Then $\frac{f_{2n}(x_2)-f_{2n}(x_1)}{x_2-x_1}=\frac{x_2-x_1}{x_2-x_1}=1.$ According to the Mean Value Theorem, this would imply that there must be a number $c$ in between $x_1$ and $x_2$ such that $f_{2n}'(c)=1$ , but this would contradict the inequality (2).

So the answer to the question in the problem is that $f_{2016}(x)=x$ has a unique real solution.

This could be solved graphically, I guess. Looking at the recursive function it can be deduced that, $\large f_{2016}{(x)} = \sqrt{7-\sqrt{\,7+\sqrt{\,7-\sqrt{\,7+... \sqrt{\,7+x}}}}}$ where 7 is repeated 4030 times. It's important to note that the pattern is repeated a finite number of times.

Now if we try to work backwards. The graph of $\sqrt{\,7+x}$ intersects x at exactly one point. The graph of $\sqrt{\,7-\sqrt{\,7+x}}$ becomes steeper but still intersects x at exactly one point. As we continue with pattern the graph becomes ever so steeper but there's no reason why it would change its behavior suddenly.

I acknowledge the pretense of this solution is superficial. Nevertheless.

How about a direct simple solution?

We know that $f_{2016} (x)$ $\rightarrow$ 2 or just extremely near to 2 by repeated substitutions of an evaluation.

$f_{2016} (x)$ = 2+ XOR 2-.

No matter how the case cannot be exactly 2 instead of 2+ or 2-, we can always have one leveling which enables us to get either $f_{2016} (2+)$ = 2+ XOR $f_{2016} (2-)$ = 2- to be true.

Since $f_{2016} (x)$ is a one value constant, $x$ can always choose to equate the constant.

Therefore, the answer is "Only one real solution."!