Consider f(x) = <img src="http://www.qbyte.org/puzzles/p088s1.gif" alt="sqrt(4 + sqrt(4 - x))" height="15" width="70">.
Then f(f(x)) = <img src="http://www.qbyte.org/puzzles/p088.gif" alt="sqrt(4 + sqrt(4 - sqrt(4 + sqrt(4 - x))))" height="19" width="134"> = x.
A solution to f(x) = x, if it exists, will also be a solution to f(f(x)) = x.
<h2>
Solving f(x) = x</h2>
Consider, then, f(x) = <img src="http://www.qbyte.org/puzzles/p088s1.gif" alt="sqrt(4 + sqrt(4 - x))" height="15" width="70"> = x.
Let y = <img src="http://www.qbyte.org/puzzles/p088s2.gif" alt="sqrt(4 - x)" height="14" width="39">. Then y2 = 4 − x.
We also have x = <img src="http://www.qbyte.org/puzzles/p088s3.gif" alt="sqrt(4 + y)" height="16" width="42">, from which x2 = 4 + y.
Subtracting, we have x2 − y2 = x + y.
Hence (x + y)(x − y − 1) = 0.
Since x <img src="http://www.qbyte.org/puzzles/ge.gif" alt="greater than or equal to" height="13" width="13"> 0 and y <img src="http://www.qbyte.org/puzzles/ge.gif" alt="greater than or equal to" height="13" width="13"> 0, x + y = 0 <img src="http://www.qbyte.org/puzzles/imp.gif" alt="implies" height="9" width="16"> x = 0, which does not satisfy f(x) = x.
Therefore we take x − y − 1 = 0, or y = x − 1.
Substituting into x2 = 4 + y, we obtain x2 = x + 3, or x2 − x − 3 = 0.
Rejecting the negative root, we have x = <img src="http://www.qbyte.org/puzzles/p088s4.gif" alt="(1 + root 13)/2" class="w" height="36" width="49">
Pure Solution
