Question:
$$4 + \dfrac{1}{2 + \dfrac{1}{1 + \dfrac{1}{3 + \dfrac{1}{1+\dfrac{1}{2 + \dfrac{1}{8 + \dfrac{1}{\ddots}}}}}}} = \sqrt{A}$$
Find the positive integer $A$ in the equation above.
Details and assumptions
- The pattern repeats 2, 1, 3, 1, 2, 8 infinitely, but 4 comes only once, i.e., at the beginning.
My Approach:
I was thinking whether I should calculate each fractional part of the denominators or is there any other shorter method to solve this problem.
Let's take a simpler example first: say, $$\sqrt{A}=1+{2\over 3+{2\over 3+...}}.$$ Here the repetition $2,3,2,3, ...$ continues forever after the $1$.
The problem is that this looks on the face of it like some sort of "infinitary" equation (or rather, it's a normal equation with an "infinitary" term in it). We want to somehow "tame" the right-hand side.
This is where the repetition, or self-similarity, comes in. Let $x={2\over 3+{2\over 3+...}}$. Then if we take the reciprocal of both sides, we see something interesting: $${1\over x}={3+x\over 2}.$$ Solving for $x$ yields $$x^2+3x-2=0,$$ which has as its solutions $$-3\pm\sqrt{9+8}\over 2,$$ that is, $$x={\sqrt{17}-3\over 2}\quad\mbox{or}\quad x={-\sqrt{17}-3\over 2}.$$ Since $x$ is clearly positive$^*$, it has to be the former.
Plugging back into the original expression, we have $$\sqrt{A}=1+x={\sqrt{17}-1\over 2}.$$ Squaring both sides finishes the problem.
Can you see how to follow a similar argument here? (It's definitely going to be messier since your patter takes longer to repeat, but the basic idea still applies.)
An interesting feature of this argument is: it totally breaks down if we don't have repetition! To see what I mean, try to calculate $$1\over 2+{3\over 4+{5\over 6+...}}.$$ In general, things are much more complicated and the "naive" argument above simply doesn't work.
There is a subtlety here; namely, that we haven't really defined what an "infinitary" fraction (called a "continued fraction") like this actually is. But it should be clear that any reasonable definition will yield a positive answer here, since no piece involved is negative.