My Algorithm

I didn't know anything about continued fractions before I saw this problem. The Wikipedia article is pretty long: en.wikipedia.org/wiki/Continued_fraction
Even more important, there is another Wikipedia article about the relationship of square roots and continued fractions: en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion

The whole algorithm can be found in my function getPeriodLength.
My first step is to find the integer part of the square root. That's pretty easy:
root = sqrt(x) → if root*root == x then x is a perfect square and we can abort.

In the example above, x=4 and root=4. Even though a_0 was the first step, there is a step before it (kind of a_{-1}):
sqrt{23} = 0 + dfrac{sqrt{23} - 0}{1}
The variable numerator = 0 refers to the zero which is subtracted in the numerator.
The variable denominator = 1 refers to the 1 in the denominator.
The variable a = root will be a_0.

I do the following to compute a_1:
numerator = denominator * a - numerator → numerator_1 = 1 * 4 - 0 = 4

denominator = \lfloor dfrac{x - numerator^2}{denominator} \rfloor → denominator_1 = \lfloor dfrac{23 - 4^2}{1} \rfloor = 7

a = \lfloor dfrac{root + numerator}{denominator} \rfloor → a_1 = \lfloor dfrac{4 + 4}{7} \rfloor = 1

As you can see in the problem statement, the first line contained the fraction dfrac{sqrt{23}+4}{7}

And for a_2:
numerator_2 = 7 * 1 - 4 = 3

denominator_2 = \lfloor dfrac{23 - 3^2}{7} \rfloor = 2

a_2 = \lfloor dfrac{4 + 3}{2} \rfloor = 3

The mathematical reasoning is based on the general concept that (a - b) * (a + b) = a^2 - b^2.
If a = sqrt{x} and b = y then (sqrt{x} - y) * (sqrt{x} + y) = x - y^2

For x = 23 and y = \lfloor sqrt{x} \rfloor = \lfloor sqrt{23} \rfloor = 4:
dfrac{1}{(sqrt{23} - 4)}

= dfrac{sqrt{23} + 4}{(sqrt{23} - 4) * (sqrt{23} + 4)}

= dfrac{sqrt{23} + 4}{23 - 4^2}

= dfrac{sqrt{23} + 4}{7}

Then the largest possible integer such that 0 < numerator < denominator:
= dfrac{sqrt{23} + 4 - 7 + 7}{7}

= 1 + dfrac{sqrt{23} - 3}{7}

A loop can be identified by keeping track of the tupel (a, numerator, denominator). If it appears a second time, we have entered a loop.
We could create a data structure - something like std::set<std::tupel> - but Wikipedia mentions a neat trick that I don't understand:
The equation a == 2 * root becomes true as soon as we enter a loop.