A Generalization

There's one last point that I'd like to address. You may have noticed that I don't always consider a proof of something to be a reason for it. That's the case here. We started with T(u), performed two sequences of operations on it, saw that the results were equal, and deduced that the middle part had reflection symmetry. But why? What is it about T(u) that forces the symmetry to exist? For once, I actually have an answer.

The key is to generalize. For the operations to make sense and the final expansions to be equal (except for the middle part), the initial expansion needs to have the following form.

a₀a₁ABCD(a₁−1)1

Here a₀ can be anything, positive, negative, or zero; a₁ is greater than or equal to 2; and ABCD can also be anything, not just pairs of 1s and 2s.

Then, for the final matrices to be equal, it turns out that there's only one condition: the trace of the initial matrix has to be a₀ + 1 times the entry in the lower left corner!

Let's call the matrices in GL(2,Z) that satisfy that condition generalized Cohn matrices. We can parametrize them in just the same way as regular Cohn matrices.

a0u + t		b
u		u − t

Here u and a₀ are arbitrary integers. Since the matrix is in GL(2,Z), the determinant is ±1, and b has the following value.

b = a₀(u − t) + t − (t² ± 1)/u

Then, for b to be an integer, t² ± 1 must be divisible by u, so the parameter t must be a square root of  ∓1 mod u. To be perfectly clear, I mean that for every distinct square root in the group U_u, we get an infinite family of parameter values, a primary one in the range (0,u) and an infinite number of others that are congruent to it mod u.

Finally, based on the determinant, we can compute the square roots of −1 as described in Square Roots of −1, or we can compute the square roots of 1 in much the same way.

That takes care of the normal case where u ≥ 3, but let's also take a brief look at all the other cases.

• u = 2. The difference here is that the group U₂ is trivial and has only one element, 1. −1 is 1. −1 and 1 have only one square root, 1. The one infinite family of parameter values consists of all the odd numbers.
• u = 1. The group U₁ doesn't even exist, but if we back up a step, we want t² ± 1 to be divisible by 1. That's not a constraint, so the one infinite family of parameter values consists of all the integers.
• u = 0. Here the matrix parametrization is still technically correct, but three of the six terms are zero. The determinant is −t², so it can only be −1, not 1, and t can only be ±1. b is decoupled from the determinant, so it's free to take on any value, and a₀ is decoupled from the whole matrix, so it's meaningless.
• u < 0. Being divisible by the negative number u is the same as being divisible by the positive number |u|, so we already know all about this case.

Now let's see if we can work backward. Suppose we have a matrix and all we know about it is that it's a generalized Cohn matrix. Does the corresponding expansion have the required form? Does the matrix even represent a continued fraction? We can answer those questions.