Let’s generalize the model above by replacing 3 with y and 2 with y-1:

Now, we can provide any value for y. The question is, do any values of x exist where y is not equal to 3? In other words, does this “loop” exist anywhere else in the same form?
This is a tricky equation to work with because involves two variables, one of which is raised to the power of the other. Instead of trying to isolate y directly, let’s work from a conjecture. When I originally tackled this problem, I observed this relationship between y and x from the original model in the previous section:

The value x above can be substituted for y-1 to the root of y-1., analogous to 2 to the root of 2. My conjecture is that for all y, the following holds for x:

This equation isolates x in terms of y, and if true, would prove that there are infinitely many such “loops” of this form.
Let’s try to prove it by substituting it back into the generalized model above:

In the first step above, we substituted the occurrence of x on the right side with the assignment made in the conjecture. Then, we reduced it until the last step, where we found that
we’ve derived the statement we’ve made in the conjecture, thereby proving the conjecture true. This means that you can pick any value of y, find it’s corresponding x value, and thereby create your own “loop.” For example, if we say that y is 4, then x is the cube root of 3, and we have a loop:
- Take the cube root of three, roughly 1.4422…
- Raise it to the fourth power, which is roughly 4.3627…
- Divide it by three, which brings you back to 1.4422…
Let’s make one with 5,627:
- The 5,626th root of 5,626 is roughly 1.0015…
- Raise it to the 5,627 power, which is roughly: 5634.64…
- Divide it by 5626, which brings you back to 1.0015…
With that, we have infinitely many solutions for x and y, which means we have infinitely many “loops.” As a note, you can further generalize to replace (y-1) with (y-z) and generate infinitely many more solutions.