# Find the diophantine equation $x^2(y^2-1)=z^2-1$ solution

How can I solve (find all the solutions) the nonlinear Diophantine equation

Let $x,y,z$ be postive integers ,and $x,y,z\ge 2$,find this following equation solution $$x^2=\dfrac{z^2-1}{y^2-1}$$

I included here what I had done so far. such $z=7,y=2,x=4(\dfrac{7^2-1}{2^2-1}=16=4^2)$ is one solution

thanks for any help.

## 1 Answer

Every $y \geq 2$ works. For each $y,$ we get an infinite sequence of solutions $(z_n, x_n)$ beginning with $$(1,0),$$ $$(y,1),$$ $$(2y^2 - 1,2y),$$ $$(4y^3 - 3y,4y^2 - 1),$$ $$(8y^4 - 8y^2+1,8y^3 - 4y),$$ continuing forever with $$z_{n+2} = 2 y z_{n+1} - z_n,$$ $$x_{n+2} = 2 y x_{n+1} - x_n.$$ The two separate recurrences come from a single combined recurrence by Cayley-Hamiton, $$(z_{n+1}, \, x_{n+1}) = \left( y \, z_n + (y^2-1) \, x_n, \; \; z_n + y \, x_n \right)$$

I have not yet found these polynomials as named sequences, although it is quite likely that they have names. There are, for example, the named https://en.wikipedia.org/wiki/Fibonacci_polynomials although we are not using those. Alright, from comment below, these are the Chebyshev polynomials, the $z_n$ are the FIRST KIND, while the $x_n$ are the SECOND KIND

However, take a real number $t > 0$ so that $$y = \cosh t,$$ or $$t = \log \left( y + \sqrt{y^2 - 1} \right).$$ Then $$(z_n, x_n) = \left( \cosh nt, \; \; \frac{\sinh nt}{\sinh t} \; \right)$$

Here are $z \leq 1000$

    z    y    x
7    2    4
17    3    6
26    2   15
31    4    8
49    5   10
71    6   12
97    2   56
97    7   14
99    3   35
127    8   16
161    9   18
199   10   20
241   11   22
244    4   63
287   12   24
337   13   26
362    2  209
391   14   28
449   15   30
485    5   99
511   16   32
577    3  204
577   17   34
647   18   36
721   19   38
799   20   40
846    6  143
881   21   42
967   22   44


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