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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 Answers1

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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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Will Jagy
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