Diophantine $\frac{a^2 + b^2}{ab + 1} = \frac{c^2}{d^2} $

Question

Consider the Diophantine equation with $a,b,c,d > 0$ :

$$\frac{a^2 + b^2}{ab + 1} = \frac{c^2}{d^2} $$

For the case $d=1$ , this is a Classic ; we know that

Diophantine $\frac{a^2 + b^2}{ab + 1} = K $ implies $K$ is a square. This is the standard example of the so-called vièta jumping.

Although that is not a parametric solution yet. I think this is pretty well understood since it is similar to Pell’s equation. I think there is A fibonacci like recursion that solves the parametric problem.

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I considered using modular aritmetica ( mod prime ) but failed. I think the problem is division is iso to a product mod p.

Like $ c^2 / d^2 = s^2 \mod p $.

Also going from 3 to 4 variables makes me dizzy :)

Is this related to elliptic curves ?

Is there A solution $a,b$ for Every $c,d$ ?

Is there A closed form ( parametric ) solution ?

Is vièta jumping still relevant here ? Can it be used as a proof ? And is it the best method ?

Answer 1

Above equation shown below:

$\frac{a^2 + b^2}{ab + 1} = \frac{c^2}{d^2}$ ----$(1)$

For known, $(c,d) =(2,1)$ there is a parametric solution & is shown below:

$a=4wk(2k-1)$

$b=2w(3k-1)(5k-1)$

Where, $w= [(1)/(k^2-4k+1)]$

For $k=4$, we have, $(a,b,c,d)= (112,418,2,1)$