Pell equation in ${\mathbb Q}(x)$

Question

Is it known whether the equation $A^2-(x^2+3)B^2=1$ has a solution $A,B\in{\mathbb Q}(x)$ with $B\neq 0$ ?

My thoughts : I think that there is no solution, as the fundamental solution of $A^2-(x^2+3)B^2=1$ for $x\in {\mathbb Z}$ seems to vary uncontrollably.

Answer 1

$$ \left( 2x^2 + 3 \right)^2 - \left( x^2 + 3 \right) \left( 2x \right)^2 = 9 $$ $$ \left( \frac{2}{3}x^2 + 1 \right)^2 - \left( x^2 + 3 \right) \left( \frac{2}{3}x \right)^2 = 1 $$

From $ \color{red}{\mbox{Jyrki's}} $ comment, material I had not realized applied here: we get a matrix that takes a solution ( as a column vector) to the next one, in matrix form $$ M = \left( \begin{array}{cc} \frac{2}{3} x^2 + 1 & \frac{2}{3} x^3 + 2x \\ \frac{2}{3} x & \frac{2}{3} x^2 + 1 \end{array} \right) $$ In turn, we have solutions $(A_n, B_n)$ with $A_0 = 1, A_1 = \frac{2}{3} x^2 + 1,$ then $B_0 = 0, B_1 = \frac{2}{3} x,$ finally from Cayley-Hamilton $$ A_{n+2} = \left(\frac{4}{3} x^2 + 2 \right) A_{n+1} - A_n $$ and $$ B_{n+2} = \left(\frac{4}{3} x^2 + 2 \right) B_{n+1} - B_n. $$

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$$ \left( \frac{8}{163}x^2 + \frac{8}{163}x + \frac{165}{163} \right)^2 - \left( x^2 + x + 41 \right) \left( \frac{8}{163}x + \frac{4}{163} \right)^2 = 1 $$

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$$ \left( x^2 + \frac{k}{2} \right)^2 - \left( x^2 + k \right) \left( x \right)^2 = \frac{k^2}{4} $$

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$$ \left( x^2 + \frac{4k+1}{8} \right)^2 - \left( x^2 + x + k \right) \left( x + \frac{1}{2} \right)^2 = \frac{(4k-1)^2}{64} $$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

For these principal forms, we can always adjust by multiplying the appropriate terms by a rational constant to get $1$ on the right hand side, either $2/k$ or $8/(4k-1).$

Answer 2

having trouble doing anything in the ring $\mathbb Q[x]$ for quartics. On the other hand, if you really meant the field of rational functions $\mathbb Q(x),$ we get an easy one. Given some $h(x)$ which could be a polynomial, take your favorite polynomials $f,g.$ Then $$ \left( \frac{h g^2 + f^2}{h g^2 - f^2} \right)^2 - h \cdot \left( \frac{2fg}{h g^2 - f^2} \right)^2 = 1 $$