TITLE: Diophantine equations for polynomials
QUESTION [9 upvotes]: I know that there has been work on diophanitine equations with solutions in poynomials ( rather than integers ) of the Fermat and Catalan type $x(t)^n+y(t)^n=z(t)^n$ ; $x(t)^m-y(t)^n=1$ and these have been completely solved (For Fermat's equation with $n>2$ and in $\mathbb C[t]$ by Greeneaf and for Catalan's equation in $\mathbb C[t]$ by Nathanson ) . I would like to know whether there has been similar work on solutions in $\mathbb C[x]$ or $\mathbb Z[x]$ for Pell type equations $f(x)^2-ng(x)^2=1$ , where $n$ is given positive integer ; or similarly for say Erdos-Strauss conjecture $4f(x)g(x)h(x)=n(f(x)g(x)+g(x)h(x)+h(x)f(x))$,  where $n>1$ is given integer , or say concerning Ramanujan-Nagell-Lebesgue type equation $f^2+D=Ag^n$, where $D$,$A$ are given integers and we have to find polynomials $f,g$ and positive integer $n$  . Any reference or link  concerning these and other types of Diophantine equations with solutions in polynomials will be highly appreciated . Thanks in advance

REPLY [2 votes]: If $n$ is constant (integer or not) then
$f^2 - ng^2 = 1$ has no solution in nonconstant polynomials $f,g$:
over $\bf C$ we can extract a square root $m$ of $n$,
factor $f^2 - ng^2$ as $(f-mg)(f+mg)$, and observe that
both factors must have degree zero because their product does.
If $D,A$ are constants with $D \neq 0$ then there is no solution to
$f^2 + D = Ag^n$ in nonconstant polynomials except in the trivial case $n=1$.
This is a consequence of the
Mason(-Stothers)
theorem (polynomial analogue of the
abc conjecture), 
because $f^2$, $D$, and $Ag^n$ would be relatively prime as polynomials, and would have too many repeated factors.
(The exponent 2 case also gives
an alternative proof of the result on $f^2-ng^2=1$.)