I am trying to solve the following quadratic diophantine equation in $\mathbb C[T]$, but I did not manage. I hope someone could give some hints or solutions to my problem. Here is the equation


$$\left[(t+1)X+tY-1\right]^2=24XY,$$
where the unknowns are $X,Y$.
If $\deg(X)>\deg(Y)$ this is impossible. Same for $\deg(X)<\deg(Y)$ Hence, $\deg(X)=\deg(Y)$. So now I am stuck here.


Thanks in advance for any help.



Quadratic diophantine equation in $\mathbb C[T]$
Clearly $X$, $Y$ are coprime, hence both are squares of polynomials. Write $X=3f^2$, $y=2g^2$, then $3(t+1)f^2+2tg^2-1=\pm 12 fg$, without loss of generality sign is plus, else change sign of $g$. This rewrites as $$(3(t+1)f-6g)^2+6(t^2+t-6)g^2=3(t+1).$$ Denote $3(t+1)f-6g=h$, then $h^2+6(t^2+t-6)g^2=3(t+1)$. This is Pell-type equation $h^2-Ng^2=3(t+1)$, where $N=6(t^2+t-6)=6(t-2)(t+3)$. There is a solution $(h\_0,g\_0)$ of $h\_0^2-Ng\_0^2=1$ with linear $h\_0$ and constant $g\_0$. Namely, $h\_0+1=2(t+3)/5$, $h\_0-1=2(t-2)/5$, $g\_0=2/5\sqrt{6}$. Then if $(h,g)$ is a solution of $h^2-Ng^2=3(t+1)$ so are $(hh\_0\pm Ngg\_0,hg\_0\pm gh\_0)$. Without loss of generality $\deg h=\deg g+1$, $h=ct^{d+1}+\dots$, $g=\sqrt{6}ct^d+\dots$. Then I claim that if $d>0$ then $(hh\_0-Ngg\_0,hg\_0-gh\_0)$ is a solution with lower value of degree. This may be checked by direct computation of two leading coefficients or by studying asymptotics of $h\pm \sqrt{N} g$ for large values of $t$. That is, any solution reduces to case $d=0$, corresponding to constant $f$, $g$.


