# Quadratic diophantine equation in $\mathbb C[T]$

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.

• Here is a hint. There is a solution where both $X$ and $Y$ are constant in $t$. You can use that solution to find many other solutions in the usual way. Commented Jul 19, 2015 at 13:09

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 $f$, $g$.