Let $n>3$ be an integer. What are the rational solutions of
$$y^2 = 4x^n + z^{n-1}$$
?
$\begingroup$
$\endgroup$
1
1 Answer
$\begingroup$
$\endgroup$
4
We consider two cases: $n$ is odd or $n$ is even. If $n=2k+1$ then put $$ x=uz, \quad y=vz^{k}, $$ where $u, v$ are non-zero rational parameters. Thus $v^2z^{2k}=4u^{2k+1}z^{2k+1}+z^{2k}$ and after division by $z^{2k}$ we get linear equation in $z$ which can be easily solved, i.e., $z=(v^2-1)/(4u^{2k+1})$. We thus get $$ x=\frac{v^2-1}{4u^{2k}},\quad y=v\left(\frac{v^2-1}{4u^{2k+1}}\right)^{k}. $$
If $n=2k$ one can use exactly the same type of reasoning, i.e., we put $$ y=ux^{k},\quad z=vx. $$ After necessary simplifications, we get $x=v^{2k-1}/(4-u^2)$ and thus $$ y=u\left(\frac{v^{2k-1}}{4-u^2}\right)^{k},\quad z=v^{2k}/(4-u^2). $$
-
1$\begingroup$ Are there other solutions besides yours? $\endgroup$– joroCommented Aug 28, 2020 at 13:01
-
1$\begingroup$ Some solutions are omitted. If $n$ is odd and $x=0$ then we need to take $v=\pm 1$. However, then we get $y=0$ from our parametrization. This is not a real problem because we can parametrize $y^2=z^{n-1}$ easily. The same situation occurs in the case $n$ even and $v=0$. In both cases the reason for this is simple: the maps constructed are rational and not defined everywhere. $\endgroup$ Commented Aug 28, 2020 at 13:39
-
$\begingroup$ So you gave rational parametrization of two families of rational surfaces? $\endgroup$– joroCommented Aug 29, 2020 at 11:20
-
$\begingroup$ @joro Exactly. That is what I have done. $\endgroup$ Commented Aug 30, 2020 at 17:43
elliptic-curvesnorhyperelliptic-curves. It is a surface rather than a curve. Nevertheless the usual attack to this kind of problems is as in ABC conjecture or Fermat's last theorem, which involves considering an associated Frey curve. I don't know how deep one can go with this equation. $\endgroup$