Rational zeros of polynomials Let's have the polynomial $(x+y)^n+(y-x)^n=z^n$. Does anyone know when this polynomial has rational solutions? $x,y,z,n$ positive integers and $n>2$.
 A: Suppose $x=\frac{a}{b},y=\frac{c}{d},z=\frac{e}{f} $ is a rational solution to $(x+y)^n+(y-x)^n=z^n$ with $n>2$. Then 
$$\left(\frac{ad+bc}{bd}\right)^n+\left(\frac{ad-bc}{bd}\right)^n=\left(\frac{e}{f}\right)^n$$
hence
$$(f(ad+bc))^n+(f(ad-bc))^n=(ebd)^n$$
is a counterexample to Fermat's Last Theorem unless $ebd=0$ (hence $e=0$, hence $z=0$), or $ad-bc=0$ (hence $x=y$), or $ad+bc=0$ (hence $x=-y$). 
If $x=y$, then $z=\pm2x=\pm 2y$ if $n$ even, and $z=2x=2y$ if $n$ odd. 
If $x=-y$, then $z=\pm 2x=\mp 2y$ if $n$ even, and $z=-2x=2y$ if $n$ odd.
If $z=0$, then $n$ is odd and $x+y=-(y-x)$, hence $y=0$. 
Thus, if $n$ is odd, the only rational solutions are those of the form
$$x=t,\quad y=t,\quad z=2t$$
$$x=t,\quad y=-t,\quad z=2t$$
$$x=t,\quad y=0,\quad z=0$$
for a rational number $t$, and if $n$ is even, the only rational solutions are those of the form
$$x=t,\quad y=t,\quad z=2t$$
$$x=t,\quad y=t,\quad z=-2t$$
$$x=t,\quad y=-t,\quad z=2t$$
$$x=t,\quad y=-t,\quad z=-2t$$
for a rational number $t$.
