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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 $n>2$.

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Question as it stands is ambiguous. You want rational solutions of $x$,$y$ and $z$ but you already say that they are in $\mathbb{Z}_+$. So, Aren't you just searching for its roots? –  user21436 Dec 25 '11 at 18:50

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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 $t$.

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