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Probably this is known, but doesn't show in searches.

If a certain hyperelliptic curve has only trivial rational points, FLT-like curve also has only trivial rationals points for fixed $n$.

Working over the rationals for integer $a$ define:

$$ u^n-v^n=a \qquad (1)$$

From $(u^n-v^n)^2 + 4 (uv)^n=(u^n+v^n)^2$ we get

$$ x^n + a^2/4=y^2 \qquad (2)$$

From $(uv)^n=v^n(v^n+a)$ we get $$ x^n=y(y+a) \qquad (3)$$

Non trivial rational point on (1) leads to non trivial $x \ne 0$ rational point on (2) and (3). The converse need not hold.

Only trivial points on the hyperelliptic curves means only trivial on (1).

On MO answers find all rational points on a given hyperelliptic with the help of computer.

Is finding the rational points on the hyperelliptic curves easier than on the FLT-like curve?

Fix $a$. Is it feasible to find all rational points on the hyperelliptic curves for $n$ up to say $50$?

Added 1 Is it known or open that for rational $y$, the solutions to $y(a+y)=x^k$ are finite for $k>4$ and nonzero $x$? Basically quadratic is perfect power at rationals

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    $\begingroup$ This is a relatively standard reduction that can be done more generally (as for example in Henri's answer). The point is that the generalized Fermat curve $F_n:ax^n+by^n+cz^n=0$ maps to quite a few curves of lower genus. You've rediscovered hyperelliptic quotients. The underlying idea is that Aut$(F_n)$ is large, and taking the quotient of $F_n$ by a finite subgroup of Aut$(F_n)$ gives a quotient curve, usually of lower genus. In some cases, those curves still have genus $\ge2$, and then yes, people have used them to study rational points, both theoretically and computationally. $\endgroup$
    – Joe Silverman
    Commented Aug 25, 2018 at 15:02
  • $\begingroup$ @JoeSilverman Thanks. Is it known or open that for rational $y$, the solutions to $y(a+y)=x^k$ are finite for $k>4$ and nonzero $x$? Basically quadratic is perfect power at rationals. $\endgroup$
    – joro
    Commented Aug 25, 2018 at 15:27

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Let $p$ be an odd prime, and assume $ax^p+by^p+cz^p=0$ with $x\ne0$; then $$Y^2=X^p+a^2(bc)^{p-1}/4$$ has a nontrivial point with $$X=-bcyz/x^2\quad\,\quad Y=(-bc)^{(p-1)/2}(by^p-cz^p)/(2x^p)\;.$$

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  • $\begingroup$ Thanks. How is this answer, I don't have $b,c$? Is this a generalization I am not asking for? $\endgroup$
    – joro
    Commented Aug 25, 2018 at 14:55
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A reference for all quotients of the Fermat curve is Lang's book "Introduction to Algebraic and Abelian Functions". There, you'll find your maps (up to twist) and several others.

If you have a map $X \to Y$ and $X$ has a rational point, so does $Y$. At first sight, you'd think it would be easier to prove that $X$ has no points than $Y$ (assuming that's the case). But, in some situations, showing that $Y$ has no points can be easier. A famous instance is Mazur's work on modular curves (and torsion on elliptic curves), where he shows that certain quotients of modular curves have Mordell-Weil rank zero and thus a describable set of rational points. Maybe this is possible with the Fermat curves also, I don't know. Another possibility is to use Chabauty's method if the Mordell-Weil rank of $Y$ is less than its genus. These days, Chabauty's method is almost automated for hyperelliptic curves of low genus.

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  • $\begingroup$ Thanks. Magma efficiently solves n=5. Is the computer doing the dirty job easy or elementary? $\endgroup$
    – joro
    Commented Aug 26, 2018 at 12:52
  • $\begingroup$ @joro I don't understand the question. $\endgroup$
    – Felipe Voloch
    Commented Aug 26, 2018 at 22:40

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