I suspect that the curve $x^5 + y^5=7$ has no $\mathbb Q$ points, and a brief computer search verifies this hypothesis for denominators up to $10^4$. What techniques can be used to show that there are no solutions?



Rational points on the "quintic circle" $x^5 + y^5 = 7$
There is an action of $\mu\_5$, the group of fifth roots of unity, on your curve,
given by $\zeta \cdot (x,y) = (\zeta x, \zeta^{-1} y)$. The quotient by this
group action is the hyperelliptic curve
$$C \colon Y^2 = X^5 + \frac{49}{4},$$
the map being given by $(X, Y) = (-xy, x^5 - \frac{7}{2})$. So it is enough
to find all the rational points on $C$. $C$ is isomorphic to
$C' \colon Y^2 = 4 X^5 + 49$. By a 2-descent, one can show that the Jacobian
variety of $C'$, $J'$, has Mordell-Weil rank (at most) 1, and since one
finds a point of infinite order
($(x^2 - \frac{10}{9} x - \frac{10}{9}, \frac{200}{27} x - \frac{61}{27})$
in Mumford representation), Chabauty's method is applicable and shows
that $\infty$, $(0, \pm \frac{7}{2})$ are the only rational points on $C$.
This implies that there are no rational points on your affine curve.


Here is Magma code to check this:



> 
> P<x> := PolynomialRing(Rationals());
> 
> 
> C := HyperellipticCurve(4\*x^5 + 49);
> 
> 
> J := Jacobian(C);
> 
> 
> RankBound(J); // --> 1
> 
> 
> ptsJ := Points(J : Bound := 500); // the first 5 are torsion
> 
> 
> Chabauty(ptsJ[6]); // --> { (0 : -7 : 1), (1 : 0 : 0), (0 : 7 : 1) }
> 
> 
> 


(If you do not have direct access to Magma, you can try this out
with the online Magma calculator at <http://magma.maths.usyd.edu.au/calc/> .)


