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?

1$\begingroup$ It helps to instead consider the corresponding projective curve $X^5 + Y^5 = 7Z^5$ in $\mathbb{P}^2$ of genus $6$. This has a rational point, namely $(X:Y:Z) = (1:1:0)$. Your conjecture is that this is the only rational point. The similarity here to the $n=5$ case of Fermat's last theorem should be very clear. Of course the proof of FLT for all $n$ was very hard, however there exist elementary proofs for the $n=5$ case (en.wikipedia.org/wiki/…). Perhaps these could be adapted to your case. $\endgroup$ – Daniel Loughran Nov 22 '15 at 8:58

1$\begingroup$ The modular techniques and Frey curve machinery used to prove FLT could also possibly be made to apply in your case, but I am not an expert in these methods. $\endgroup$ – Daniel Loughran Nov 22 '15 at 9:03
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 2descent, one can show that the Jacobian variety of $C'$, $J'$, has MordellWeil 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/ .)

1$\begingroup$ Could you explain 'By a 2descent, one can show that the Jacobian variety of C′'? $\endgroup$ – user76479 Nov 22 '15 at 12:31

3$\begingroup$ @Arul : This is explained in detail in my paper "Implementing 2descent for Jacobians of hyperelliptic curves" (Acta Arith. 98, 245277, 2001). If you are familiar with 2descent on elliptic curves, then it can be seen as a generalization of that: one computes the 2Selmer group of the Jacobian as a subgroup of ${\mathbb Q}(\theta)^\times$ mod squares, where $\theta = \sqrt[5]{49/4}$. There is an injection of $J({\mathbb Q})/2 J({\mathbb Q})$ into the 2Selmer group, so we obtain an upper bound for the rank. $\endgroup$ – Michael Stoll Nov 22 '15 at 12:42

$\begingroup$ It is worth pointing out that passage from $x^5+y^5=7$ to the hyperelliptic curve is achieved by rewriting as $x^{10}+(xy)^5=7x^5$ changing variables to $u=x^5$ and $v=xy$ and completing the square. $\endgroup$ – prekidney Jun 30 at 3:52
In his 1825 paper, Lejeune Dirichlet proved that the equation $x^5 + y^5 = cz^5$ has no nontrivial solution with x and y coprime integers and z integer for a rather large class of integers c. His paper can be read on the site Gallica : http://gallica.bnf.fr/ark:/12148/bpt6k5842885h/f2.item.zoom but the visualization is very bad.

$\begingroup$ It's actually not so bad already, but here is an improved OCRed file: dropbox.com/s/jkbl5gtzdyuzktt/dirichlet.pdf?dl=0 $\endgroup$ – Igor Rivin Nov 22 '15 at 12:32

$\begingroup$ Looking through Dirichlet's paper, it seems that he always assumes $c$ to be divisible by 2 or 5 (unless $c = 1$). $\endgroup$ – Michael Stoll Nov 22 '15 at 12:37

$\begingroup$ You are right. Dirichlet doesn't solve the problem for $c = 5$. My answer was rather a comment. $\endgroup$ – Panurge Mar 20 '16 at 10:57

$\begingroup$ @Panurge meant to write "Dirichlet doesn't solve the problem for $c=7$". $\endgroup$ – prekidney Jun 30 at 3:46