The motivation for this question is the Selmer curve, given by
$$\displaystyle 3x^3 + 4y^3 + 5z^3 = 0.$$
One can show that this curve has no rational integer solutions, despite having a solution modulo $p$ for any prime $p$ and a solution over $\mathbb{R}$ (in other words, the Selmer curve fails the Hasse principle). However, as was pointed out to me in this question ([Hasse principle and Brauer-Manin obstruction for forms of large degree](https://mathoverflow.net/questions/163055/hasse-principle-and-brauer-manin-obstruction-for-forms-of-large-degree)) that no example of any surface of general type for $n > 2$ is known to fail the Hasse principle.


The obvious next case to consider is to pick $n = 3$ and $d = 5$ (since for a surface to be of general type, the degree $d$ must satisfy $d > n+1$), and examining the surface
$$\displaystyle a\_1 x\_1^5 + a\_2 x\_2^5 + a\_3 x^5 + a\_4 x^5 = 0$$
where $a\_i$ are rational integers for $1 \leq i \leq 4$. Given the fact cited above, what is the main difficulty in obtaining the analogous 'Selmer surface'? Do we expect most Fermat surfaces of degree 5 to fail the Hasse principle? What heuristics are there to support (or dispel) this hypothesis?



Fermat surface known to have very few rational integer solutions
These answers:


[Are most curves over Q pointless?](https://mathoverflow.net/questions/138581/are-most-curves-over-q-pointless/138592#138592)


[Are most cubic plane curves over the rationals elliptic?](https://mathoverflow.net/questions/11349/are-most-cubic-plane-curves-over-the-rationals-elliptic?rq=1)


both mention my paper with Bjorn Poonen (also mentioned in the answer by Daniel to your other question). There we show that a positive proportion of all hypersurfaces are everywhere locally solvable. The same argument should apply to diagonal quintic surfaces. One expects that, on the other hand, most diagonal quintics do not have a rational point, but this has not been proved. We give an heuristic in our paper which should apply here too. Unfortunately, this does not lead to an explicit counterexample to the Hasse principle.


The corresponding statement for diagonal curves has been proved by Browning and Dietmann (Contemporary Math. 493 (2009))


