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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) 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?

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What do you mean by "rational integer solutions"? Any integer solution is necessarily rational; and any rational solution can immediately be converted to an integer solution. –  John Bentin Apr 12 at 9:01
    
"Rational integer" means an element of $\mathbb{Z}$ as opposed to an element in the ring of integers in some number field; this is a necessary clarification in this subject area as frequently integral solutions over number fields are sought as well. –  Stanley Yao Xiao Apr 12 at 15:02

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up vote 3 down vote accepted

These answers:

Are most curves over Q pointless?

Are most cubic plane curves over the rationals elliptic?

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))

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Thanks for the response! –  Stanley Yao Xiao Apr 12 at 15:03

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