# Fermat surface known to have very few rational integer solutions

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 '14 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 '14 at 15:02