<s>[INST] Failing of heuristics from circle method The heuristic from circle method for integral points on diagonal cubic surfaces $x^3+y^3+z^3=a$ ($a$ is a cubic-free integer) seems to fit well with numerical computations by ANDREAS-STEPHAN ELSENHANS AND JORG JAHNEL. The only known exception is the surface $x^3+y^3+z^3=2$. Circle method predicts that the number of integral points $(x,y,z)$ with $\max(\vert x\vert,\vert y \vert,\vert z\vert)<N$ is $\approx 0.16\log N$. But parametric solutions $(1+6t^3,1-6t^3,-6t^2)$ are missing.
Question: Why the heuristic fails for $x^3+y^3+z^3=2$? 
 [/INST] Note as a caveat that the heuristics can fail in the other direction as well: Dietmann and Elsholtz ( http://www.math.tugraz.at/~elsholtz/WWW/papers/papers26de08.pdf ) have shown that if $p\equiv 7\pmod{8}$ is prime, then $p^2$ cannot be represented as $x^2+y^2+z^4$. This strongly contradicts the heuristic of the circle method, since $\frac{1}{2}+\frac{1}{2}+\frac{1}{4}>1$ should imply that every sufficiently large integer $n$ is either representable as $n=x^2+y^2+z^4$, or there is some local constraint, i.e. a modulus $q$ such that $n\equiv x^2+y^2+z^4\pmod{q}$ is unsolvable.
This theorem is one of those cases where formulating the theorem is much more difficult than proving it. Write $x^2+y^2+z^4=p^2$ as $x^2+y^2=(p-z^2)(p+z^2)$. Now $p-z^2$ is $3\pmod{4}$ or $6\pmod{8}$, hence divisible by some prime $q\equiv 3\pmod{4}$ to the first power. But a sum of two squares cannot be divisible by such a prime with odd exponent, thus $q$ divides $p+z^2$ as well. Now $q|(p-z^2, p+z^2)$, thus $q$ divides $2p$ and $2z^2$, hence $p=q$ and $p|z$. But then $z^4$ is larger than $p^2$.
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