Failing of heuristics from circle method

Question

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 $x^3+y^3+z^3=2$?

Answer 1

The heuristic fails for precisely the reason you state; there is a parametric family of solutions which makes it fail.

People often use heuristic arguments to predict the number of integral/rational solutions of bounded height to diophantine equations. These heuristics are often true, provided one allows a small margin of error given by ignoring possible parametric families of solutions which make things go wrong (often called "special subvarieties"). This is exactly what is happening here.

This philosophy is present in many conjectures and research problems in the modern study of diophantine equations, such as Manin's conjecture, the Bombieri-Lang conjecture, the André–Oort conjecture,.....

The prediction from the circle method comes here from the major arc analysis. One hopes in general that the minor arcs are smaller than the major arcs, however this is sometimes not true and in such cases it is expected that there are special subvarieties which contribute to the minor arcs and make things go wrong. I expect this is what is happening in your case.

Another example close to my heart where this happens (and closely related to your case), is the case of cubic surfaces. Take for example the Fermat cubic surface.

$$x_0^3 + x_1^3 = x_2^3 + x_3^3.$$

Here the circle method predicts that there are around $N(\log N)^3$ primitive integral solutions of bounded height. However this is wrong, as there is a parametric family of "trivial" solutions $$x_0=x_2, \quad x_1=x_3,$$ which yields around $N^2$ solutions of bounded height.

However, once one removes these trivial solutions, plus other obvious ones, it is expected that the circle method prediction should be right. This is a special case of the Manin conjecture. Unfortunately one does not know this for the Fermat cubic surface, but a lower bound of the correct order of magnitude in this case is a recent nice result of Sofos.

Answer 2

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 $p^2$.