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We got infinitely many solutions to $a^4+b^4+c^4=18$ over $\mathbb{Z}[i],i^2=-1$. Probably we can get infinitely many solutions to $a^5+b^5+c^5=N$ over $\mathbb{Z}[\alpha]$ for algebraic $\alpha$.

Are these results known and/or trivial?

There is some chance an improvement of the method to solve similar equations over the integers.

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    $\begingroup$ Do you mean the ring of integers $\mathcal{O}_K$ in a number field, rather than the number field $K$? $\endgroup$
    – Adam P. Goucher
    Commented Jan 1, 2018 at 16:16
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    $\begingroup$ Do the solutions you produce lie in a closed (not necessarily irreducible) subvariety? I believe that Vojta's conjecture would imply that they would have to in both examples. $\endgroup$
    – dhy
    Commented Jan 1, 2018 at 19:28
  • $\begingroup$ @AdamP.Goucher Thanks, number field was too broad, edited. $\endgroup$
    – joro
    Commented Jan 2, 2018 at 7:14
  • $\begingroup$ @dhy In both cases they are on a subvariety. I believe Vojta applies only for the second example. $\endgroup$
    – joro
    Commented Jan 2, 2018 at 7:14
  • $\begingroup$ @joro: I think Vojta will apply to the first because you are looking at integral points, which corresponds to looking at a quartic surface with a log divisor of the hyperplane section at infinity instead of just a quartic surface. The quartic surface is Calabi-Yau, but the log quartic surface is log general type. $\endgroup$
    – dhy
    Commented Jan 2, 2018 at 10:08

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Let $K$ be a number field. Then the Bombieri–Lang–Vojta–... conjecture asserts that for $d \geq 5$, all but finitely many of the $K$-points of the smooth projective surface of general type $$X = V(x_0^d+x_1^d+x_2^d-Nx_3^d) \subseteq \mathbb P^3$$ are contained in some strict subvariety $Z \subsetneq X$. A stronger version asserts that we can choose $Z$ independent of $K$, and that we may take it to be the union of all rational and elliptic curves in $X$. This is a good thing to try.

Over $\mathbb C$ and if $d$ is odd, we have a map \begin{align*} \mathbb P^1 &\to X\\ [x:y] &\mapsto [x:-x:y\sqrt[d]{N}:y]. \end{align*} There are obvious modifications one can make for $d$ even. This exhibits a specific $\mathbb P^1$ on $X$, which is actually defined over $\mathbb Q[\sqrt[d]{N}]$.

This already gives infinitely many $\mathbb Q[\sqrt[d]{N}]$-points on $X$. Setting $y = 1$ and varying $x$ over $\mathbb Z[\sqrt[d]{N}]$, we actually get infinitely many integral points. Explicitly, I have exhibited for you the points $$(a,b,c) = (x,-x,\sqrt[d]{N}),$$ where $x \in \mathbb Z[\sqrt[d]{N}]$ is arbitrary.

One could probably have found these without referring to fancy conjectures. On the other hand, the conjecture can serve as a guide: conjecturally, this type of construction is the only way in which we can produce infinitely many rational points. In general, however, trying to produce maps $\mathbb P^1 \to X$ is probably not your winning strategy for producing rational or integral points...

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  • $\begingroup$ Do all of your solutions have cancellation of the form $a^d= - b^d$? $\endgroup$
    – joro
    Commented Jan 2, 2018 at 9:22
  • $\begingroup$ @joro: the ones I just constructed do. I have not been able to construct other maps $\mathbb P^1 \to X$ or $E \to X$, but they might exist. But even if you can prove that they don't exist, you still need the Bombieri–Lang conjecture to conclude that we have found all but finitely many of the rational points. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 2, 2018 at 19:30
  • $\begingroup$ To give an idea of how hard this is: as I understand it, we don't know a single non-trivial example where we can prove Bombieri–Lang. Trivial ones are ones where we can prove $X(k) = \varnothing$, or $X$ has a map to another variety with few rational points (e.g. a genus $g \geq 2$ curve). Using this paper by Poonen, even knowing that for our particular variety $X$ the rational points over $\mathbb Q(i,\sqrt[d]{N})$ are not Zariski dense has profound implications! $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 2, 2018 at 19:35
  • $\begingroup$ The solution: mathoverflow.net/questions/289840/… $\endgroup$
    – joro
    Commented Jan 3, 2018 at 14:22
  • $\begingroup$ @joro: ah, for $n = 4$ the story is completely different. The smooth projective surface you get is a K3 surface, which conjecturally contains a $\mathbb P^1$ through every $\bar {\mathbb Q}$-point on it; see for example Conjecture 12 of this note. Although your new question is also very different in that you ask for solutions over $\mathbb Z$ only. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 3, 2018 at 19:21

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