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For $x$ and $y$ in $\mathbf{Z}$, not both zero, let $cfp(x,y)$ be the cube-free part of $x^3 -3xy^2 + y^3$ (normalized to be $> 0$). One sees:

(#) $cfp(x,y)$ is either a product of primes $p$, with each $p$ being 1 or 8 mod 9, or 3 times such a product.

I'd like information about which cube-free integers satisfying (#) can be written as $cfp(x,y)$. More specifically:

Questions:

(1) Is every cube-free integer satisfying (#) so representable? What about $p$ or $3p$ when $p$ is a prime that's 1 or 8 mod 9?

(2) Consider the proportion of cube-free elements of $N$ that are $< M$ and can be written as $cfp(x,y)$ to cube-free elements of $N$ that are $< M$ and satisfy (#). What happens to this proportion as $M\rightarrow\infty$? (Again the same question can be asked when attention is restricted to elements of $N$ that are $p$ or $3p$ as in (1).)

(3) I have doubts that much is known about (2), but can anyone provide guesses, based on computer calculation, as to what the situation might be?

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  • $\begingroup$ Why is p equal 1 or 8 mod 9, or 3 times that? $\endgroup$
    – Dabed
    Commented Jul 18, 2023 at 17:47
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    $\begingroup$ @Dabed You can assume x and y are coprime. If either is divisible by 3, or x and y are both 1 mod 3, or both 2 mod 3, x^3 - 3*xy^2 + y^3 is prime to 3. When 3 divides x+y, 9 divides x^3 +y^3, and so x^3 -3*xy^2 +y^3 is 3 or 6 mod 9. $\endgroup$
    – paul Monsky
    Commented Jul 18, 2023 at 20:03
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    $\begingroup$ @Dabed I was cut short--I'll finish up. It remains to show that no p that's 7,4,2 or 5 mod 9 divides our F(x,y). If p is 7 or 4 mod 9, then (Z/p)* contains a subgroup {1,u,v} of order 3. Also, u is not a cube in Z/p--adjoining a cube root w of u to Z/p gives a degree 3 extension. Let t be the element (w) + (1/w) of this extension. Then t^3 = u + v + 3*t, and so t^3 -3*t +1 =0. If this last equation had a root in Z/p, then w would satisfy a degree 2 equation over Z/p, which is impossible. The argument is the same when p is 2 or 5 mod 9, but now we use the field of p^2 elements ; not Z/p. $\endgroup$
    – paul Monsky
    Commented Jul 18, 2023 at 20:33

2 Answers 2

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The approach taken by Bhargava, Alpoge, and Shnidman (Integers expressible as the sum of two rational cubes) seems to be able to answer your question. In particular, they show that a positive portion of integers can be expressed as the sum of two rational cubes, and a positive proportion cannot be so expressed. Their argument basically boils down to a good estimate on the average rank of genus 1 curves of the shape $x^3 + y^3 = n$ as $n$ varies. Levent gave a talk on this paper last year at Banff, but I seem to recall that they did have to make use the shape of $x^3 + y^3$ significantly, and even small perturbations in the coefficients of the binary cubic form seems to make the problem hard enough as to necessitate some new ideas. I am not sure how hard it would be to make their arguments work for $x^3 - 3xy^2 - y^3$.

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Here are the first several primes $p$ and numbers $3p$ for which I haven't found an expression yet.

$$ 89 , 107 , 219 , 271 , 327 , 397 , 431 , 433 , 449 , 467 , 523 , 537 , 543 , 557 , 593 , 597 , 647 , 683, 719 , 773 , 807 , 829 , 863 , 881 , 919 , 921 , 991 , 1077 , 1117 , $$

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    $\begingroup$ In fact, $89$ and $107$ and probably most of the others on this list do not occur. The curve $x^{3} - 3xy^{2} - y^3 = 89z^{3}$ represents a $3$-torsion element of the Shafarevich-Tate group of the rank zero elliptic curve $y^{2} - 801y = x^{3}$. (Note that $863$ and $881$ do occur. We get $863$ with $x = 89$, $y = 305$ and we get $881$ with $x = 49$, $y = 228$.) $\endgroup$
    – Jeremy Rouse
    Commented Jul 17, 2023 at 1:18
  • $\begingroup$ @JeremyRouse thanks. $\endgroup$
    – Will Jagy
    Commented Jul 17, 2023 at 1:21
  • $\begingroup$ @Jeremy Rouse I discovered that part 1 of my question is an old one--Selmer calls the equation I was looking at "Sylvester's" equation. After reading your comment I thought that Will hadn't searched far enough in looking for solutions for 3p when p is 179, 269 or 359--after all each of these p is a sum of two rational cubes. (When one consults Elkies tables one finds that the fundamental solution to x^3 + y^3 =p has large height in all three cases). My suspicions were confirmed when I looked at table 3 on page 351 of Selmer's opus--he lists the solutions. Will was right about the rest. $\endgroup$
    – paul Monsky
    Commented Jul 18, 2023 at 17:46
  • $\begingroup$ @Paul, which book by Selmer do you mean? There seem to be a few. $\endgroup$
    – Will Jagy
    Commented Jul 18, 2023 at 18:19
  • $\begingroup$ @Will I meant the 1951 Acta Arithmetica article--not a book, but more than 100 pages. The tables all come at the end. $\endgroup$
    – paul Monsky
    Commented Jul 18, 2023 at 19:54

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