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In this question, Bogdan Grechuk found the "Smallest" open Diophantine equation, where size is determined by taking absolute values of all coefficients, then substituting 2 into all variables.

The equation is $y(x^3-y)=z^3+3$. There are a couple of different reformulations of this problem, for example "Is there any two numbers whose sum is a cube and whose product is three more than a cube?"

However, one direction where I have begun to make progress is to reframe the equation as: $yx^3=z^3+y^2+3$. Taking this equation modulo 27, it can be shown that (proof on request):

1) y is not 0, 1, or 8 mod 9 (note in particular this forces y to not be a cube)

2) If y is divisible by 3, then y is 6 or 21 mod 27

Let v be the square-free part of $y^2+3$. Modulo v, the equation becomes $yx^3=z^3$, so either v divides x and z, or $y=(\frac zx)^3$.

In the former case, this implies $v^3|y^2+3$, which seems extremely unlikely due to how v was defined (though I'm not sure how to prove that).

In the latter case, we have a third condition on y:

3) y must be a cube mod $v$.

I have tested all values of y up to 1000 so far, and no values of y satisfy all three conditions. So my question is:

Can it be proven that there is no integer y which satisfies all three constraints?

And a follow up question for the extra case: Can $y^2+3$ be a 3-Powerful number?

Thomas Blok
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  • So, with $y^2+3=\prod_j p_j^{e_j}$ for distinct primes $p_j$, if $p_j\mid v$, then $e_j$ is odd by definition of $v$, and $e_j>2$. So provided that $e_j>1$ for all $j$, $y$ qualifies. $y^2+3$ need only be 2-powerful: if $p^2||y^2+3$ then $p\nmid v$. I find that this true for $(y,v)=(1,1); (37,7); (79196,7\cdot37); (177833,13)$ and no more for $y<1000000$. – Rosie F Jul 03 '22 at 08:24
  • So for any prime p dividing $y^2+3$, you can show that either $p^3$ divides $y^2+3$ or y is a cubic residue mod $y^2+3$. I had a realization about the latter condition, that it is equivalent to 3 being a cubic residue mod $y^2+3$, which might be easier to work with. – Thomas Blok Jul 04 '22 at 05:30

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