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I should have asked this question before the other one since Moore's tables for $5$th powers have radius $<17700$, while Wrobleski's tables for $3$rd powers go up to one million, hence more data to work with. And since this is Diophantine, then it is not required that the $z_k$ be positive integers.

I. Form $x_1^3+x_2^3 = x_3^3+x_4^3$

We focus on primitive solutions and consider $(x_1,x_2) = (x_3,x_4)$ as trivial. The first ones are,

\begin{align} 1^3 & + 12^3 \,=\, 10^3 + 9^3\\ 2^3 & + 16^3 \,=\, 15^3 + 9^3\\ 3^3 & + 60^3 \,=\, 59^3 + 22^3\\ 4^3 & + 110^3 = 101^3 + 67^3\\ 5^3 & + 76^3 \,=\, 69^3 + 48^3\\ 6^3 & + 552^3 = 551^3 + 97^3 \end{align}

and so on, for all $x_1<3000$ (which is quite a long stretch). Note that for some $x_1 = 13m$, it seems terms are a bit larger,

$$13^3 + 5288^3 = 5148^3 + 2253^3$$

II. Form $x_1^3+x_2^3+x_3^3=x_4^3$

Likewise, we disregard trivial solutions $x_i = x_4$. Hence,

\begin{align} 1^3 & + \,6^3\, + \,8^3 \,=\, 9^3\\ 2^3 & + 17^3 + 40^3 = 41^3\\ 3^3 & + 10^3 + 18^3 = 19^3\\ 4^3 & + 17^3 + 22^3 = 25^3\\ 5^3 & + \,4^3\, + \,3^3\, =\, 6^3\\ 6^3 & + 32^3 + 33^3 = 41^3 \end{align}

and so on, also for all $x_1<3000$. Again, for some $x_1 = 13m$, terms are a bit larger,

$$(4\times13)^3 + 321^3 + 3327^3 = 3328^3$$

P.S. Note that for squares, we have the identity,

$$n^2 + (n + 1)^2 + (n^2 + n)^2 = (n^2 + n + 1)^2$$

which proves every positive integer $n$ appears at least once. It may be true for cubes as well.

III. Questions

  1. For any positive integer $x_1$, is it true there are primitive positive solutions to both forms?
  2. If we can't prove it in general, how high can Wroblewski's tables extend the confirmed range to a bound $x_1>3000$? (Less than $10^6$ for sure.)
Tito Piezas III
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  • What about the form $x_1^3=x_2^3+x_3^3+x_4^3$? It is not equivalent to form II because you're quantifying over $x_1$, not $x_4$. – mr_e_man Aug 30 '23 at 12:44
  • Oh, I guess you didn't include it because it's obvious there isn't always a non-trivial solution. E.g. $2^3$ is not a sum of three positive cubes. – mr_e_man Aug 30 '23 at 12:49
  • @mr_e_man Yes. If we forgo positivity, the most general form would be $x_1^3+x_2^3+x_3^3+x_4^3 = 0$ and to ask if all integers (except zero) would appear at least once. Bailey's answer proves it is YES. My post on $x_1^5+x_2^5+x_3^5+x_4^5+x_5^5+x_6^5 = 0$ essentially asks the same, but the answer is yet unknown (though it has a non-trivial solution with $x_1=0$). – Tito Piezas III Aug 30 '23 at 13:17
  • What would you consider a trivial solution when negative numbers are allowed? (Two terms are opposite?) Are you sure Bailey's solution is always non-trivial? – mr_e_man Aug 30 '23 at 13:24
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    @mr_e_man I already defined trivial for both forms. And, if we forgo positivity, then a solution like $(a,-a, b, -b)$ is included in that definition. Bailey's solution is trivial only for two cases: $m=0$ (by FLT) and $m=-1$. But it can still yield $$(-1)^3+(- 6)^3+(- 8)^3 + 9^3 = 0$$ simply by negating all terms when $m=1$. So $m>0$ would suffice, then just negate all terms if so desired. – Tito Piezas III Aug 30 '23 at 13:58
  • In Form I, the difference between the sum of the terms in the LHS and the RHS is always divisible by $6$. Like $(10+9)-(1+12)=6$. Then $(15+9)-(16+2)=6$...$(551+97)-(6+552)=90$. I don't know if it is meaningful. If it is, it can limit the search for solutions by excluding the difference that is not divisible by $6$. – user25406 Aug 31 '23 at 11:04

1 Answers1

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For form $x_1^3+x_2^3+x_3^3=x_4^3$, the following parametric form (also used in this answer) generates solutions for any $x_1$:

$\qquad x_1=m$

$\qquad x_2=3m^2+2m+1$

$\qquad x_3=3m^3+3m^2+2m$

$\qquad x_4=3m^3+3m^2+2m+1$

Adam Bailey
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    I can't believe I forgot that as it is in my collection! There are at least two identities of form, $$a^3+b^3+c^3 = (c+1)^3$$ The first is by Jandasek, and the second is by Ampon, namely, $$m^3 + (3m^2 + 2m + 1)^3 + (3m^3 + 3m^2 + 2m)^3 = (3m^3 + 3m^2 + 2m + 1)^3\ (3n^2)^3 + (6n^2 + 3n + 1)^3 + (9n^3 + 6n^2 + 3n)^3 = (9n^3 + 6n^2 + 3n + 1)^3$$ The first yields, $$1^3 + 6^3 + 8^3 = 9^3\ 2^3 + 17^3 + 40^3 = 41^3$$ which are in the list in my post. – Tito Piezas III Aug 30 '23 at 14:21
  • @TitoPiezasIII - Form I remains unanswered. – mr_e_man Aug 30 '23 at 14:25
  • @mr_e_man Hmm, usually I don't impose a positivity requirement when dealing with "equal sums of like powers" (like in my post for $5$th powers). Makes it harder to find identities. But it does raise an interesting question, if Form I can be settled by a simple identity as well. – Tito Piezas III Aug 30 '23 at 14:32
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    @TitoPiezasIII - In fact we can use these formulas with $m$ negative, to get positive solutions to I. Specifically, with $-m=k>0$, $$(k)^3+(3k^3-3k^2+2k)^3=(3k^2-2k+1)^3+(3k^3-3k^2+2k-1)^3$$ and all terms are positive. – mr_e_man Aug 30 '23 at 15:27
  • @mr_e_man Nice! I guess that takes care of third powers. Fifth powers in the orher post are proving a tad more difficult. – Tito Piezas III Aug 30 '23 at 17:38
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    @mr_e_man I suggest that your parametric form for I is worth posting as an answer. It's worth adding that it yields non-trivial solutions for all $k \geq 2$, while the case $x_1=1$ is covered by the familiar $1^3+12^3=10^3+9^3$. – Adam Bailey Aug 30 '23 at 18:01