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The goal is to understand a set of nontrivial solutions of cubic polynomials

$$ \sum_{i=1}^3 a_i ^3=\sum_{i=1}^3 b_i ^3 \Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, \tag{1} $$ $$ \sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i \Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , \tag{2} $$ for $a_i,b_i\in \mathbb{Z}$ where we demand $(a_1,a_2,a_3)\neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).

  • Question: Are there any nontrivial solutions? (e.g. $a_i,b_i\in \mathbb{Z}$ where we demand $(a_1,a_2,a_3)\neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)

  • A non-trivial example or a proof of non-existence is required to be accepted as a final answer.


Note add:

My trials/attempts: Since it is encouraged to show one's own attempt.

First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $\sum_{i=1}^2 a_i ^3=\sum_{i=1}^2 b_i ^3 $ and $\sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.

Let me share a few comments on what these two equations boil down to:

If we take (2)$^3$-(1), we get $$ (a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), \tag{3} $$ we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.

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Some solutions: $$\matrix{a_1 &= 1,\; a_2 &= 5,\; a_3 &= 5,\; b_1 &= 2,\; b_2 &= 3,\; b_3 &= 6\cr a_1 &= 1,\; a_2 &= 9,\; a_3 &= 9,\; b_1 &= 4,\; b_2 &= 4,\; b_3 &= 11\cr a_1 &= 1,\; a_2 &= 10,\; a_3 &= 14,\; b_1 &= 3,\; b_2 &= 7,\; b_3 &= 15\cr a_1 &= 1,\; a_2 &= 10,\; a_3 &= 15,\; b_1 &= 4,\; b_2 &= 6,\; b_3 &= 16\cr a_1 &= 2,\; a_2 &= 8,\; a_3 &= 10,\; b_1 &= 4,\; b_2 &= 5,\; b_3 &= 11\cr a_1 &= 2,\; a_2 &= 10,\; a_3 &= 12,\; b_1 &= 3,\; b_2 &= 8,\; b_3 &= 13\cr a_1 &= 2,\; a_2 &= 10,\; a_3 &= 10,\; b_1 &= 4,\; b_2 &= 6,\; b_3 &= 12\cr a_1 &= 5,\; a_2 &= 10,\; a_3 &= 11,\; b_1 &= 6,\; b_2 &= 8,\; b_3 &= 12\cr }$$

EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).

$$a_1 = t, \; a_2 = 5 t, \; a_3 = 7 t - 2, \; b_1 = t + 1, \; b_2 = 5 t - 2,\; b_3 = 7 t - 1 $$

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  • $\begingroup$ any generating formula or just brute force? $\endgroup$ – Momo Jul 17 '18 at 2:59
  • $\begingroup$ Semi-brute. I looped over $1 \le a_1 \le a_2 \le 10$ and $a_1 < b_1 \le b_2 \le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 \ge a_2$ and $b_3 \ge b_2$ then output the result. $\endgroup$ – Robert Israel Jul 17 '18 at 3:02
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Here is one non-trivial solution:

$$ 1 + 5 + 5 = 2 + 3 + 6 $$

$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$

The first line sums are $11$ and the second line sums are $251$.

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  • $\begingroup$ This is good - how do you obtain it? +1 (hopefully not just brutal force!) $\endgroup$ – wonderich Jul 17 '18 at 2:35
  • $\begingroup$ I like this solution - this looks really interesting. $\endgroup$ – wonderich Jul 17 '18 at 2:36
  • $\begingroup$ I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found. $\endgroup$ – hardmath Jul 17 '18 at 2:39
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if you put

$\alpha_1=a_2+a_3$

$\alpha_2=a_1+a_3$

$\alpha_3=a_1+a_2$

$\beta_1=b_2+b_3$

$\beta_2=b_1+b_3$

$\beta_3=b_1+b_2$

Then your (3) becomes $\alpha_1\alpha_2\alpha_3=\beta_1\beta_2\beta_3$ and (2) becomes $\alpha_1+\alpha_2+\alpha_3=\beta_1+\beta_2+\beta_3$

Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.

The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $\alpha_i,\beta_i$

For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.

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  • $\begingroup$ yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers. $\endgroup$ – wonderich Jul 17 '18 at 2:29
  • $\begingroup$ thanks for the link $\endgroup$ – wonderich Jul 17 '18 at 2:37
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$$\left\{\begin{aligned}&X_1+X_2+X_3=Y_1+Y_2+Y_3\\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3\end{aligned}\right.$$

$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$

$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$

$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$

$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$

$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$

$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$

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