Nontrivial integer solutions of $\sum_{i=1}^3 a_i ^3=\sum_{i=1}^3 b_i ^3$ and $\sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i$

Question

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.

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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.

Answer 1

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.

Answer 2

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 $251$.

Answer 3

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).

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

Answer 4

$$\left\{\begin{aligned}&X_1+X_2+X_3=Y_1+Y_2+Y_3\\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3\end{aligned}\right.$$

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

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

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

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

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

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