Prove (or disprove) 2 equation of 6 variables

Question

Given

$$(s-x_{1})(s-y_{1})(s-z_{1}) = (s-x_{2})(s-y_{2})(s-z_{2})$$

Where $s=x_1+y_1+z_1$ and all variables are positive non-zero integers.

I need to prove that such values of $x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2}$ can or can't exist.

I have tried to perform an expansion of both sides but this produced a complex expression.

Please use simple terms to describe the solution.

Note: This is not a homework. I am too old for those now...

Thank you for your help.

Problem Origins

A question was asked where there was a need to create a database which rows could be stored but only when there are no permutations in the database for 3 given columns. That is, if a row such as $3,5,1$ exists you can't store $1,5,3$.

I thought this can be solved if I could find a function $f(x,y,z) = v$ such that when you use any of the 6 permutations of $x,y,z$, the function result is always the same. That is, $v=f(x,y,z)=f(x,z,y)=f(y,x,z)=...$

I though since I have 3 values, I may use the area of a triangle using the points $(x,p), (y,p+1)$ and $(z,p+2)$ where $p$ is an arbitrary integer. The area can be easily calculated this way and it satisfies the above requirement. However, another requirement was to prove that the function does not produce the same value $v$ for any set of numbers other than $x,y,z$ or their permutations. We know that we can have 2 triangles of the same area and different side lengths. So I tried to add that restriction in and got into above question. If we can prove that the above equality does not hold, the the area of a triangle can be used to provide the desired function.

EDIT-1 Thank you for your all your help. As @André Nicolas mentioned, the constraints are:

$s1=x_1+y_1+z_1 = s2=x_2+y_2+z_2$

Also, the points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ can't be permutations of one another. For example the points (1,1,2) and (1,2,1) don't represent a solution.

Please accept my apologies for missing this in the original version.

Edit-2 - Answer

I used brute force and found that indeed some points satisfy the requirement of an equality.

For example:

(1,5,5) and (2,3,6)

for (1,5,5): s1=11, lhs product=10*6*6=360

for (2,3,6): s2=11, rhs product=9*8*5=360

and the 2 points are not permutations of the same set of numbers.

Thanks all for your help.

Answer 1

Here's what I meant in my comment.

Assume $x,y,z \in \mathbb{Q}$. The tuple $(x,y,z)$ is determined uniquely up to order by the polynomial $$(X - x)(X - y)(X - z) = X^3 - (x + y + z)X^2 + (xy + xz + yz)X - (xyz).$$ Since $1, \sqrt{2}, \sqrt{3}$ are linearly independent in $\mathbb{R}$ over $\mathbb{Q}$, the function $$f(x,y,z) = (x+y+z) + \sqrt{2} (xy + xz + yz) + \sqrt{3}(xyz)$$ can never take the same value for different values of $x+y+z, xy+xz+yz, xyz \in \mathbb{Q}$. So $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ if and only if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are permutations of each other.

Answer 2

Above equation(given below) has parametrization:

$(s-x_{1})(s-y_{1})(s-z_{1}) = (s-x_{2})(s-y_{2})(s-z_{2})$

$(x_{1},y_{1},z_{1})$=$((30k^2+5k-5),(6k^2-5k-25),(30k^2-25))$

$(x_{2},y_{2},z_{2})$=$((18k^2+24k-10),(12k^2-24k-15),(36k^2-30))$

$s=11(6k^2-5)$

For $k=3$: we get $(x_{1},y_{1},z_{1})$=$(40,2,35)$

$(x_{2},y_{2},z_{2})$=$(32,3,42)$ and $s=77$

Answer 3

Regarding the original problem: Can't you just arrange the 3 values in ascending order? In that way f(x,y,z) = f(x,z,y) = ...etc always. And it's injective, 1-1, as you needed it.

I'm afraid I have no idea about the equation though.

Answer 4

Surely you take any number that can be represented as the product of $3$ discrete numbers in $2$ different ways, and set $s$ as the sum of one set of products.

For example:

$48 = (1 \times 4 \times 12) = (2 \times 3 \times 8)$.
Let $s$ = $(2+3+8)=13$.
$(13-12)(13-9)(13-1) = (13-11)(13-10)(13-5)$

Or have I misunderstood something?