$a+b+c=x+y+z$ and $abc=xyz$ , in which each two of them unequal.

Question

$a,b,c,x,y,z\in \mathbb Z$ ,they are all positive. And not equal to each other.

Let $a>b>c>0,x>y>z>0$

$$ \begin{cases} a+b+c=x+y+z\\ abc=xyz \end{cases} $$

now I have try out $1+8+12=2+3+16$ ,but is there more solutions?

comment tells me :$k(a,b,c,x,y,z)$ also works. So now the question is , how many essentially different solutions can be found?

With this Mathematica program, I get these solutions.

l = Table[{i, j, k, i j k, i + j + k}, {i, 1, 20}, {j, i + 1, 20}, {k,
  j + 1, 20}] // Flatten[#, 2] &;
Select[GatherBy[l, #[[{-2, -1}]] &], Length[#] > 1 &] // Column

Show part of them.

{1, 8, 12, 96, 21}, {2, 3, 16, 96, 21},

{1, 9, 10, 90, 20}, {2, 3, 15, 90, 20},

{2, 7, 12, 168, 21}, {3, 4, 14, 168, 21},

{2, 8, 9, 144, 19}, {3, 4, 12, 144, 19},

{2, 9, 15, 270, 26}, {3, 5, 18, 270, 26},

{2, 10, 12, 240, 24}, {3, 5, 16, 240, 24},

{2, 12, 15, 360, 29}, {3, 6, 20, 360, 29},

{3, 8, 10, 240, 21}, {4, 5, 12, 240, 21},

{3, 9, 16, 432, 28}, {4, 6, 18, 432, 28},

{3, 10, 12, 360, 25}, {4, 6, 15, 360, 25}

Answer 1

Here is a simple parametrization of a family of solutions.

Let $R,S,T,U$ be any four positive integers such that $RS=TU$.

Then a solution is given by $$\{a,b,c\}=\{R+1,S+1,TU+T+U+1\},\{x,y,z\}=\{T+1,U+1,RS+R+S+1\}.$$

Example

$2\times6=3\times4$ so let $R=2,S=6,T=3,U=4$.

This gives the solution $\{3,7,20\},\{4,5,21\}$.

All solutions can be generated by the following rather more involved procedure. This was obtained by noting that since $a$ is a factor of $xyz$ we can let $a=x_1y_1z_1$ where $x_1$ divides $x$, ... etc.

Choose the numbers $x_2,x_3,y_1,y_3,z_1,z_2$ arbitrarily.

Define $A=y_1z_1-x_2x_3,B=x_2z_2-y_1y_3,C=x_3y_3-z_1z_2$.

Then let $x_1,y_2,z_3$ be any solution of $$Ax_1+By_2+Cz_3=0.$$

The required triples of numbers are then $\{x_1y_1z_1,x_2y_2z_2,x_3y_3z_3\},\{x_1x_2x_3,y_1y_2y_3,z_1z_2z_3\}$.

Example

Let $x_2=x_3=y_1=y_3=1,z_1=2,z_2=4$

Then $A=1,B=3,C=-7$.

The general solution of $x_1+3y_2-7z_3=0$ is $x_1=7r-3s,y_2=s,z_3=r$.

The required triples of numbers are then $\{r,4s,14r-6s\},\{8r,s,7r-3s\}$ for 'any' choice of $r$ and $s$.

Answer 2

Given a sum $S$ and a product $P$, we are looking for solutions to $a+b+c = S$ and $abc = P$. We substitute $a = S - b - c$ into the product equation to obtain

$$Sbc - b^2c - bc^2 = P \iff bc^2 + (b^2-Sb)c + P = 0.$$

This implies

$$\begin{align} c &= \frac{Sb - b^2\pm\sqrt{{(b^2-Sb)}^2-4bP}}{2b} \tag{$*$} \end{align}$$

In order for $c$ to be integral, we need $\Delta = {(b^2-Sb)}^2-4bP$ to be a perfect square. Notice that $\frac Pb = ac$ must be an integer, so $b^2\mid \Delta$ and hence there must be some integer $r$ with

$${(b-S)}^2 - 4\frac Pb = r^2.$$

But ${(b-S)}^2 = {(a+c)}^2$ and $\frac Pb = ac$ and hence we must have

$$\begin{align} r^2 &= {(a+c)}^2- 4ac \\&= a^2+2ac+c^2-4ac \\&= a^2-2ac+c^2 = (a-c)^2. \end{align}$$

It follows that $\Delta$ must be $b^2(a-c)^2$ and hence

$$\begin{align} c &= \frac{Sb - b^2\pm b(a-c)}{2b} \\&= \frac{S - b\pm (a-c)}{2} \end{align}$$

and because $S-b = a+c$, we see that we must take the minus sign. We also see that taking the plus sign in the RHS of $(*)$ yields $a$.

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Let's get what we know back. Given a sum value $S$ and product value $P$, we consider $1<b<S/2$ which must be a proper divisor of $P$. A solution to $a+b+c = S$ and $abc = P$ exists if and only if

$$|r|^2 = (S-b)^2 - 4\frac Pb\tag{1}$$

is a perfect square, and in this case we have

$$c = \frac{S-b - |r|}{2}$$ $$a = \frac{S-b + |r|}{2}$$

With this, it should be fairly easy to look for solutions to the equation. Pick some numbers $a$, $b$ and $c$ from which we obtain $S$ and $P$, and then test for when divisors of $P$ that are less than $S/2$ make $(1)$ true. If you find some other value of $b$ that makes it work, you're good.

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I was hoping to find a parametrization of some family of non-multiple solutions, but I haven't managed so far. Perhaps this can help someone else on getting to a more complete answer, or perhaps I'll find something out and edit this later.

Answer 3

Solution is: For (a+b+c)=(x+y+z) & abc=xyz

a=8w(w-1) b=(16w+11) c=11(4w-1)

x=22(w-1) y=2w(4w-1) z=2(16w+11)