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Let us suppose $3$ integers are selected at random from a large range, say $$-1000\leq x\leq y\leq z\leq 1000$$

Now, we define the sum and product: $$\begin{align*}s&=x+y+z \\p&=xyz\end{align*}$$

($s$ and $p$ will not be equal in most cases, sorry for the confusion)

What is the probability that there exists another solution for $(x,y,z)$ that satisfies above 3 equations? (reordering of x, y and z not allowed)

My friend gave me this question, and I have no idea where to start. If we limit ourselves to positive integers, is there a unique solution, or not?

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    $\begingroup$ If reordering is not allowed then you could write the first constraint as $-1000\le x \le y \le z \le 1000$. $\endgroup$ – Jimmy R. Mar 17 '15 at 8:55
  • $\begingroup$ Among positive integers $1+5+8=2+2+10$ and $1\times5\times8 =2 \times 2 \times 10$ and there are more examples $\endgroup$ – Henry Mar 17 '15 at 8:58
  • $\begingroup$ @Henry Unfortunately not quite what the question is asking. It's asking for $xyz=x+y+z$. $\endgroup$ – Aza Mar 17 '15 at 8:58
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    $\begingroup$ @Emrakul You may be correct but I read the question as looking for duplicate pairs of $s$ and $p$ but not requiring $s=p$ $\endgroup$ – Henry Mar 17 '15 at 9:03
  • $\begingroup$ @Henry The title says "same sum and product", which is why I concluded that, though I agree it is mildly unclear. $\endgroup$ – Aza Mar 17 '15 at 9:17

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