# Probability of another 3 integers with same sum and product as the first 3 integers

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?

• If reordering is not allowed then you could write the first constraint as $-1000\le x \le y \le z \le 1000$. Mar 17, 2015 at 8:55
• Among positive integers $1+5+8=2+2+10$ and $1\times5\times8 =2 \times 2 \times 10$ and there are more examples Mar 17, 2015 at 8:58
• @Henry Unfortunately not quite what the question is asking. It's asking for $xyz=x+y+z$.
– user66698
Mar 17, 2015 at 8:58
• @Emrakul You may be correct but I read the question as looking for duplicate pairs of $s$ and $p$ but not requiring $s=p$ Mar 17, 2015 at 9:03
• @Henry The title says "same sum and product", which is why I concluded that, though I agree it is mildly unclear.
– user66698
Mar 17, 2015 at 9:17