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Can we uniquely identify the set of k different natural numbers (no two are the same) by knowing only their sum and product (and the value of k itself)?

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    $\begingroup$ I suggest you try examples, with k equal to 3. $\endgroup$ – Mariano Suárez-Álvarez Mar 25 '16 at 2:02
  • $\begingroup$ @MarianoSuárez-Alvarez: I don't see an obvious 3-number counter-example if $0$ is not a natural number. $\endgroup$ – user21820 Mar 25 '16 at 3:14
  • $\begingroup$ @user21820 Agreed. Though there are 3-number counterexamples, they take a while to find by hand. $\endgroup$ – Bamboo Mar 25 '16 at 4:09
  • $\begingroup$ It's possible for $k=2$. Not for $k \ge 3$. $\endgroup$ – TheRandomGuy Mar 25 '16 at 4:58
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    $\begingroup$ Think about polynomials and their coefficients... This would imply that you could find all the roots of a polynomial by simply looking at the first and last coefficient. $\endgroup$ – Aleks J Mar 25 '16 at 11:14
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{4,9,10} and {5,6,12} have the same sum (23) and product (360) so the answer is no.

As such then {4k,9k,10k} and {5k,6k,12k} also have the same sum and product so there is a multitude of counter examples. I'm imagine there are others based on different triples as well.

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  • $\begingroup$ First (I think) triple example is: $$6\times8\times25=1200\text{ and }6+8+25=39$$ $$5\times10\times24=1200\text{ and }5+10+24=39$$ $$4\times15\times20=1200\text{ and }4+15+20=39$$ $\endgroup$ – Ian Miller Mar 26 '16 at 15:50
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I think the minimal counterexample (in the sense of minimal sum) are the triples $\{2,8,9\}$ and $\{ 3,4,12 \}$ for which we have $$ 2+8+9 = 19 \qquad 2\cdot 8\cdot 9 = 144$$ and $$ 3 + 4 + 12 = 19 \qquad 3 \cdot 4 \cdot 12 = 144.$$

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    $\begingroup$ Also, the minimal counterexample in the sense of minimal product is {1, 9, 10} and {2, 3, 15}. $\endgroup$ – user200783 Mar 25 '16 at 8:36

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