Suppose $k$ is say $3$. Let $A,B$ be a sets of $3$ natural numbers. $A$ not equal to $B$. Can sum and product of the elements in $A$ be same as that of $B$.

If the numbers are primes then the conditions should hold. But I couldn't prove it if they are not primes. Is there a proof or counter example?


What if every number is greater than 1?

  • $\begingroup$ Have you made any attempt to solve this yourself? $\endgroup$ Jul 17, 2013 at 16:47
  • 3
    $\begingroup$ $$1 \cdot 7 \cdot 18=126=2 \cdot 3 \cdot 21$$ and $$1+7+18=26=2+3+21.$$ $\endgroup$ Jul 17, 2013 at 16:56
  • $\begingroup$ @Douglas Thanks $\endgroup$
    – banarun
    Jul 17, 2013 at 16:58
  • $\begingroup$ Please check the updated question $\endgroup$
    – banarun
    Jul 17, 2013 at 17:06

1 Answer 1


The solutions of $$X^3+aX^2+bX+c=0$$ have sum $-a$ and product $-c$. Since $b$ can still vary, your conjecture is definitely not true with complex or real numbers instead of naturals. On the other hand, the same argment with a quadratic polynomial shows that the case with $2$ numbers is true even if we allow real or complex numbers.

For naturals start with an arbitrary almost-solution (e.g. found by pushing around prime factors) such as $1\cdot 8\cdot 9=3\cdot 4\cdot 6$ but $1+8+9\ne 3+4+6$. Then try to find $r,s$ such that $1r+8+9s=3r+4+6s$, i.e. $3s=2r-4$, say $s=2, r=5$. This gives us a solution $(1r,8,9s,3r,3,6s)$, i.e. $$ 5\cdot 8\cdot 18 = 15\cdot 4\cdot 12\qquad \text{and}\qquad 5+8+18=15+4+12.$$


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