Im looking at pairs of triples of natural numbers without repititions such that the sums of the two triples are equal and the products of the two triples are equal.

To be precise: Let $x<y<z$ and $x'<y'<z'$ be positive integers such that $x+y+z=x'+y'+z'$ and $xyz=x'y'z'$.

Is it true that the maximum of these numbers, i.e. $\max(z,z')$, cannot be a prime number?

Experiments up to $\max(z,z')=43$ confirm this, but I didn't give it any more thought.

  • 2
    $\begingroup$ Sure. Say $z=p$ was the maximum. Since the products are equal $p|x'y'z'$ but that's not possible since all of those factors are $<p$. $\endgroup$
    – lulu
    Mar 25 '16 at 10:38
  • $\begingroup$ This is inspired by <math.stackexchange.com/questions/1712445/…>, but probably not otherwise related. $\endgroup$ Mar 25 '16 at 10:38
  • $\begingroup$ @lulu Ha, took you less time to figure this out than it took me to post the question. But you forgot the case $z=z'$. Then you can repeat the same argument for $y$ and $y'$, concluding that $y=y'$ and, in the end, $x=x'$. $\endgroup$ Mar 25 '16 at 10:41
  • $\begingroup$ They can't be equal. if $z=z'$ then $x+y=x'+y'$ and $xy=x'y'$ but that quickly implies that $x=x'$ and $y=y'$. $\endgroup$
    – lulu
    Mar 25 '16 at 10:42
  • $\begingroup$ @lulu and Uli, one of you should post an answer so this question gets moved out of the unanswered questions list :) $\endgroup$ Mar 25 '16 at 11:15

First point is that you can't have $z=z'$. To see that, note that equality of the $z's$ would imply that $$x+y=x'+y'\;\;\&\;\;xy=x'y'$$ But that would imply $x=x',y=y'$ (if $(x+y)=A$ and $xy=B$ with $y>x$ then $y-x=\sqrt{A^2-4AB}$ so you can solve for $x,y$.)

Now suppose that $z=p$ was the maximum. Then $p>z'>y'>x'$ but $p\,|\,x'y'z'$, an impossibility if $p$ is prime.

  • $\begingroup$ By induction, one can then show that the same statement holds for arbitrary $n$-tuples instead of triples: if $x_1<\dotsm<x_n$ and $x_1'<\dotsm<x_n'$ are positive integers such that $x_1+\dotsm+x_n=x_1'+\dotsm+x_n'$ and $x_1\dotsm x_n=x_1'\dotsm x_n'$, then $\max(x_n,x_n')$ is not a prime number. $\endgroup$ Mar 25 '16 at 20:24
  • $\begingroup$ I don't think so. My argument depends on knowing that, for $n=2$, knowing the sum and product determine the pair. This is false for $n>2$. Of course if you assume that $x_n\neq x_n'$ then you are ok (and don't need induction). $\endgroup$
    – lulu
    Mar 25 '16 at 20:27
  • 1
    $\begingroup$ Just to be clear: take any two triples with the same sum, product. Let's say $\{2,8,9\}$ and $\{3,4,12\}$. Now take any prime $p$ bigger than any of the six numbers. $p=13$ works. Then we have the two $4$-tuples $\{2,8,9,13\}$ and $\{3,4,12,13\}$. They have the same sum, same product and the maximum term is $13$. $\endgroup$
    – lulu
    Mar 25 '16 at 20:44

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