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Given two sequences of nondecreasing distinct positive integers such that $$x_1 + x_2 + ... + x_i = y_1 + y_2 + ... + y_i , i>0$$ and that $$x_1x_2 ... x_i = y_1y_2 ... y_i$$ Prove/disprove that the sequences are equal i.e. $$x_1 = y_1, x_2 = y_2, ... , x_i = y_i$$

I started with
Let $x_1x_2 ... x_i$ be $A$. If $A$ is prime, $x_1 = A = y_1$ (since $A$ cannot be factored any more) and we are done.

What I don't know is what happens when $A$ is not prime. Intuitively, it sounds true, and I cannot find any counter examples.

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  • $\begingroup$ $x_1=1, x_i=0 \forall i \ne 1$ and $y_2=1, y_i=0 \forall i\ne 2$ $\endgroup$
    – user563311
    Oct 29, 2018 at 8:40
  • $\begingroup$ @JoeyDoey the Xs and Ys are positive integers $\endgroup$
    – Donald
    Oct 29, 2018 at 9:17
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    $\begingroup$ You should state in your question that $x_i \gt 0$. Positive vs non-negative is a source of confusion. I'm pretty sure I've been told in school (France) that positive is $\ge 0$ while in English, people use "non-negative" to describe this. This is a case where symbols actually make thing clearer. Anyway, your statement is at least false because of permutation. $y_0 = x_i \land \forall i \ne 0, y_i = x_{i-1}$ $\endgroup$
    – F.Carette
    Oct 29, 2018 at 9:36
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    $\begingroup$ Donald, you should at least require that the sequences $(x_i),(y_i)$ are non-decreasing. That way, William Elliot's counterexample is ruled out. $\endgroup$
    – TonyK
    Oct 29, 2018 at 10:00
  • $\begingroup$ @TonyK thank you for the tip, edited. $\endgroup$
    – Donald
    Oct 29, 2018 at 16:59

2 Answers 2

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Counterexample:

$12+4+3 \ =\ 9+8+2$

$12\cdot4\cdot3 \ = \ 9\cdot8\cdot2$

Moreover, for $\ i>2\ ,\ $you can always find infinitely many counterexamples.

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  • $\begingroup$ Welcome to Math.SE! It would improve your answer to explain the existence of "infinitely many counterexamples" in more detail (and perhaps add a remark about solutions with $i=2$). $\endgroup$
    – hardmath
    Oct 30, 2018 at 15:25
  • $\begingroup$ This answer saves me the trouble of trying to find a proof for $i > 2$. I removed my answer because the missing full proof I'm trying to find doesn't exist. +1. $\endgroup$
    – Ronald
    Oct 30, 2018 at 23:48
  • $\begingroup$ Very good really. $\endgroup$
    – Piquito
    Oct 31, 2018 at 0:53
  • $\begingroup$ @hardmath with $12, 4, 3$ and $9, 8, 2$ a counterexample, $12a, 4a, 3a$ and $9a, 8a, 2a$ with $a > 0$ is a counterexample too, since $a(12 + 4 + 3) = a(9 + 8 + 2)$ and $a^3 \cdot 12 \cdot 4 \cdot 3 = a^3 \cdot 9 \cdot 8 \cdot 2$. That makes an infinite number of counterexamples in case $i = 3$ $\endgroup$
    – Ronald
    Oct 31, 2018 at 7:34
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Let $x_1 = y_2, x_2 = y_1$ for a quick counter example.

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  • $\begingroup$ Since the sequences must be non-decreasing, this substitution only works if $x_1=x_2$ and $y_1=y_2$, in which case it is not a counter-example. $\endgroup$
    – M. Nestor
    Oct 29, 2018 at 17:17
  • $\begingroup$ @M.Nestor: the question has changed. $\endgroup$
    – TonyK
    Oct 29, 2018 at 18:35

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