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Let's start from the beginning. about a week ago, I asked a question in MSE with(almost) the same title.

After looking again at everything(the conjecture, the article, the different questions and answers, and more), I've come to the conclusion that for what I'm trying to do, it is enough to prove the conjecture with the condition that each two terms in the finite sequence are NOT co-prime.
i.e $\gcd(x_j,x_l)>1$ for every $1\le j\neq l\le n$

Let $\begin{align} \{ x_i \}_{i=1}^n \end{align}$ be a finite sequence such that $x_i\in\mathbb{N}$ that satisfies $\gcd(x_j,x_l)>1$ for every $1\le j\neq l \le n$.
Prove that if $x_i$ divides $\begin{align}\sum_{i=1}^n x_i\end{align}$, $\forall{ 1\le i\le n}$ then there are $1\le h\neq k\le n$ such that $x_h$ divides $x_k$.

At my previous question, I've tried to prove the statement by induction, we managed to prove the base cases, for $n=2,3$. Obviously, those still hold for this version of the question, because it is a weaker variant.
EDIT - After it was brought to my attention, this claim is also false. But a new claim that can be interesting is if that is true for $\gcd(x_1,x_2,\cdots x_n)=1$

Cusp Connoisseur
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    As currently stated, this new conjecture is also false. With the counter-example (mentioned in your third bullet point) to your original conjecture, multiply each of its $x_i$ by some factor $\gt 1$, e.g., $2$. We then have $\gcd(x_j,x_l)\gt 1$ for every $1\le j\neq l\le n$, but there are still no $x_h$ dividing $x_k$. I suggest you add another condition such as that $\gcd(x_1,x_2,\ldots,x_n)=1$ to avoid this possibility. With a change like that, I'm then currently unsure of whether or not your new conjecture is true. – John Omielan Dec 13 '23 at 21:43
  • Can you explain further/ give a real counter example? – Cusp Connoisseur Dec 13 '23 at 21:49
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    What sort of further explanation would you like, e.g., is there something in my comment you didn't understand? Also, I'm unsure of what mean by giving a "real counter example". What I stated is a valid counter-example, and thus it's "real", based on the currently stated conditions for your conjecture (note I also provided a fairly simple change you can make to avoid that issue). If by "real" you mean a counter-example where $\gcd(x_1,x_2,\ldots,x_n)=1$ then, as I stated earlier, I currently don't have one, and also I'm even unsure of whether or not any exist. – John Omielan Dec 13 '23 at 21:56
  • By multiplying all of the numbers by some number , the sum of their recipients is no longer equal to one. Also, the conjectures were equivalent before but aren’t necessarily now. @JohnOmielan – Cusp Connoisseur Dec 13 '23 at 22:05
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    I should have been more clear in my earlier comment. I meant the $x_i$ being the "transformed" version to your version of the counter-example to the conjecture where the reciprocals sum to $1$. To avoid confusion, let $y_i$ be the value of that original counter-example, i.e., $\sum_{i=1}^{n}\frac{1}{y_i}=1$. Then with $m=\operatorname{lcm}(y_1,y_2,\ldots,y_n)$, multiply both sides by $m$ to get $\sum_{i=1}^{n}x_i=m$ where $x_i=\frac{m}{y_i}$. If any $x_h\mid x_k$ then there's an integer $j$ where $x_k=nx_h;\to;\frac{m}{y_k}=\frac{jm}{y_h};\to;y_h=jy_k;\to;y_k\mid y_h$, which is not ... – John Omielan Dec 13 '23 at 22:17
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    (cont.) permitted in that counter-example. Now apply the reasoning in my original comment to this transformed version, i.e., multiply each term by some factor $\gt 1$, to get a counter-example to your question conjecture as it's currently stated. Finally, note that dividing the adjusted sum (i.e., where each term is multiplied by a new factor) by the new sum, we get back the original counter-example of the sum of fractions being $1$ so, in that sense, the result is still basically equivalent to the other conjecture. – John Omielan Dec 13 '23 at 22:20
  • As for removing the question, that's up to you. However, with the relatively small change I suggested, i.e., adding the condition that $\gcd(x_1,x_2,\ldots,x_n)=1$, then I think it's an interesting question, with no obvious (at least to me) solution. Also, I will probably work on trying to solve it myself when I have some spare time. In addition, I'm curious regarding what sorts of answers, in particular techniques used in them, you might get. – John Omielan Dec 13 '23 at 22:35
  • @JohnOmielan I did think about it as well, but it won't help me with my work, because I'm trying to use this proof on something else – Cusp Connoisseur Dec 13 '23 at 22:38
  • Thanks for the extra information. Nonetheless, I still think it could be an interesting question, just on its own, even if it doesn't help you with your work. – John Omielan Dec 13 '23 at 22:40
  • @JohnOmielan Would you suggest me to edit my question, to the new topic/ – Cusp Connoisseur Dec 13 '23 at 22:41
  • @Chessplayer Once again, whether to delete the question or update it such as how I suggested, is up to you. As I've already stated it would be an interesting question to me, and I would like to know what the solution(s), if any, to it are. – John Omielan Dec 13 '23 at 22:42

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