Do these $n$ cyclic conditions imply every two numbers are pairwise coprime?

Question

The question probably sounds vague in the title, so here is the question. Suppose that for a set of $n$ (not necessarily distinct) integers $S=\{x_1,x_2,x_3,\ldots,x_n\}$, it is true that all integers in $S\setminus\{x_k\}$ are coprime for each $k\in\{1,2,\ldots,n\}$ (in the sense that no integer $d>1$ divides every element in $S$). Is it true that $\operatorname{gcd}(x_i,x_j)=1$ for all $i\neq j$ ?

Forgive me but I do not know how to proceed. I think this claim is true, since the $n$ sets of criteria seem to be sufficient to imply this. But I have no idea how to actually prove it. I thought that since no integer $d>1$ divides all of $\{x_1,\ldots,x_{n-1}\}$, then surely there must be two numbers in this set which are coprime. This would immediately imply the conclusion, but I have no clue how to prove that claim either. Any help is appreciated, thanks!

Answer 1

It is false for $n=2$ as in the other answer - take $S=\{2,4\}$.

It is true for $n=3$ because your assumptions just mean all pairs are coprime.

It is false for $n\geq 4$. Take $S=\{p_1p_2,p_2p_3,\ldots,p_np_1\}$ where $p_i$ are distinct primes.

Answer 2

This is wrong if $n=2$ - for instance, consider $S = \lbrace 2, 4 \rbrace$.

Otherwise, given $i \neq j$, we can find $k \notin \lbrace i, j \rbrace$ and since all integers in $S \setminus \lbrace x_k \rbrace$ are coprime, $x_i$ and $x_j$ are.