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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 $Ssetminus{x_k}$ are coprime for each $kin{1,2,ldots,n}$ (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 $ineq j$ ?

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!

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 $ngeq 4$. Take $S={p_1p_2,p_2p_3,ldots,p_np_1}$ where $p_i$ are distinct primes.

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.