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I am trying to prove that

let $n_1,\ldots,n_k \in \Bbb Z\setminus\{0\}$. then $\gcd(n_i,n_j)=1 \forall i\neq j$ iff $\operatorname{lcm}(n_1,\ldots,n_k)=n_1\cdots n_k$

I can prove "$\Rightarrow$" this direction by the fact that $\gcd(n_1,n_1)\operatorname{lcm}(n_1,n_2)=n_1n_2$ and by induction on $k.$

But I do not know if the converse is true or not, it is obvious when $k=1$, as $\gcd(n_1,n_1)\operatorname{lcm}(n_1,n_2)=n_1n_2$. But I got stuck at extend $k$ from $2$ to any natural number.

Any suggestion will be appreciated

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  • $\begingroup$ Be careful! "Coprime" and "pairwise coprime" are two different things when you have more than two numbers. You want pairwise coprime here -- let $p$ be a common prime factor of $n_j,n_{j'}$, then lcm of the two is strictly less than the absolute value of their product, so the same is true when you add back the other $n_i$s. $\endgroup$ Jul 19 '20 at 6:01
  • $\begingroup$ @ user10354138 thank you. I will edit it. $\endgroup$
    – user809800
    Jul 19 '20 at 6:03
  • $\begingroup$ @JohnOmielan thank you very much. i edited it. $\endgroup$
    – user809800
    Jul 19 '20 at 6:18
  • $\begingroup$ @BXY You're welcome. As for solving the problem, have you tried using the unique prime factorizations, in particular that $\gcd(n_i,n_j) = 1$ means there's no overlap of primes used among the $n_i$? $\endgroup$ Jul 19 '20 at 6:19
  • $\begingroup$ @JohnOmielan ohohoh, i got the idea, let me try. I'ill come back later. $\endgroup$
    – user809800
    Jul 19 '20 at 6:33
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If $g:=\gcd(n_i,n_j)>1$ for some $i\neq j$.

Note that $\frac {n_1 \cdots n_k} {g} < n_1 \cdots n_k$ is a common multiplier of $n_1, \ldots ,n_k$, which implies $\text{lcm}(n_1, \ldots ,n_k)\leq\frac {n_1 \cdots n_k} {g}<n_1 \cdots n_k$

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  • $\begingroup$ Thank you for your proof. very simple and clear. $\endgroup$
    – user809800
    Jul 19 '20 at 6:50
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Let $k\ge 2$ and suppose $n_1,...,n_k$ are nonzero integers which are not pairwise coprime.

Without loss of generality, assume $n_1,n_2$ have a common factor $d > 1$.

Let $N=\left|\prod_{i=1}^k n_i\right|$, and let $M={\large{\frac{N}{d}}}$.

Since $n_1$ is a multiple of $d$, so is $N$, hence $M$ is positive integer, and since $d > 1$, we have $M < N$.

Then from $$M=\Bigl(\frac{n_1}{d_1}\Bigr)(n_2\cdots n_k)$$ it follows that $M$ is a multiple of $n_2,...,n_k$, and from $$M=\Bigl(\frac{n_2}{d}\Bigr)(n_1)(n_3\cdots n_k)$$ it follows that $M$ is a multiple of $n_1$.

Thus $M$ is a common multiple of $n_1,...,n_k$.

Therefore $N$ is not the least common multiple of $n_1,...,n_k$.

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  • $\begingroup$ (Psst. The questioner wanted to try it him- or herself.) $\endgroup$
    – k.stm
    Jul 19 '20 at 6:40
  • $\begingroup$ @k.stm: Sure, but the comment stream gave the clear impression that the OP was planning an approach via unique factorization, so my answer doesn't in any way interfere with that. $\endgroup$
    – quasi
    Jul 19 '20 at 6:46
  • $\begingroup$ @quasi thank you for your proof. $\endgroup$
    – user809800
    Jul 19 '20 at 6:48
  • $\begingroup$ @k.stm thank you for your kindness. I proved it myself. $\endgroup$
    – user809800
    Jul 19 '20 at 6:49

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