if the lcm is simply the product, then the integers are pairwise prime

Question

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

Answer 1

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 $\text{lcm}(n_1, \ldots ,n_k)\leq\frac {n_1 \cdots n_k} {g}<n_1 \cdots n_k$

Answer 2

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 $n_1,...,n_k$.