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

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

• 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. Jul 19 '20 at 6:01
• @ user10354138 thank you. I will edit it.
– user809800
Jul 19 '20 at 6:03
• @JohnOmielan thank you very much. i edited it.
– user809800
Jul 19 '20 at 6:18
• @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 $n_i$? Jul 19 '20 at 6:19
• @JohnOmielan ohohoh, i got the idea, let me try. I'ill come back later.
– user809800
Jul 19 '20 at 6:33

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}

• Thank you for your proof. very simple and clear.
– user809800
Jul 19 '20 at 6:50

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 $$n_1,...,n_k$$.

• (Psst. The questioner wanted to try it him- or herself.) Jul 19 '20 at 6:40
• @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. Jul 19 '20 at 6:46
• @quasi thank you for your proof.
– user809800
Jul 19 '20 at 6:48
• @k.stm thank you for your kindness. I proved it myself.
– user809800
Jul 19 '20 at 6:49