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

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. — user10354138, Jul 19 '20 at 06:01
@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 $n_i$? — John Omielan, Jul 19 '20 at 06:19
@JohnOmielan ohohoh, i got the idea, let me try. I'ill come back later. — , Jul 19 '20 at 06:33
Does this answer your question? If $\gcd(a,b)=1$, then $\operatorname{lcm}(a,b)=ab$ — Chrystomath, Jul 31 '20 at 15:06

score 3 · Accepted Answer · edited Jul 19 '20 at 07:14

3

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

edited Jul 19 '20 at 07:14

John Omielan

47,976

answered Jul 19 '20 at 06:46

xyz

1,457

Thank you for your proof. very simple and clear. – Jul 19 '20 at 06:50

quasi · Answer 2 · 2020-07-19T07:34:46.320

1

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 $n_1,...,n_k$.

edited Jul 19 '20 at 07:34

answered Jul 19 '20 at 06:40

quasi

58,772

(Psst. The questioner wanted to try it him- or herself.) – k.stm Jul 19 '20 at 06:40
@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. – quasi Jul 19 '20 at 06:46
@quasi thank you for your proof. – Jul 19 '20 at 06:48
@k.stm thank you for your kindness. I proved it myself. – Jul 19 '20 at 06:49

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

2 Answers2