A sufficient condition for the formula $\frac{a_1 \cdots a_n}{gcd(a_1,\ldots,a_n)} = lcm(a_1,\ldots,a_n)$

Question

I am trying to find a sufficient condition so that

$$\frac{a_1 \cdots a_n}{gcd(a_1,\ldots,a_n)} = lcm(a_1,\ldots,a_n)$$

holds.

My idea would be to require that $gcd(a_1,\ldots,a_n) = 1$, i.e. the $a_i$ are pairwise coprime, which is clearly a necessary condition.

However, I am having trouble proving that $gcd(a_1,\ldots,a_n) = 1$ is indeed also sufficient. I tried to start by assuming that

$$\frac{a_1 \cdots a_n}{gcd(a_1,\ldots,a_n)} = lcm(a_1,\ldots,a_n)$$

holds, but I do not see how I can conlcude from there that $gcd(a_1,\ldots,a_n) = 1$. Could you give me a hint?

See also https://math.stackexchange.com/questions/319297/gcd-to-lcm-of-multiple-numbers — lhf, May 18 '21 at 23:29

Calvin Lin · Accepted Answer · 2021-05-19T14:30:11.577

2

Since the statement is true for $n = 2$, henceforth we focus on $ n \geq 3$.

Note: $ \gcd (a_i) = 1 $ is not a sufficient condition, which might explain why you're having trouble proving a false statement.

Find a counter example.

Hint: If $ a_i = \prod {p_k} ^ {i_k}$, then $\gcd(a_i) = \prod p_k ^ { \min i_k } , lcm (a_i) = \prod p_k ^ { \max i_k } $.

So, for the equation to hold, we must have for every prime $p_k$.

$$ \sum i_k = \min i_k + \max i_k $$

When can this hold?

Show that for each $p_k$, at most one $i_k$ is non-zero.

Thus, what is the necessary and sufficient condition?

Each prime $p_k$ divides at most one $a_i$.

If you're stuck even after reading the hints, show your work and what you've tried.

edited May 19 '21 at 14:30

answered May 18 '21 at 23:34

Calvin Lin

1

The argument is incorrect. Your claim is sufficient but not necessary (e.g. it is true for all $a_i$ when $n = 2)\ \ $ – Bill Dubuque May 19 '21 at 00:33
@BillDubuque Yes indeed. I made the implicit assumption that $ n \geq 3$. Let me add that in. – Calvin Lin May 19 '21 at 14:29

1 Answers1