I’ve been working on generalizing the classic identity:
$$\def\lcm{\operatorname{lcm}} \lcm(a, b) \cdot \gcd(a, b) = ab$$
to more than two integers. I came up with the following result, which seems to hold for all ($ n \geq 3 $):
$$ \frac{ \displaystyle \prod_{i=1}^n a_i}{\lcm(a_1, \dots, a_n)} = \frac{ \displaystyle \prod_{1 \le i < j \le n} \gcd(a_i, a_j) }{ \displaystyle \prod_{i=1}^{n-2} \left( \max_{\substack{S \subset \{a_1, \dots, a_n\} \\ |S| = n - i + 1}} \gcd(S) \right)^{n - i - 1} } $$
I’ve proven it using $p$-adic valuations (included below), but I’m curious:
Has this identity appeared in the literature before?
Does this form suggest any deeper algebraic or number-theoretic structure?
Is there a cleaner or more symmetric way to express the denominator?
Is the identity even valuable, or is it trivial? Because I’ve noticed that $n=4$ doesn’t give out any cool/useful results (for me at least), it just gives me a messy bunch.
Proof Sketch
We begin by defining:
Let $a_1, a_2, \dots, a_n \in \mathbb{N}^* $. Define:
$$ Q= \{ p\in P \mid \forall\ i= \overline{1,n}, p \mid a_i\} $$
For each $p \in Q$, consider the $p$-adic valuations $v_p(a_i)$. Reorder the $a_i$ as $b_{1,p}, b_{2,p}, \dots, b_{n,p}$ such that:
$$v_p(b_{1,p}) \le v_p(b_{2,p}) \le \dots \le v_p(b_{n,p})$$
The left-hand side of the identity becomes: $$ \frac{\displaystyle \prod_{i=1}^n a_i}{\lcm(a_1, \ldots, a_n)} $$
Taking $( v_p )$ on both sides, we get: $$ v_p\left( \prod_{i=1}^n a_i \right) - \max_{i} v_p(a_i) = \sum_{i=1}^n v_p(b_{i,p}) - v_p(b_{n,p}) $$
Now consider the right-hand side of the identity, which consists of:
The product of all $( \gcd(a_i, a_j) )$, which contributes $( (n - i) v_p(b_{i,p}) )$ over all pairs. A denominator consisting of powers of max-GCDs over subsets of size $k$, for $k = n, n-1, \ldots, 3$. These are shown to correspond to multiplicities of $v_p(b_{i,p}$, with exponents $n - i - 1$ for the $i$-th smallest valuation.
Putting everything together, both sides reduce to: $$ \sum_{i=1}^n v_p(b_{i,p}) - v_p(b_{n,p}) = \sum_{1 \le i < j \le n} \min(v_p(a_i), v_p(a_j)) - \sum_{k=2}^{n} \left( \max_{|S|=k} v_p(\gcd(S)) \cdot w_k \right) $$ By careful bookkeeping and using the ordered $p$-adic valuations, we find the two sides match exactly for each prime $p$, and thus, multiplying across all $p$, the full identity holds.
Example: Let $(a_1, a_2, a_3, a_4) = (6, 9, 15, 10)$, then:
LHS = $\dfrac{6 \cdot 9 \cdot 15 \cdot 10}{\lcm(6,9,15,10)}=90$
RHS = $\dfrac{3 \cdot 3 \cdot 2 \cdot 3 \cdot 1 \cdot 5}{\gcd(6,9,15,10)^2 \cdot \max(\gcd(6,9,15), \gcd(6,9,10), \gcd(6,15,10), \gcd(9,15,10))}=90$
Also, I’ve noticed that the entire process can be reversed (lesser $\to$ greater), (min $\to$ max), and thus we have a second equation that holds, just by switching GCD() and LCM(), and min $\to$ max. It looks something like this: $$ \frac{\displaystyle \prod_{i=1}^n a_i}{\gcd(a_1, \ldots, a_n)} = \frac{ \displaystyle \prod_{1 \le i < j \le n} \lcm(a_i, a_j) }{ \displaystyle \prod_{i=1}^{n-2} \left( \min_{\substack{S \subset \{a_1, \ldots, a_n\} \\ |S| = n - i + 1}} \lcm(S) \right)^{n - i - 1} } $$
I’d appreciate any insights, simplifications, or known references!
