Least Common Multiple Identity Proof (Andrews Number Theory, Problem 2.4.12)

Question

I'm having some trouble manipulating these expressions to reach the following conclusion:

Let $d_1 = \gcd(a,b), d_2 = \gcd(b,c), d_3 =\gcd(c,a), D = \gcd(a,b,c), L=\lcm(a,b,c)$. Prove that:

$$L = \frac{abcD}{d_1d_2d_3}$$

Using some previous exercises in the book, I can rewrite $d_1, d_2, d_3, D$ and $L$ as:

$$d_1 = t_1^{\alpha_1}...t_v^{\alpha_v}$$ $$d_2 = t_1^{\beta_1}...t_v^{\beta_v}$$ $$d_3 = t_1^{\gamma_1}...t_v^{\gamma_v}$$ $$D = t_1^{\zeta_1}...t_v^{\zeta_v}$$ $$L = t_1^{\mu_1}...t_v^{\mu_v}$$

where each $t_i$ is a factor of at least one of $a,b,c$, and $\alpha_i$, $\beta_i$, $\gamma_i$, and $\zeta_i$ are the smallest powers for their corresponding $\gcd$ (e.g. $\alpha_1$ is the smaller of the two powers of factor $t_i$ that $a$ and $b$ have). Conversely, $\mu_i$ is the greatest exponent between $a,b,c$. It is here that I'm getting stuck. I've substituted these expressions into the equation for $L$, but something about the manipulation is throwing me off. I'm trying to logically work out what, say, $t_i^{\frac{\alpha_i+\beta_i+\gamma_i}{\mu_i}}$ is, but going from that to demonstrating the equality is not computing. Any suggestions would be sincerely appreciated.

I think that maybe drawing a Venn diagram for three sets, which will represent the numbers $a,b,c$ would help you. Interpret intersections in the diagram as taking $\gcd$. Choose variable names for each piece of the diagram that doesn't contain any piece inside. Write the variables in the problem as products of these variables. — plop, Aug 19 '22 at 01:16
Indeed, the identity in question is a multiplicative example of the inclusion-exclusion principle. — anon, Aug 19 '22 at 01:18
Does this answer your question? Prove $\text{lcm}(a,b,c) = \frac{a \cdot b \cdot c \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}$ - found using an Approach0 search. Note the accepted answer there uses the prime factorizations, similar to what you're doing. ... — John Omielan, Aug 19 '22 at 01:19
(cont.) Note I found that question as being listed as a duplicate of Prove that $\operatorname{lcm}(a,b,c)=\frac{abc \operatorname{gcd}(a,b,c)} {\operatorname{gcd}(a,b)\operatorname{gcd}(b,c)\operatorname{gcd}(c,a)}$, with this also listing Prove $\gcd(a,b) \gcd(a,c) \gcd(b,c) ,\text{lcm} (a,b,c)^2=$ $\text{lcm}(a,b),\text{lcm}(a,c) ,\text{lcm}(b,c) \gcd(a,b,c)^2$ and Prove that: $\gcd[a,b,c]=\frac{abc.\operatorname{lcm}(a,b,c)}{\operatorname{lcm}(a,b)\operatorname{lcm}(a,c)\operatorname{lcm}(b,c)}$ as additional duplicates. ... — John Omielan, Aug 19 '22 at 01:25
(cont.) In addition, there's the AoPS thread How to prove this identity ?. — John Omielan, Aug 19 '22 at 01:25

score 0 · Answer 1 · answered Aug 19 '22 at 01:21

Shortcut:

It is sufficient to show that the equality holds, for any prime number $p$ that happens to occur in the prime factorizations of any of $a,b,c.$

Assume that :

$a$ contains $p^A$ in it's prime factorization.
$b$ contains $p^B$ in it's prime factorization.
$c$ contains $p^C$ in it's prime factorization.

Note that one or more of $A,B,C$ may equal $(0)$.

Since the computations of $d_1, d_2,$ and $d_3$ are cyclic, you can assume, without loss of generality that $A \leq B \leq C.$

Then, solely with respect to the prime number $p$, the assertion is equivalent to asserting that

$$\frac{p^A \times p^B \times p^C \times p^A}{p^A \times p^B \times p^A} = p^C.$$

Please strive not to post more dupe answers to dupes of FAQs, cf. recent site policy announcement here. — Bill Dubuque, Aug 19 '22 at 02:33

Least Common Multiple Identity Proof (Andrews Number Theory, Problem 2.4.12)

1 Answers1