# prove (a,b)(a,c)(b,c) [a,b,c]^2=[a,b][a,c][b,c] (a,b,c)^2 [closed]

prove gcd(a,b).gcd(a,c).gcd(b,c).lcm [a,b,c]^2=lcm[a,b].lcm[a,c].lcm[b,c] .gcd(a,b,c)^2 it is a part of solution of another problem I try a lot to prove it but I could not be succeed

## closed as off-topic by Xam, user370967, Simply Beautiful Art, Namaste, GlorfindelJul 21 '17 at 19:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xam, Community, Simply Beautiful Art, Namaste, Glorfindel
If this question can be reworded to fit the rules in the help center, please edit the question.

• I'd use the prime factorization of $a,b,c$. Then you need to consider the exponents and their min (gcd) and max (lcm). – Wuestenfux Jul 21 '17 at 7:48
• I tried it but it didn't help me . – math enthusiastic Jul 21 '17 at 7:52
• How far did you get? – Arthur Jul 21 '17 at 8:07

By here $\ [a,b,c]\, =\, \dfrac{abc}{(ab,bc,ca)} =\dfrac{abc(a,b,c)}{(a,b)(b,c)(c,a)}$ by $\ (a,b)(b,c)(c,a) = (a,b,c)(ab,bc,ca)$

Squaring that yields your equation (after replacing $ab/(a,b)$ by $[a,b]$ etc).

Let's write $$a=p_1^{a_1}\cdots p_r^{a_r},$$ $$b=p_1^{b_1}\cdots p_r^{b_r},$$ $$c=p_1^{c_1}\cdots p_r^{c_r},$$ where $p_i$ are primes. Now

$$\displaystyle \rm{gdc}(a,b)gcd(a,c)gcd(b,c)lmc(a,b,c)^2=\Pi_{i=1}^rp_i^{\min\{a_i,b_i\}+\min\{a_i,c_i\}+\min\{b_i,c_i\}+2\max\{a_i,b_i,c_i\}}$$ and

$$\displaystyle \rm{lcm}(a,b)lcm(a,c)lcm(b,c)gcd(a,b,c)^2=\Pi_{i=1}^rp_i^{\max\{a_i,b_i\}+\max\{a_i,c_i\}+\max\{b_i,c_i\}+2\min\{a_i,b_i,c_i\}}$$

Just check that $$\min\{a_i,b_i\}+\min\{a_i,c_i\}+\min\{b_i,c_i\}+2\max\{a_i,b_i,c_i\}=\max\{a_i,b_i\}+\max\{a_i,c_i\}+\max\{b_i,c_i\}+2\min\{a_i,b_i,c_i\}.$$ WLOG assume $a_i\le b_i\le c_i$ and see that both terms are $2a_i+b_i+2c_i.$