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, Glorfindel Jul 21 '17 at 19:13

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  • 3
    $\begingroup$ I'd use the prime factorization of $a,b,c$. Then you need to consider the exponents and their min (gcd) and max (lcm). $\endgroup$ – Wuestenfux Jul 21 '17 at 7:48
  • $\begingroup$ I tried it but it didn't help me . $\endgroup$ – math enthusiastic Jul 21 '17 at 7:52
  • 1
    $\begingroup$ How far did you get? $\endgroup$ – 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 $2a_i+b_i+2c_i.$


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