Prove that $(m,n)(n,k)(k,n)[m,n,k]^2=[m,n][n,k][k,n](m,n,k)^2$

Question

Here (a,b) represents GCD of a and b

[a,b] represents LCM of a and b

I deduced the relation $(m,n,k)^2[m,n,k]=mnk$

But I am not able to proceed further. Any suggestions??

This question was posted here before (with a minor variation): Identity involving LCM and GCD: $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$. (I found that question by searching in Approach0 first. One of the posts found in this way contained a link to the question post I've mentioned here.) — Martin Sleziak, Dec 01 '19 at 08:43
I will add that some useful advice on searching (which also mentions Approach0) can be found here: How to search on this site? (I will leave my answer here for a bit - in case some of the stuff mentioned there is useful to you. But since Bill Dubuque's answer in the linked post says basically the same thing, my answer probably should be deleted at some point.) — Martin Sleziak, Dec 01 '19 at 08:46
Also the LHS contained $(m,n)$ twice - I have edited it. I assume that the current version of the title is what you actually wanted to ask. — Martin Sleziak, Dec 01 '19 at 08:51

score 0 · Answer 1 · answered Dec 01 '19 at 08:20

0

Hint

WLOG the highest exponent of prime $p$ that divides $m,n,k$ be $M,N,K$

WLOG $M\ge N\ge K\ge0$

Now find each LCM and GCD separately

e.g. $[m,n,k]=M$ and $(m,n,k)=K$

This holds true for any prime that divides $mnk$

answered Dec 01 '19 at 08:20

lab bhattacharjee

1 Answers1