I thought about it, and this is what I got so far:
lcm[a,b] = ab/(a,b), and so on.
(a,c)=ax+cy
(a,b)=ax+by
(b,c)=bx+cy
(a,b)(b,c)=(ax+by)(bx+cy)=(abx^2 + acxy+ b^2 xy + bcy^2)
I am not sure how to factor a ax+cy out of that...
For every choice of non zero integers a,b,c, the integer lcm[a,c] divides
the integer lcm[a,b] * lcm[b,c] / b.
I was thinking of letting a=p/q, b=r/s, c=t/u. Then, somehow, the integer lcm[a,b] * lcm[b,c] / b will factor out an lcm[a,c], but I am not sure how to prove it that way.
Let be the set of primes which divide either one of . Write
be the canonical factorizations of , where some of the exponents may be 0. Then, letting ,
and hence
Now you just need to show that, for every , we have . Hint : consider separately the case where and the case where .
Flawless! Good job.
There may be a shorter way... but when you have proofs with greatest common divisors and least common multiples, a guaranteed path to success is to translate the identity you wish to prove as a statement about the canonical factorization of the numbers involved. It might be hard work but it always takes you where you want to go - just be careful choosing a good notation so you don't hit a brick wall.
I corrected my post above - thanks for the heads up
I know this is done, but I just wanted to point something out that could help in the future ...
This was written down:
(a,c)=ax+cy
(a,b)=ax+by
(b,c)=bx+cy
But we can not assume that all the x and y are the same ... that's the bit that could help in the future ... make sure you distinguish your variables:
Another way to prove this is:
definition of lcm[a,b]=d is the d is the smallest integer such that a|d and b|d
let [a,b] = d, [a,c] = e, and [b,c] = f
since b|d and b|f, let bx=d and by=f
We are trying to show that [a,c]|[a,b][b,c]/b ... in other words, e|df/b
df/b = bxf/b = xf and since c|f, c|df/b
df/b = dby/b = dy and since a|d, a|df/b
So a|df/b and c|df/b ... so df/b is a common multiple of a and c ... can you show that the [a,c] divides this common multiple?