Let (a, b, ... , k) denote the greatest common divisor of the integers a, b, ... k and [a, b, ... , k] denote their least common multiple. Show that for any positive integers a, b, c we have (a, b, c)2 [a, b] [b, c] [c, a] = [a, b, c]2 (a, b) (b, c) (c, a).
Solution
If we express a, b, c as a product of primes then the gcd has each prime to the smallest power and the lcm has each prime to the largest power. So the equation given is equivalent to showing that 2 min(r, s, t) + max(r, s) + max(s, t) + max(t, r) = 2 max(r, s, t) + min(r, s) + min(s, t) + min(t, r) for non-negative integers r, s, t. Assume r ≤ s ≤ t. Then each side is 2r + s + 2t.
© John Scholes
jscholes@kalva.demon.co.uk
13 Jul 2002