Prove that (2m)!(2n)! is a multiple of m!n!(m+n)! for any non-negative integers m and n.
Solution
The trick is to find a recurrence relation for f(m,n) = (2m)!(2n)!/(m!n!(m+n)!). In fact, f(m,n) = 4 f(m,n-1) - f(m+1,n-1), which is sufficient to generate all the f(m,n), given that f(m,0) = (2m)!/(m!m!), which we know to be integeral.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998
Last updated/corrected 14 Mar 03