Show that for any non-negative reals x, y, [5x] + [5y] ≥ [3x+y] + [x+3y]. Hence or otherwise show that (5a)! (5b)!/( a! b! (3a+b)! (a+3b)! ) is integral for any positive integers a, b.
Solution
If is obviously sufficient to prove that [5x] + [5y] ≥ [3x+y] + [x+3y] for 0 < x < y < 1. If 2x ≥ y, then [5x] ≥ [3x+y] and [5y] ≥ [x+3y], so the result holds. So assume 2x < y. It is now a question of examining a lot of cases.
If y < 2/5, then 3x + y < 5y/2 < 1, so [3x+y] = 0, and [3y+x] <= [5y], so the result holds. If 2/5 ≤ y < 3/5 and x < 1/5, then [5y] = 2, [3x+y] = 0 or 1 and [x+3y] = 1. If 2/5 ≤ y < 3/5 and x ≥ 1/5, the [5x]+[5y] = 3, [3x+y] = 1, [x+3y] = 1 or 2. If 3/5 ≤ y < 4/5, then [5y] = 3, [3x+y] = 0 or 1, [x+3y] = 2 or 3. If y ≥ 4/5 and x < 1/5, then [5x] + [5y] = 4, [3x+y] = 0 or 1, [x+3y]= 2 or 3. If y ≥ 4/5 and 1/5 ≤ x, then [5x] + [5y] = 5, [3x+y] = 2, [x+3y] = 2 or 3.
The highest power of a prime p dividing n! is [n/p] + [n/p2] + [n/p3] + ... . Thus it is sufficient to show that [5m/p] + [5n/p] ≥ [ (3m+n)/p ] + [ (m+3n)/p ], which follows immediately from the previous result, putting x = m/p, y = n/p.
© John Scholes
jscholes@kalva.demon.co.uk
11 Jul 2002