Let f(n) = ∑1n [r/2]. Show that f(m + n) - f(m - n) = mn for m > n > 0.
It is a trivial induction to show that f(2k) = k2, f(2k+1) = k2 + k. So if m and n have the same parity, then f(m+n) + f(m-n) = (m+n)2/4 + (m-n)2/4 = mn. If m and n have opposite parity, then f(m+n) + f(m-n) = (m+n-1)(m+n+1)/4 + (m-n-1)(m-n+1)/4 = mn.
27th Putnam 1966
© John Scholes
25 Jan 2002