49th Putnam 1988 Problem B3 α n is the smallest element of the set { |a - b√3| : a, b non-negative integers with sum n}. Find sup α n . Solution
: (√3 + 1)/2. Evidently the elements of { |a - b√3| : a, b non-negative integers with sum n} are spaced a distance (√3 + 1) apart, so either smallest positive or the largest negative lies in the interval ( -(√3 + 1)/2, (√3 + 1)/2). Thus (√3 + 1)/2 is certainly an upper bound. Let a n , b n be the pair with |a n - b n √3| = α n . All the values (a n - b n √3) lie in the interval ( - (√3 + 1)/2, (√3 + 1)/2) ), so for any ε > 0, we can find two distinct values, corresponding to say m and n, which differ by less than ε. In other words α m-n < ε. But now consider the sequence of values α m-n , α 2(m-n) , α 3(m-n) , ... For kα m-n < (√3 + 1)/2 we have α k(m-n) = k α m-n . So we get a sequence of increasing values with spacing less than ε. Thus the sequence gives a value within ε of (√3 + 1)/2, which is therefore the least upper bound. 49th Putnam 1988 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001