IMO 2001 SL A1

Origin: IND | Category: Algebra

Problem

Let T denote the set of all ordered triples (p, q, r) of nonneg- ative integers. Find all functions f : T \toR such that f(p, q, r) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ if pqr = 0, 1 + 1 6 (f(p + 1, q −1, r) + f(p −1, q + 1, r) +f(p −1, q, r + 1) + f(p + 1, q, r −1) +f(p, q + 1, r −1) + f(p, q −1, r + 1)) otherwise.

Solution

First, let us show that such a function is at most unique. Suppose that f1 and f2 are two such functions, and consider g = f1 −f2. Then g is zero on the boundary and satisfies g(p, q, r) = 1 6[g(p + 1, q −1, r) + \cdot \cdot \cdot + g(p, q −1, r + 1)], i.e., g(p, q, r) is equal to the average of the values of g at six points (p + 1, q −1, r), . . . that lie in the plane \pi given by x + y + z = p + q + r. Suppose that (p, q, r) is the point at which g attains its maximum in absolute value on \pi \capT . The averaging property of g implies that the values of g at (p + 1, q −1, r) etc. are all equal to g(p, q, r). Repeating this argument we obtain that g is constant on the whole of \pi \capT , and hence it equals 0 everywhere. Therefore f1 \equivf2. It remains to guess f. It is natural to try f(p, q, r) = pqr first: it satisfies f(p, q, r) = 1 6[f(p + 1, q −1, r) + \cdot \cdot \cdot + f(p, q −1, r + 1)] + p+q+r . Thus we simply take f(p, q, r) = p + q + rf(p, q, r) = 3pqr p + q + r and directly check that it satisfies the required property. Hence this is the unique solution.