IMO 1991 SL 20

Origin: IRE

Problem

Let \alpha be the positive root of the equation x2 = 1991x + 1. For natural numbers m, n define m ∗n = mn + [\alpham][\alphan], where [x] is the greatest integer not exceeding x. Prove that for all natural numbers p, q, r, (p ∗q) ∗r = p ∗(q ∗r).

Solution

We prove the result with 1991 replaced by any positive integer k. For natural numbers p, q, let ϵ = (\alphap −[\alphap])(\alphaq −[\alphaq]). Then 0 < ϵ < 1 and

ϵ = \alpha2pq −\alpha(p[\alphaq] + q[\alphap]) + [\alphap][\alphaq]. Multiplying this equality by \alpha−k and using \alpha2 = k\alpha+1, i.e. \alpha(\alpha−k) = 1, we get (\alpha −k)ϵ = \alpha(pq + [\alphap][\alphaq]) −(p[\alphaq] + q[\alphap] + k[\alphap][\alphaq]). Since 0 < (\alpha−k)ϵ < 1, we have [\alpha(p∗q)] = p[\alphaq]+q[\alphap]+k[\alphap][\alphaq]. Now (p ∗q) ∗r = (p ∗q)r + [\alpha(p ∗q)][\alphar] = = pqr + [\alphap][\alphaq]r + [\alphaq][\alphar]p + [\alphar][\alphap]q + k[\alphap][\alphaq][\alphar]. Since the last expression is symmetric, the same formula is obtained for p ∗(q ∗r).