IMO 1984 SL 11

Origin: CAN

Problem

Let n be a natural number and a1, a2, . . . , a2n mutually distinct integers. Find all integers x satisfying (x −a1) \cdot (x −a2) \cdot \cdot \cdot (x −a2n) = (−1)n(n!)2.

Solution

Suppose that an integer x satisfies the equation. Then the numbers x − a1, x −a2, . . . , x −a2n are 2n distinct integers whose product is 1 \cdot (−1) \cdot 2 \cdot (−2) \cdot \cdot \cdot n \cdot (−n). From here it is obvious that the numbers x−a1, x−a2, . . . , x−a2n are some reordering of the numbers −n, −n + 1, . . . , −1, 1, . . ., n −1, n. It follows that their sum is 0, and therefore x = (a1 + a2 + \cdot \cdot \cdot + a2n)/2n. This is

the only solution if {a1, a2, . . . , a2n} = {x −n, . . . , x −1, x + 1, . . . , x + n} for some x \inN. Otherwise there is no solution.