IMO Shortlist 1987 problem 23
Dodao/la: arhiva2. travnja 2012.
Prove that for every natural number
) there exists an irrational number
such that for every natural number
Remark. An easier variant: Find
as a root of a polynomial of second degree with integer coefficients.
Proposed by Yugoslavia.
Izvor: Međunarodna matematička olimpijada, shortlist 1987