Correct solutions were received from Juan Carlos Marivela, Philippe Fondanaiche, Jens Vob, Paul Botham.

First note that x²⁰⁰³ + 1 = (x + 1)(x²⁰⁰² - x²⁰⁰¹ + ... + x² - x + 1).  There are 2003 terms in the sum within the second set of parentheses, so this sum is odd.  Therefore 2ⁿ divides x²⁰⁰³ + 1, then 2ⁿ must divide x + 1.  Clearly, the smallest value for this to hold is x = 2ⁿ - 1. 

You are visitor number 2315 to this page.