Now, What Year Is It? 11125

Abstract

Solution by Andrew D. Loveless (student), Washington State University, Pullman, WA. Note that 2003 is prime. Since 1001 = (2003 l)/2, Euler's criterion yields 19971001 EEE (^|) = 1 (mod 2003), where we have used the Legendre symbol. Like wise it is well known that p divides (p ) when p is prime, k > 0, and 0 < n < pk, so (200f ?4) = 0 (mod 2003) for 0 < n < 20032004. Thus when 0 < n < 20032004, the equation requires that m2 1 2m + 5 = 0 (mod 2003), or equivalent^, that (m l)2 = -3 (mod 2003). Since (?^) = -1, there are no such solutions. On the other hand, when n is 0 or 20032004, the only integral solution to the resulting quadratic equation is m = 1. Thus the only solutions are (m, h) = (1, 0) and (m, n) = (1, 20032004).