Problem:
Positive integers a,b,c, and d satisfy a>b>c>d,a+b+c+d=2010, and a2−b2+c2−d2=2010. Find the number of possible values of a.
Solution:
Note that
a2−b2+c2−d2=(a−b)(a+b)+(c−d)(c+d)=a+b+c+d
and thus a−b=c−d=1. Hence 2010=a+(a−1)+c+(c−1), so a+c= 1006. The condition a>c implies that a≥504, and the condition c>d implies that c≥2, so that a≤1004. For each integer k with 0≤k≤500, the ordered quadruple (a,b,c,d)=(504+k,503+k,502−k,501−k) satisfies the required conditions, and thus the number of possible values of a is 501.
The problems on this page are the property of the MAA's American Mathematics Competitions