Problem:
The sum of the first m positive odd integers is 212 more than the sum of the first n positive even integers. What is the sum of all possible values of n?
Answer Choices:
A. 255
B. 256
C. 257
D. 258
E. 259
Solution:
The sum of the first k positive integers is 2k(k+1). Therefore the sum of the first k even integers is
2+4+6+⋯+2k=2(1+2+3+⋯+k)=2⋅2k(k+1)=k(k+1).
The sum of the first k odd integers is
(1+2+3+⋯+2k)−(2+4+6+⋯+2k)=22k(2k+1)−k(k+1)=k2.
The given conditions imply that m2−212=n(n+1), which may be rewritten as n2+n+(212−m2)=0. The discriminant for n in this quadratic equation is 1−4(212−m2)=4m2−847, and this must be the square of an odd integer. Let p2=4m2−847, and rearrange this equation so that (2m+p)(2m−p)=847.
The only factor pairs for 847 are 847⋅1,121⋅7, and 77⋅11. Equating these pairs to 2m+p and 2m−p yields (m,p)=(212,423),(32,57), and (22,33). Note that the corresponding values of n are found using n=2−1+p, which yields 211,28, and 16, respectively. The sum of the possible values of n is (A)255.
The problems on this page are the property of the MAA's American Mathematics Competitions