Problem:
For each positive integer n, let f(n)=n4−360n2+400. What is the sum of all values of f(n) that are prime numbers?
Answer Choices:
A. 794
B. 796
C. 798
D. 800
E. 802
Solution:
Note that f(n)=n4+40n2+400−400n2=(n2+20)2−(20n)2= (n2+20n+20)(n2−20n+20). Because the first factor is greater than 1 , the product cannot be prime unless the second factor is 1 . The solutions of the equation n2−20n+20=1 are 1 and 19. The values of f(1)=12+20⋅1+20=41 and f(19)=192+20⋅19+20=761 are prime, and the requested sum is 41+761=802​.
The problems on this page are the property of the MAA's American Mathematics Competitions