Problem:
Find the largest integer n satisfying the following conditions:
(i) n2 can be expressed as the difference of two consecutive cubes;
(ii) 2n+79 is a perfect square.
Solution:
Let m be an integer such that (m+1)3−m3=n2, which implies that 3(2m+1)2=(2n−1)(2n+1). Because 2n−1 and 2n+1 are consecutive odd integers, they are relatively prime. Thus 3 can only divide one of 2n−1 and 2n+1. Therefore either 2n−1=3k2 and 2n+1=j2, or 2n−1=k2 and 2n+1=3j2. The first case implies that j2−3k2=2, which can be shown to be impossible by examining the equation modulo 3. The second case implies that 4n=3j2+k2 and 3j2−k2=2. Let k=2a+1 for some integer a. Then 3j2=k2+2=(2a+1)2+2. Therefore 4n=(2a+1)2+(2a+1)2+2=8a2+8a+4, or n=2a2+2a+1. Furthermore, because 2n+79=d2 for some integer d, then 2n+79= d2=2(2a2+2a+1)+79=4a2+4a+81. This equation is equivalent to 80=d2−(4a2+4a+1)=d2−(2a+1)2=(d−2a−1)(d+2a+1). Both factors on the right side are of the same parity, so they both must be even. Then the two factors on the right are either (2,40),(4,20), or (8,10), and (d,a)=(21,9),(12,27),(9,0). The first solution gives n=2441−79=181, and the last solution gives n=281−79=1. Thus the largest such n is 181 (with m=104).
The problems on this page are the property of the MAA's American Mathematics Competitions