Problem:
Let a and b be positive integers satisfying a+bab+1<23. The maximum possible value of a3+b3a3b3+1 is qp, where p and q are relatively prime positive integers. Find p+q.
Solution:
First observe that if a=1 or b=1, then a3+b3a3b3+1=1. Assume that a≥2 and b≥2. The inequality a+bab+1<23 implies that 2ab+2<3a+3b, and hence 3b−2>a(2b−3) giving 3b−2>4b−6 which implies that b<4; by symmetry a<4. The pair (a,b)=(3,3) does not satisfy a+bab+1<23, but checking the pairs (a,b)=(2,2) and (a,b)=(2,3), it is seen that the maximum value of a3+b3a3b3+1 is 531, which occurs at (a,b)=(2,3). The requested sum is 31+5=36.
OR
From 2ab+2<3a+3b it follows that 2ab−3a−3b+2<0 implying 4ab− 6a−6b+4+5<5 and (2a−3)(2b−3)<5. Thus 2a−3 and 2b−3 are odd integers whose product is less than 5. This shows that either a or b is 1, or {a,b}⊆{2,3}, and the analysis proceeds as above.
The problems on this page are the property of the MAA's American Mathematics Competitions