Problem:
How many ordered four-tuples of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc−ad=93?
Solution:
Since a+d=b+c, we may take (a,b,c,d)=(a,a+x,a+y,a+x+y), where x and y are integers with 0<x<y. Then
93=bc−ad=(a+x)(a+y)−a(a+x+y)=xy
from which either (x,y)=(1,93) or (x,y)=(3,31). In the first case,
(a,b,c,d)=(a,a+1,a+93,a+94)
is in the desired range for a=1,2,…,405. In the second case,
(a,b,c,d)=(a,a+3,a+31,a+34)
is in the desired range for a=1,2,…,465. These two sets of four-tuples are disjoint, so a total of 405+465=870​ four-tuples of integers (a,b,c,d) satisfy the given conditions.
The problems on this page are the property of the MAA's American Mathematics Competitions