Problem:
Assume that a,b,c and d are positive integers such that a5=b4,c3=d2 and c−a=19. Determine d−b.
Solution:
Since the prime factorization of positive integers is unique, and since 4 is relatively prime to 5 and 2 is relatively prime to 3, one may conclude that there exist positive integers m and n such that
a=m4,b=m5,c=n2 and d=n3
Then
19=c−a=n2−m4=(n−m2)(n+m2)
Since 19 is a prime, and since n−m2<n+m2, it follows that
n−m2=1 and n+m2=19
Therefore, m=3,n=10,d=1000,b=243 and d−b=1000−243=757.
The problems on this page are the property of the MAA's American Mathematics Competitions