Problem:
How many pairs of positive integers (a,b) are there such that gcd(a,b)=1 and
ba+9a14b
is an integer?
Answer Choices:
A. 4
B. 6
C. 9
D. 12
E. infinitelymany Solution:
Let u=a/b. Then the problem is equivalent to finding all positive rational numbers u such that
u+9u14=k
for some integer k. This equation is equivalent to 9u2−9uk+14=0, whose solutions are
u=189k±81k2−504=2k±619k2−56
Hence u is rational if and only if 9k2−56 is rational, which is true if and only if 9k2−56 is a perfect square. Suppose that 9k2−56=s2 for some positive integer s. Then (3k−s)(3k+s)=56. The only factors of 56 are 1,2,4,7, 8,14,28, and 56 , so (3k−s,3k+s) is one of the ordered pairs (1,56),(2,28), (4,14), or (7,8). The cases (1,56) and (7,8) yield no integer solutions. The cases (2,28) and (4,14) yield k=5 and k=3, respectively. If k=5, then u=1/3 or u=14/3. If k=3, then u=2/3 or u=7/3. Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3),(2,3),(7,3), and (14,3).
Rewrite the equation
ba+9a14b=k
in two different forms. First, multiply both sides by b and subtract a to obtain
9a14b2=bk−a
Because a,b, and k are integers, 14b2 must be a multiple of a, and because a and b have no common factors greater than 1 , it follows that 14 is divisible by a. Next, multiply both sides of the original equation by 9a and subtract 14b to obtain
b9a2=9ak−14b
This shows that 9a2 is a multiple of b, so 9 must be divisible by b. Thus if (a,b) is a solution, then b=1,3, or 9 , and a=1,2,7, or 14 . This gives a total of twelve possible solutions (a,b), each of which can be checked quickly. The only such pairs for which
ba+9a14b
is an integer are when (a,b) is (1,3),(2,3),(7,3), or (14,3).