How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and ab+14b9a is an integer?
How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and ab+14b9a is an integer?
The correct option is D 4
Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that
u+149u=k, for some integer k.
This equation is equivalent to 9u2−9uk+14=0
9u2−9uk+14=0 whose solutions are,
u=9k±√81k2−50418
=k2±16√9k2−56
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.
(3k+S)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56.
So, (3k-S) and (3k+S) is one of the ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no integral solutions.
The cases (2, 28) and (4, 14) yield k = 5 and k = 3 respectively.
If k=5, then u=13 or u=143.
If k=3, then u=23 or u=73.
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3).
4
Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that
u+149u=k, for some integer k.
This equation is equivalent to 9u2−9uk+14=0
9u2−9uk+14=0 whose solutions are,
u=9k±√81k2−50418
=k2±16√9k2−56
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.
(3k+S)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56.
So, (3k-S) and (3k+S) is one of the ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no integral solutions.
The cases (2, 28) and (4, 14) yield k = 5 and k = 3 respectively.
If k=5, then u=13 or u=143.
If k=3, then u=23 or u=73.
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3).
