Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
--> (x^6 - y^2) = 127varun289 wrote:Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127
--> (x^3 - y)(x^3 + y) = 127
As 127 is a prime number, it can be expressed as the product of two integers in only one way, i.e. 1*127
Hence, either
- (x^3 - y) = 1 and (x^3 + y) = 127
--> x = 4 and y = 63
- (x^3 - y) = 127 and (x^3 + y) = 1
--> x = 4 and y = -63 ---> Not possible
The correct answer is B.
