Question

2. If x 2 – 5y 2 = 1232, how many pairs are possible for (x, y)

Answer 1

LHS (X^2-5y^2) must have unit digit 2 .

unit digit of 5Y^2 will either be 5 or 0. Hence unit digit of X^2 must be 7 or 2.

There is no integer , square of which has 7 or 2 at its unit place.Hence there is no value of X and Y possible.

Answer 2

No possible pair of x and y can satisfy this equation if x and y are both limited to be integers. But if x and y can be real numbers, then there are infinite solutions possible.

1) For real numbers, let x be any number say 20, then we can calculate y as 12.89.

Similarly, we can assume x can be anything and compute y giving us an Infinite set of possible values.

2) But if x and y are meant to be only integers, then this is not possible.

Since RHS = 1232, then the difference of x^2- 5 y^2 should have 2 on unit place

Since Y^2 is multiplied by 5, the resulting quantity would end with either 5 or 0.

Hence the last digit of X^2 should be 7 or 2. which is not possible as no Number has its square ending with 7 or 2