2014/026) How many ordered pairs of positive integers (x,y) satisfy:

x^2 + 10! = y^2

Solution

10! = 2 * 3 * 2^2 * 5 * (2 * 3) * 7 * (2^3) * 3^2 * ( 2* 5)
= 2^8 * 3 ^ 4 * 5^2 * 7

(x+y)(x-y) = 2^8 * 3 ^ 4 * 5^2 * 7

x and y both are integers when x + y and x -y both are even or odd

both cannot be odd

so both are even

no of factors of (x+y) *(x-y) = (8+1)(4+1)(2+1)(1+1) = 270

from this we need to remove the number of odd factors

(4+1)(2+1)(1+1) = 30
no solutions = 240