The roots are a and b: a + b = p and ab = 12 (a + b)^2 = p^2 (a - b)^2 = (a + b)^2 - 4ab => (a - b)^2 = p^2 - 12 * 4 = p^2 - 48 If |a - b| ≥ 12 { Difference between the roots is at least 12} then, (a - b)^2 ≥ 144 p^2 - 48 ≥ 144 p^2 ≥ 192 P ≥ 8√3 or P ≤ -8√3

1, 3, 6, 10, 15 ..... Let nth term = an^2 + bn + c Put n = 1 a + b + c = 1 4a + 2b + c = 3 9a + 3b + c = 6 5a + b = 3 3a + b = 2 2a = 1 a= 1/2 b = 2-3/2= 1/2 C = 0 So (1/2)n^2 + 1/2(n) Nth term = 1/1/2(n^2+n)= 2/n(n+1) => 2 [ 1/n - 1/(n+1)] 2 [ 1/1-1/2+1/2-1/3+.....+1/9-1/10] 2[ 1-1/10] 2 * 9/10 = 9/5

In keynote 4, example 2, total non-negative unordered integral solutions is just 1 (4,0) and total non-negative ordered integral solutions are 2 (4,0 and 0,4). Please Correct me if I'm wrong.

@anurag_chauhan 10

@vikas_saini can u plz explain 2nd and 3rd+4th step again. 2nd step:what is -1-what are we reducing in it 3rd step: 3rd+4th step doesnt giv value in 2nd step. what are we missing in it. 3rd+4th= 2T-2

x^2 + y^2 - xy = x + y (x - y)^2 + (x -1)^2 + (y -1)^2 = 2 (0,0) (0,1) (1,0) (2,2) (1,2) (2,1) 6 integral solutions 3 positive integer solutions [credits : @Raman_Sharma]

@vishwa2017 for question no. 2, 256 is a perfect square, so i think answer should be 14