In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1
if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76
if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511
if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10
Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2).
You can try out with various numbers (may be smaller numbers) so that this can be verified.
F(x) = x^4-360x^2+400F(x) = (x^2+20)^2 - 400x^2F(x) = (x^2-20x+20)(x^2+20x+20) = Prime NumberSince the value of second factor is always positive and greater than 1 therefore, 1st factor hasto be equals to 1.x^2-20x+20 = 1x^2-20x+19= 0On solving,x = 1, 19.Putting back the values in the original equation we get f(x) = 41 and f(x) = 761.
So, the sum will be 41+761 = 802