1/x + 1/y = 1/n=> xy = nx + ny=> (x - n)(y - n) = n²
Now, n² must have 5 factors to get 5 ordered pairs=> n must be of form p², where p is prime number
Hence all primes less than 20

@sumit-agarwal
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.

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