Please explain steps with as much detail as possible Solutio

Please explain steps with as much detail as possible :]

Solution

IF N IS PRIME IT IS NOT POSSIBLE

SINCE

FOR A PRIME N

sigma(n) = n+1

and d(n) = 2

now lets take cases

1) d(n) = 3 means n is a prime square

now n+97 = sigma(n)

sigma(n) = p.p^2.1 = p^3

and n = p^2

not possible

2) d(n) = 4

n = pq

sigma(n) = n^2

n^2 = n+96

n^2-n-96 = 0

No integer solution

anther subcase n= p^3

sigma(n) = n^2 in that case too

so no such number is possible

after that

for n having exactly 5 divisors it is p^4 for some prime and the smallest number is 16 but sigma(16) is huge compared to 100. for the 6 divisors smallest number is 12 and sigma(12) is also huge. So actually sigma(n) [at least 2^{d(n)} times n ] will be huge compared to n+100