Fermat's S(n) problem:
n+S(n) is simply the sum of all the divisors.
If n is the cube of a prime, say n = p^3, then n+S(n) = p^3 + p^2 + p
+ 1 = (p+1)(p^2+1). It is also possible to use prime powers, so n=p^6
or p^9 or whatever, but these tend to not factor so well.
Suppose n is the cube of a product of two relatively prime numbers,
say n=(pq)^3. The factors of n are simply the products of all factors
of p^3 with all products of n^3.

Very quickly, I find that n=7^3 works; the sum of its factors is 400.

My strategy is to write out the square-free prime factorizations of
(p+1)(p^2+1) for many primes. I combine these in various ways to
complete squares and thus cancel out factors. When I start combining
the combinations, I cannot repeat a factor in n, because this violates
the relatively prime condition, but this is not a problem, because I
can simply drop the factor, for the same effect on the square-free
factorization.

So I found the following independent solutions:
n=7^3
n=(5x23x41x83)^3
n=(5x13x43x191x307)^3
n=(3x17x31x41x43x239x307)^3
n=(2x3x17x31x239x293x313x443x463)^3
n=(3x11x31x443x499)^3

These are all independent combinations of factors, so all 2^6
combinations (including the null combination which gives the trivial
solution n=1) yield solutions.
Joseph DeVincentis
--------------------------------------------------------------------
I assume that the requirement is to do it without calculating aids.

  I have found five (or six if you allow n=1) solutions in an hour.
  The following n all have the factors of n^3 summing to a square

n=1,7,2*3*5*13*41*47,2*3*5*7*13*41*47,5*11*13*27*41*47,5*7*11*13*27*41*47.

Will continue, and will describe method later.

Having moved my method to a computer, I have found that there are exactly 
2^13 numbers N such that:
  i) The sum of the factors of N^3 is a square
  ii) If p^k|N and p is prime, k an integer, then p^k<1000.

The smallest are
  1,7,751530,4370879,

The largest is
  28267118845351289604953424330730236774643814325667330932127905
The sum of the factors of its cube is the square of

2741377889597887358488522157253188638703535264733490362101001948122763493376
*10^17
Dr. Luke Pebody
----------------------------------------------------------------------