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 ----------------------------------------------------------------------