Determine all possible positive whole numbers t such that t is equal to the sum of the squares of its four smallest positive divisors including 1.

For example, 1, 2, 4 and 5 are the four smallest positive divisors of 40.
But, 1² + 2² + 4²+ 5² = 46. So, t = 40 is NOT a valid solution.

The smallest four divisors of t can be one of the following, where p1, p2, p3 are distinct primes:
1, p1, p2, p3
1, p1, (p1^2), p2
1, p1, p2, (p1*p2)
1, p1, (p1^2),(p1^3) 

Except for 2, all primes and their squares and cubes. are odd.  The sum of the four divisors would be even if all four divisors were odd.  Yet, if the sum were even then t must be divisible by 2 – which it is not an odd number, therefore one of the divisors must be 2 and t must be even. 

As t is even, one of the other two remaining smallest divisors of t must be a multiple of 2 and the other not.  This leaves the divisors as:

1, 2, (2^2), p2; or

1, 2, p2, (2*p2)

Where t is the sum of the squares of it four smallest divisors, for the first set of divisors, {1, 2, 4, p2},  t Mod 4 is 2.  Therefore, this set of divisors is not a valid solution.

For the second set, {1, 2, p2, 2*p2}, except for where p2 is 3 or 5, t Mod p2 is 5.  For p2 = 3, t mod p2 is 2, thus 3 is not one of the divisors, leaving only 5.  And as t Mod 5 is 0 it is a valid divisor for the solution. The solution is therefore 130 with the four smallest divisors as 1, 2, 5, and 10.

Edited on May 4, 2007, 12:34 am

Posted by Dej Mar on 2007-05-03 16:04:40