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.

Right away, it can be deducted that t must be even.  If t were odd, all of its factors - as well as their squares - would be odd, yet the sum of four odd numbers is always even.  Therefore, t is a multiple of 2.

For t = a² + b² + c² + d², we already know that a=1 and b=2.  We also know that between c and d, one is odd while the other is even.  This means that either c=4 and d is prime OR c is prime and d=2*c.

Consider the former:  t = 1 + 4 + 16 + d² = d²+21

If d is a prime factor of t, it must also be a prime factor of 21(either 3 or 7).  Neither 1, 2, 3, 4 nor 1, 2, 4, 7 fit the equation.

Now consider the latter:  t = 1 + 4 + c² + 4c² = 5(1+c²)

t must be a multiple of 5, so c=5 and d=10.

t = 1² + 2² + 5² + 10² = 130

This is the only solution.

Posted by hoodat on 2007-05-03 23:47:40