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.

x^2 + y^2 + z^2 + t^2 = n

x=1 (given)

If x, y, z, and t are all odd then n is even. If n is even then y=2.

So not all of x, y, z, t are odd. Then n is even. Therefore y=2.

(1, 2, z, t)

Either z or t is odd (the other even) to make n even.

if z = 3 then n = 1^2 + 2^2 + 3^2 + t^2 and t is even

n = 14 + t^2

This will only work if t = 14 (n = 210)

If z = 4 then n = 1^2 +2^2 + 4^2 +t^2 and t is odd

n = 21 + t^2

Then t would be 3, 7, or 21. 3 is too small and 7 and 21 do not create an n that is divisible by 4.

If z = 5 then n = 30 + t^2 (t is even)

Then t would be 6 or 10. If t=6 then n=66 and is not divible by 5. Leading to t = 10 and Jer's answer of n = 130

(1, 2, 5, 10, 130)

 

 

Edited on May 3, 2007, 4:09 pm

Posted by Leming on 2007-05-03 15:56:33