I thought of three numbers.
Their sum is 6.
The sum of their squares is 8.
The sum of their cubes is 5.
What is the sum of their fourth powers?

An imaginary number is x + y*i  where i= sqrt(-1)
   The real part is x, the imaginary part is y
That number squared is (x^2 - y^2) + 2*x*y*i
... and  cubed is (x^3 - 3*x*y^2) + (3*x^2*y  - y^3)*i

Start with 3 numbers, each with a Real and Imaginary parts:
a+bi, c+di, e+fi     so there are 6 variables

And we have 3 sums:
6+0i, 8+0i, 5+0i

So we can make 6 equations with 6 unknowns:
1 Sum Realpart N = 6
2 Sum Imaginary N =0
3 Sum Realpart Nsquared = 8
4 Sum Imaginary Nsquared =0
5 Sum Realpart Ncubed =5
6 Sum Imaginary Ncubed =0

1: a+c+e=6
2: b+d+f=0
3: a^2 - b^2 +c^2 - d^2 +e^2 - f^2 =8
4: 2*a*b +2*c*d +2*e*f =0
5: a^3 - 3*a*b^2 + c^3 - 3*c*d^2 + e^3 - 3*e*f^2 =5
6: 3*a^2*b - b^3 + 3*c^2*d  - d^3 + 3*e^2*f  - f^3 =0

So this suggests to me that there is a complex number solution.  Anyone care to solve these 6 equations?

Posted by Larry on 2004-07-05 02:21:13