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?

Let the numbers be a, b, and c. Then we have

a + b + c = 6

a^{2} + b^{2} + c^{2} = 8

a^{3} + b^{3} + c^{3} = 5

We will find the (monic) cubic equation whose roots are a, b, and c.

If cubic equation x^{3} − Ax^{2} + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find

A = a + b + c

B = ab + bc + ca

C = abc

Then B = ab + bc + ca = ½ [(a + b + c)^{2} − (a^{2} + b^{2} + c^{2})] = 14.

Hence a, b, c are roots of x^{3} − 6x^{2} + 14x − C = 0, and we have

a^{3} − 6a^{2} + 14a − C = 0

b^{3} − 6b^{2} + 14b − C = 0

c^{3} − 6c^{2} + 14c − C = 0

Adding, we have (a^{3} + b^{3} + c^{3}) − 6(a^{2} + b^{2} + c^{2}) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0.

Hence C = 41/3, and x^{3} − 6x^{2} + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x^{4} − 6x^{3} + 14x^{2} − 41x/3 = 0. Then

a^{4} − 6a^{3} + 14a^{2} − 41a/3 = 0

b^{4} − 6b^{3} + 14b^{2} − 41b/3 = 0

c^{4} − 6c^{3} + 14c^{2} − 41c/3 = 0

Adding, we have (a^{4} + b^{4} + c^{4}) − 6(a^{3} + b^{3} + c^{3}) + 14(a^{2} + b^{2} + c^{2}) − 41(a + b + c)/3 = 0.

Hence a^{4} + b^{4} + c^{4} = 6×5 − 14×8 + (41/3)×6 = 0.

That is, the sum of the fourth powers of the numbers is 0.

Generalization

Using the above approach, we can show that if

a + b + c = r

a^{2} + b^{2} + c^{2} = s

a^{3} + b^{3} + c^{3} = t

then a, b, c are roots of x^{3} − rx^{2} + ½(r^{2} − s)x + (½r(3s − r^{2}) − t)/3 = 0.

It then follows that

a

^{4} + b

^{4} + c

^{4} = 4rt/3 − ½s(r

^{2} − s) + r

^{2}(r

^{2} − 3s)/6. = (r

^{4} − 6r

^{2}s + 3s

^{2} + 8rt)/6.

Let f(n) = a^{n} + b^{n} + c^{n}, where n is a positive integer.

Then, given the equation x^{3} − 6x^{2} + 14x − 41/3 = 0, we can multiply by x^{n} and sum over the three roots to yield the following recurrence relation:

f(n+3) = 6f(n+2) − 14f(n+1) + (41/3)f(n).

Successive applications of this formula allow us to calculate a^{5} + b^{5} + c^{5}, a^{6} + b^{6} + c^{6}, and so on.

*Edited on ***October 28, 2012, 4:06 am**