# Solution to puzzle 79: Sum of fourth powers

The sum of three numbers is 6, the sum of their squares is 8, and 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
a2 + b2 + c2 = 8
a3 + b3 + c3 = 5

We will find the (monic) cubic equation whose roots are a, b, and c.
If cubic equation x3 − Ax2 + 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 − (a2 + b2 + c2)] = 14.
Hence a, b, c are roots of x3 − 6x2 + 14x − C = 0, and we have

a3 − 6a2 + 14a − C = 0
b3 − 6b2 + 14b − C = 0
c3 − 6c2 + 14c − C = 0

Adding, we have (a3 + b3 + c3) − 6(a2 + b2 + c2) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0.
Hence C = 41/3, and x3 − 6x2 + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x4 − 6x3 + 14x2 − 41x/3 = 0.  Then

a4 − 6a3 + 14a2 − 41a/3 = 0
b4 − 6b3 + 14b2 − 41b/3 = 0
c4 − 6c3 + 14c2 − 41c/3 = 0

Adding, we have (a4 + b4 + c4) − 6(a3 + b3 + c3) + 14(a2 + b2 + c2) − 41(a + b + c)/3 = 0.
Hence a4 + b4 + c4 = 6×5 − 14×8 + (41/3)×6 = 0.

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

## Roots

Note that we did not need to actually calculate a, b, and c in order to determine the sum of their fourth powers.  In fact, one of the numbers is real; the other two are complex conjugates; see below.  The approximate values of the numbers are 2.67770, and 1.66115 ± 1.53116i.

## Generalization

Using the above approach, we can show that if

a + b + c = r
a2 + b2 + c2 = s
a3 + b3 + c3 = t

then a, b, c are roots of x3 − rx2 + ½(r2 − s)x + (½r(3s − r2) − t)/3 = 0.

It then follows that

 a4 + b4 + c4 = 4rt/3 − ½s(r2 − s) + r2(r2 − 3s)/6. = (r4 − 6r2s + 3s2 + 8rt)/6.

## Recurrence Relation

Let f(n) = an + bn + cn, where n is a positive integer.
Then, given the equation x3 − 6x2 + 14x − 41/3 = 0, we can multiply by xn 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  a5 + b5 + c5,  a6 + b6 + c6, and so on.