Algebra Puzzle

Date: 04/09/2001 at 18:26:47
From: Johann Matias H. L.
Subject: Hard equation

Okay, here you go... if this is true:

                       x + y + z = 1
                 x*x + y*y + z*z = 2
           x*x*x + y*y*y + z*z*z = 3

find (without solving the equation):

     x*x*x*x + y*y*y*y + z*z*z*z = ?

I've tried everything. Please help me!

Date: 04/16/2001 at 12:06:35
From: Doctor Peterson
Subject: Re: Hard equation

Hello, Johann.

This question stuck in my mind until I had the time to solve it, and I 
want to give you a hint you can try.

I will write "x^2" for x squared, and so on. I will also generalize 
the problem a little. Suppose we know that:

           x + y + z = A

     x^2 + y^2 + z^2 = B

     x^3 + y^3 + z^3 = C

Now we want to find D:

     x^4 + y^4 + z^4 = D

I tried combining the original equations in various ways to get 
something that includes D and other symmetrical expressions; here is 
one that I found useful:

    AC = (x + y + z)(x^3 + y^3 + z^3)

       = (x^4 + y^4 + z^4) + (xy^3 + xz^3 + yx^3 + yz^3 + zx^3 + zy^3)

       = (x^4 + y^4 + z^4) + (xy*x^2 + xy*y^2 + xy*z^2)
         + (yz*x^2 + yz*y^2 + yz*z^2) + (zx*x^2 + zx*y^2 + zx*z^2)
         - (xy*z^2 + yz*x^2 + zx*y^2)

       = (x^4 + y^4 + z^4) + (xy + yz + zx)*(x^2 + y^2 + z^2)
         - xyz*(x + y + z)

       = D + PB - QA

where

     P = xy + yz + zx
     Q = xyz

The fact that P and Q are symmetrical with respect to the three 
variables suggests that they might appear in other expressions 
involving A, B, and C; if we can find P and Q in terms of A, B, and C, 
we will be able to find D.

To do this, I used:

   A^2 = (x + y + z)(x + y + z)

       = (x^2 + y^2 + z^2) + 2(xy + yz + zx)

       = B + 2P

and

    AB = (x + y + z)(x^2 + y^2 + z^2)

       = (x^3 + y^3 + z^3) + (xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2)

       = (x^3 + y^3 + z^3) + (xy*x + xy*y + xy*z - xyz)
         + (yz*x + yz*y + yz*z - xyz) + (zx*x + zx*y + zx*z - xyz)

       = (x^3 + y^3 + z^3) + xy*(x + y + z) + yz*(x + y + z)
         + zx*(x + y + z) - 3xyz

       = (x^3 + y^3 + z^3) + (xy + yz + zx)*(x + y + z) - 3xyz

       = C + PA - 3Q

You'll be able to solve this for D. What I'm not sure of is whether 
there are any actual solutions in x, y, and z at all; but if there 
are, this will tell you what x^4 + y^4 + z^4 has to be. (I think 
actually solving for x, y, and z requires solving a sixth-degree 
polynomial, and they may be complex.) I'd feel a lot surer of my 
solution if I could check it!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/