Finding x 2 + y 2 + z 2 given that x + y + z = 0, x 3 + y 3 + z 3 = 3 and x 4 + y 4 + z 4 = 15

Question

Finding       x    2    +      y    2    +      z    2   given that   x  +  y  +  z  =  0,       x    3    +      y    3    +      z    3    =  3 and       x    4    +      y    4    +      z    4    =  15

Giovanna Erickson · Accepted Answer

Note that  0  =  (  x  +  y  +  z      )    2    =  (      x    2    +      y    2    +      z    2    )  +  2  (  x  y  +  x  z  +  y  z  ), so       x    2    +      y    2    +      z    2    =  −  2  (  x  y  +  x  z  +  y  z  ).And                    15                            =                  x          4                +                  y          4                +                  z          4                                                  =        (                  x          2                +                  y          2                +                  z          2                          )          2                −        2        (                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                )        .            And                    (        x        y        +        x        z        +        y        z                  )          2                                    =        (                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                )        +        2        (                  x          2                y        z        +        x                  y          2                z        +        x        y                  z          2                )                                          =        (                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                )        +        2        x        y        z        (        x        +        y        +        z        )                                          =                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                +        2        x        y        z        (        0        )                                          =                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                    hence      x    2        y    2    +      x    2        z    2    +      y    2        z    2    =  (  x  y  +  x  z  +  y  z      )    2    .So                    15                            =                  x          4                +                  y          4                +                  z          4                                                  =        (                  x          2                +                  y          2                +                  z          2                          )          2                −        2        (                  x          2                          y          2                +                  x          2                          z          2                +                  y          2                          z          2                )                                          =        (                  x          2                +                  y          2                +                  z          2                          )          2                −        2        (        x        y        +        x        z        +        y        z                  )          2                                                  =        (                  x          2                +                  y          2                +                  z          2                          )          2                −        2        (        x        y        +        x        z        +        y        z        )        (        x        y        +        x        z        +        y        z        )                                          =        (                  x          2                +                  y          2                +                  z          2                          )          2                +        (                  x          2                +                  y          2                +                  z          2                )        (        x        y        +        x        z        +        y        z        )                                          =        (                  x          2                +                  y          2                +                  z          2                          )          2                +        (                  x          2                +                  y          2                +                  z          2                )                  (          −                      1            2                    (                      x            2                    +                      y            2                    +                      z            2                    )          )                                                  =        (                  x          2                +                  y          2                +                  z          2                          )          2                −                  1          2                (                  x          2                +                  y          2                +                  z          2                          )          2                                                  =                  1          2                (                  x          2                +                  y          2                +                  z          2                          )          2                    so   (      x    2    +      y    2    +      z    2        )    2    =  30, hence       x    2    +      y    2    +      z    2    =      30

racodelitusmn · Accepted Answer

Yes, this can be solved without guessing, using Newton's identities.
Since

x + y + z = 0

, they are the roots of

t^{3} + a t - b = 0

.
Newton's identities give us (in a straightforward mechanical manner) that

x^{3} + y^{3} + z^{3} = 3 b

x^{4} + y^{4} + z^{4} = 2 a^{2}

and

x^{2} + y^{2} + z^{2} = - 2 a

This gives us

b = 1

and

2 a^{2} = 15

.
You can solve for a and find the value of

x^{2} + y^{2} + z^{2} = - 2 a

. Note that if you assume

x, y, z

are real, then you need to pick

a < 0

which gives us

x^{2} + y^{2} + z^{2} = \sqrt{30}

Note that you did not even need to use the value of

x^{3} + y^{3} + z^{3}

.

Finding x 2 </msup> + y 2 </msup> + z 2 </

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