# Solving a System of Equations of the form $\sum x^k = C$

I just ran into this:

\begin{align} x^1+y^1+z^1&=0\\ x^3+y^3+z^3&=3\\ x^4+y^4+z^4&=15\\ x^2+y^2+z^2&=\text{?} \end{align}

I know the answer, but can this be solved without guessing?

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I see that $xyz=1$. – user21436 Mar 29 '12 at 22:09
If you can afford the CPU cycles, there's Grobner basis :) – user2468 Mar 29 '12 at 23:05

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

Since $x + y + z = 0$, they are the roots of $t^3 + at -b = 0$.

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

$$x^3 + y^3 + z^3 = 3b$$ $$x^4 + y^4 + z^4 = 2a^2$$

and $$x^2 + y^2 + z^2 = -2a$$

This gives us $b=1$ and $2a^2 = 15$.

You can solve for $a$ and find the value of $x^2 + y^2 + z^2 = -2a$. Note that if you assume $x,y,z$ are real, then you need to pick $a \lt 0$ which gives us

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

Note that you did not even need to use the value of $x^3 + y^3 + z^3$.

You can also derive the above identities in a slightly tedious way, as in my answer here: Simplifying an expression

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– Aryabhata Mar 29 '12 at 22:27
Since $t^3+at-b=0$ substituting successively for $x,y,z$ and adding gives the first identity (noting $x+y+z=0$). Just a trick which helps sometimes - though the identities work best for a computer if you are programming. – Mark Bennet Mar 29 '12 at 22:36
@MarkBennet: Right! This is what I was referring to in the last link. – Aryabhata Mar 29 '12 at 22:38
please forgive my laziness ... – Mark Bennet Mar 29 '12 at 22:54
@MarkBennet: You don't have to apologize! That sentence isn't clear on what the different way is. Your comment helps clarify it a bit. In fact, I was the one being lazy... – Aryabhata Mar 29 '12 at 22:57

Note that $0 = (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+xz+yz)$, so $x^2+y^2+z^2 = -2(xy+xz+yz)$.

And \begin{align*} 15 &= x^4+y^4+z^4\\ & = (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2 + y^2z^2).\end{align*}

And \begin{align*} (xy+xz+yz)^2 &= (x^2y^2 + x^2z^2 + y^2z^2) + 2(x^2yz+xy^2z+xyz^2)\\ &= (x^2y^2+x^2z^2 + y^2z^2) + 2xyz(x+y+z)\\ &= x^2y^2+x^2z^2+y^2z^2 + 2xyz(0)\\ &= x^2y^2+x^2z^2+y^2z^2 \end{align*} hence $$x^2y^2+x^2z^2+y^2z^2 = (xy+xz+yz)^2.$$ So \begin{align*} 15 &= x^4+y^4+z^4\\ &= (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2+y^2z^2)\\ &= (x^2+y^2+z^2)^2 - 2(xy+xz+yz)^2\\ &= (x^2+y^2+z^2)^2 -2(xy+xz+yz)(xy+xz+yz)\\ &= (x^2+y^2+z^2)^2 +(x^2+y^2+z^2)(xy+xz+yz)\\ &= (x^2+y^2+z^2)^2 + (x^2+y^2+z^2)\left(-\frac{1}{2}(x^2+y^2+z^2)\right)\\ &= (x^2+y^2+z^2)^2 -\frac{1}{2}(x^2+y^2+z^2)^2\\ &=\frac{1}{2}(x^2+y^2+z^2)^2 \end{align*} so $(x^2+y^2+z^2)^2 = 30$, hence $x^2+y^2+z^2=\sqrt{30}$.

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This is just a pedestrian way of doing what Aryabhata notes. – Arturo Magidin Mar 29 '12 at 22:26
I'll +1 this answer for just the last comment. (You may know why, by now.) Amazing clarity as ever.+1 :-) – user21436 Mar 29 '12 at 22:31