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Let $x,y,z$ be complex numbers such that

$$x+y+z = x^{5}+y^{5}+z^{5} = 0, \hspace{10pt} x^3+y^3+z^3=2$$

Find all possible values of

$$x^{2007}+y^{2007}+z^{2007}$$

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I remember seeing this question and therefore did not see the missed $x^3+y^3+z^3=2$ and assumed it$ –  Kirthi Raman Mar 11 '12 at 19:45
    
I get $2^{669}/3^{668}$. –  GEdgar Mar 11 '12 at 20:25
    
@Kirthi: Where did you see this question? –  Did Mar 11 '12 at 20:33
    
I think it was one of questions several years back in usamts.org (competition for high school math) –  Kirthi Raman Mar 11 '12 at 22:40

2 Answers 2

It $x+y+z$ = 0, then $x,y,z$ are roots of $t^3 + at-b = 0$.

Then, we can show that, $x^5 + y^5 + z^5 = -5ab$ and $x^3 + y^3 + z^3 = 3b$, either using Newton's Identities or as in my answer here: http://math.stackexchange.com/a/115534/1102

Since $b \ne 0$, we must have that $a = 0$.

Thus $x,y,z$ are roots of $t^3 = b$. We know that $b = \frac{2}{3}$ and thus can compute the expression you need easily as $3b^{669} = \dfrac{2^{669}}{3^{668}}$.

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Since $x+y+z=0$,

$$x+y = -z\tag{A}.$$

If we raise to power $3$ both sides

$$ x^3+3x^{2}y+3y^{2}x+y^{3} = -z^{3} \quad \Rightarrow \quad x^{3}+y^{3}+z^{3} = -3xy(x+y).$$

Since $x^{3}+y^{3}+z^{3} =3$,

$$ x^{3}+y^{3}+z^{3} = -3xy(x+y).$$

We can conclude therefore that

$$ xy(x+y) = -1 \tag{B}.$$

If we take fifth power to $(A)$,

$$ x^5+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5} = -z^{5} $$

$$ x^{5}+y^{5}+z^{5} = -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}).$$

And since $x^{5}+y^{5}+z^{5}=0$

$$ -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}) = 0. $$

Since $xy \neq 0$,

$$ \begin{align*} x^{3}+2x^{2}y+2xy^{2}+y^{3} &= 0\\ x^{3}+y^{3}+2xy(x+y) &= 0. \end{align*} $$

From $(B)$,

$$ x^{3}+y^{3} = 2.$$ Also

$$x^{3}+y^{3}+z^{3} = 1 \quad \Rightarrow \quad z^{3} = 1.$$

By symmetry

$$x^{3}=y^{3}=z^{3} = 1.$$

Therefore

$$x^{2007}+y^{2007}+z^{2007} = 3.$$

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7  
Where did you get that $x^3 + y^3 + z^3 = 3$? –  TMM Mar 11 '12 at 19:33
    
Apparently the condition should be $x^3 + y^3 + z^3 = 2$, as in the edited question. So then $xy(x+y) = -1$ becomes $xy(x+y) = -2/3$, and you probably get a similar result as Aryabhata. –  TMM Mar 11 '12 at 21:04

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