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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 think it was one of questions several years back in usamts.org (competition for high school math)2012-03-11

2 Answers 2

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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: https://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 $3b^{669} = \dfrac{2^{669}}{3^{668}}$.

4

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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    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.2012-03-11