neka je
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, te
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postoji li
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, takav da se iz jednakosti
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moze zakljuciti da su "
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-evi" do na permutaciju jednaki "
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-ima". ako postoji, koji je najmanji takav
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.
dodatno, postoji li
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takav da se za proizvoljnih
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brojeva
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, takvih da
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vrijedi
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te odgovarajucih jednakosti
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moze zakljuciti da su "
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-evi" do na permutaciju jednaki "
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-ima". ako postoji, koji je najmanji takav
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.
takoder, uz pretpostavku
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, smijemo li uzeti
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?
%V0
neka je $n, n \in \mathbb{N}$, te $x_1, x_2, \dots , x_n, y_1, y_2, \dots, y_n \in \mathbb{C}$
postoji li $k, k \in \mathbb{N}$, takav da se iz jednakosti $$x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n$$ $$x_1^2 + x_2^2 + \dots + x_n^2 = y_1^2 + y_2^2 + \dots + y_n^2$$ $$ \vdots $$ $$x_1^k + x_2^k + \dots + x_n^k = y_1^k + y_2^k + \dots + y_n^k$$
moze zakljuciti da su "$x$-evi" do na permutaciju jednaki "$y$-ima". ako postoji, koji je najmanji takav $k$.
dodatno, postoji li $l, l \in \mathbb{N}$ takav da se za proizvoljnih $l$ brojeva $a_1, a_2, \dots , a_l \in \mathbb{Z} \setminus \{0\}$, takvih da $\forall i \neq j$ vrijedi $M(a_i, a_j) = 1$ te odgovarajucih jednakosti
$$x_1^{a_1} + x_2^{a_1} + \dots + x_n^{a_1} = y_1^{a_1} + y_2^{a_1} + \dots + y_n^{a_1}$$ $$x_1^{a_2} + x_2^{a_2} + \dots + x_n^{a_2} = y_1^{a_2} + y_2^{a_2} + \dots + y_n^{a_2}$$ $$ \vdots $$ $$x_1^{a_l} + x_2^{a_l} + \dots + x_n^{a_l} = y_1^{a_l} + y_2^{a_l} + \dots + y_n^{a_l}$$
moze zakljuciti da su "$x$-evi" do na permutaciju jednaki "$y$-ima". ako postoji, koji je najmanji takav $l$.
takoder, uz pretpostavku $x_1, x_2, \dots , x_n, y_1, y_2, \dots, y_n \in \mathbb{R}^+$, smijemo li uzeti $a_1, a_2, \dots , a_l \in \mathbb{R} \setminus \{0\}$?