Let \( a,b,c \) be complex numbers, and \( S_n = a^n+b^n+c^n\) be the sum of their \(n\)-th powers.

If \[S_1=1, \; S_1=1, \; S_2=3, \; S_3=1 \] Show that \[ S_5 + S_{11}+S_{21}=S_8. \]

Let \( a,b,c \) be complex numbers, and \( S_n = a^n+b^n+c^n\) be the sum of their \(n\)-th powers.

If \[S_1=1, \; S_1=1, \; S_2=3, \; S_3=1 \] Show that \[ S_5 + S_{11}+S_{21}=S_8. \]

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TopNewestFrom Newton Sums, one can find that \(a+b+c=1, ab+ac+bc=-1, abc=-1 \). The polynomial \(p(x)\) of roots \(a,b,c\) can thus be written as \( p(x)=x^3-x^2-x+1 \). This leads to an adequate way to show that \( 1 \) is a double root and \( -1 \) is a simple root, meaning \( a=b=1, \; c=-1\), for instance.

It is easy to see that our desired equation yields, since \[ (1^5 + 1^5 + (-1)^5) + (1^{11} + 1^{11} + (-1)^{11}) + (1^{21} + 1^{21} + (-1)^{21}) = 1^8 + 1^8 + (-1)^8 \] \[1+1+1=3\] – Guilherme Dela Corte · 2 years, 7 months ago

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