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Consider a cubic polynomial $x^3 - x - 1 = 0$ I want the sum of the fifth powers of the roots $\sum a^5$. I know that

\begin{eqnarray*} a + b + c &=& 0 \\ ab + bc + ca &=& 1 \\ -abc &=& 1 \end{eqnarray*}

but I have no way of combining this information into the answer.

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  • $\begingroup$ Please include more context. What do you know about general symmetric polynomials? $\endgroup$ – Greg Martin Nov 21 '16 at 1:15
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    $\begingroup$ I wonder if it is possible to find $a^n + b^n + c^n$ for any $n$ perhaps they follow a simple pattern. $\endgroup$ – cactus314 Nov 21 '16 at 1:17
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    $\begingroup$ Where did the problem come from? Homework in a particular course? What are you expected to try? It is true that $a^5+b^5+c^5$ is a polynomial function of $a+b+c$, $ab+bc+ca$, and $abc$, but I'm not sure how you're expected to find that polynomial. $\endgroup$ – Greg Martin Nov 21 '16 at 1:17
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    $\begingroup$ Mistake, $bc + ca + ab = -1.$ $\endgroup$ – Will Jagy Nov 21 '16 at 1:40
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    $\begingroup$ Also $abc = 1........$ $\endgroup$ – Will Jagy Nov 21 '16 at 1:53
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$a^3=a+1$ implies that $a^5=a^3+a^2=a+1+a^2$.

$a^5+b^5+c^5=3+a+b+c+a^2+b^2+c^2=3+(a+b+c)^2-2(ab+ac+bc)=5$.

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    $\begingroup$ I believe that $ab+ac+bc=-1$ (not $1$), so that the final result is $3+0^2-2(-1)=5$. $\endgroup$ – Théophile Nov 21 '16 at 2:36
  • $\begingroup$ How is $a^5=a^3+a^2$ ? $\endgroup$ – tatan Mar 21 '19 at 16:07
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About finding general $a^n + b^n + c^n,$ there is an elementary aspect to this, although not necessarily what you want. You already know that $a+b+c = 0,$ and the other answer gives enough to find $a^2 + b^2 + c^2$ and $a^3 + b^3 + c^3.$ Call those $x_1, x_2, x_3,$ then solve in perpetuity with $$ x_{n+3} = x_{n+1} + x_n $$

Two errors in the question, $$ bc + ca + ab = -1, $$ $$ abc = 1. $$

So far, I get $$ x_1 = 0,$$ $$ x_2 = 2,$$ $$ x_3 = 3,$$ $$ x_4 = 2,$$ $$ x_5 = 5,$$ $$ x_6 = 5,$$ $$ x_7 = 7,$$ $$ x_8 = 10,$$ $$ x_9 = 12,$$ $$ x_{10} = 17,$$

Since the real root of $x^3 - x - 1$ is slightly larger than $1,$ the numbers increase, are positive and so on. The two complex roots are smaller than $1$ in modulus, so, for large $n,$ we get $x_n \approx R^n$ where $R \approx 1.3247$ is the real root. For example, $R^{10} \approx 16.643$

compared with pari:

parisize = 4000000, primelimit = 500509
? p = polroots( x^3 - x - 1 )
%1 = [1.324717957244746025960908855 + 0.E-28*I, -0.6623589786223730129804544272 - 0.5622795120623012438991821449*I, -0.6623589786223730129804544272 + 0.5622795120623012438991821449*I]~
? p[1]
%2 = 1.324717957244746025960908855 + 0.E-28*I
? p[2]
%3 = -0.6623589786223730129804544272 - 0.5622795120623012438991821449*I
? a = p[2]
%4 = -0.6623589786223730129804544272 - 0.5622795120623012438991821449*I
? b = p[3]
%5 = -0.6623589786223730129804544272 + 0.5622795120623012438991821449*I
? c = p[1]
%6 = 1.324717957244746025960908855 + 0.E-28*I
? a + b + c
%7 = 0.E-28 + 0.E-28*I
? a^2 + b^2 + c^2
%8 = 2.000000000000000000000000000 + 0.E-28*I
? a^3 + b^3 + c^3
%9 = 3.000000000000000000000000000 + 0.E-28*I
? a^4 + b^4 + c^4
%10 = 2.000000000000000000000000000 + 0.E-28*I
? a^5 + b^5 + c^5
%11 = 5.000000000000000000000000000 + 0.E-27*I
? 
? a * b * c
%12 = 0.9999999999999999999999999999 + 0.E-28*I
? b * c + c * a + a * b
%13 = -1.000000000000000000000000000 + 0.E-28*I
? 
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  • $\begingroup$ Extraordinary sequence of integers becoming from three complicated numbers. $\endgroup$ – Piquito Nov 21 '16 at 1:45

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