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Consider the polynomial: $$f=X^4+4X^3+6X^2+aX+b$$ We know that $f$ has four real roots. Let $x_1,x_2,x_3,x_4$ be the roots of this polynomial. How can one compute $$x_1^{2015}+x_2^{2015}+x_3^{2015}+x_4^{2015}?$$ If $a=4$ and $b=1$, we obtain a self-reciprocal (palindromic) polynomial. We can write $f=(X+1)^4$, thus $x_1=x_2=x_3=x_4=1$. Hence the sum computes to $-4$. Are there any other cases to consider ($a,b$)? I thought using the formula for the quartic equation and paying attention to the cases where we have only real roots. Any ideas? Thank you!

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  • $\begingroup$ Note that in case $a=4, b=1$ you don't obtain $f=(X-1)^4$ but rather $f=(X+1)^4$ which yields $x_1=x_2=x_3=x_4=-1$ and hence the sum computes to $-4$. $\endgroup$ – Tintarn Aug 26 '15 at 18:24
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We know the following to be true:

$$ \tag1x_1+x_2+x_3+x_4 = -4 $$ $$ \tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6 $$ $$ \tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a $$ $$ \tag4x_1x_2x_3x_4 = b $$

Now we can square $(1)$ to get

$$ \begin{align} (x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\ &+2[x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1+x_3+x_2x_4] \end{align} $$

Substituting in values from $(1)$ gives

$$ \begin{align} (-4)^2 &= x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+2(6)\\ 16&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+12\\ 4&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4} \end{align} $$

Using the Cauchy-Schwartz Inequality, we get

$$ \left(x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}\right)(1^2+1^2+1^2+1^2)\geq (x_{1}+x_{2}+x_{3}+x_{4})^2 $$

where equality holds when

$$ \tag5x_{1} = x_{2} = x_{3} = x_{4} $$

Thus, the equality condition holds, because $4\cdot 4= (-4)^2$. So from $(1)$ and $(5)$, we have $$x_{1} = x_{2} = x_{3} = x_{4}=-1$$

Setting the values of $x_{1} = x_{2} = x_{3} = x_{4}=-1$ in $(3)$ and $(4)$, we get $(a,b)=(4,1)$, which is thus the unique solution. So we can evaluate the sum as

$$ \begin{align} x^{2015}_{1}+x^{2015}_{2}+x^{2015}_{3}+x^{2015}_{4} &= (-1)+(-1)+(-1)+(-1)\\ &=-4 \end{align} $$

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Note that $x_1+x_2+x_3+x_4=-4$ since $f=(X-x_1)(X-x_2)(X-x_3)(X-x_4)$ and we can just compare the coefficient of $X^3$. We can use this idea to get larger powers. Similarly we can get $x_1^2+x_2^2+x_3^2+x_4^2=16-(2\times 6)=4$ and use a similar trick for cubes.

Define $Tr(x_1^i)=x_1^i+x_2^i+x_3^i+x_4^i$. This function has a deeper meaning, but all we shall use from it is that is additive, ie $Tr(a+b)=Tr(a)+Tr(b)$, which you are welcome to prove if you wish.

Now for higher powers we note that $x_1^4=-4x_1^3-6x_1^2-ax_1-b$ hence using this identity we can write $x_1^{2015}$ in the form $Ax_1^3+Bx_1^2+Cx_1+D$ and use the additive property of the trace function $Tr$ I defined above and the values we worked out for these to get an exact answer.

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  • $\begingroup$ Meaning of the $Tr$ function is not clear to me. Is it $Tr(x) = x + a + b + c$, where $x$ is an argument and $a,b,c$ are parameters? $\endgroup$ – Kaster Aug 26 '15 at 18:41
  • $\begingroup$ If you plug a root $\alpha$ of $f$ in then $Tr(\alpha^i)$ gives you the sum of the roots to the same power and the additive property allows you to extend this to a full definition. Technically it is the sum of all the galois conjugates of an element over the field extension if you understand that. $\endgroup$ – Matt B Aug 26 '15 at 18:44
  • $\begingroup$ Is it a function of the root, or root in the power of $i$, or all roots? $\endgroup$ – Kaster Aug 26 '15 at 18:46
  • $\begingroup$ When we adjoin all the roots of the polynomial to our field $\mathbb{Q}$, we get a new field $K$, and then $Tr:K \rightarrow \mathbb{Q}$ as a function. $\endgroup$ – Matt B Aug 26 '15 at 18:48
  • $\begingroup$ So in particular, it is a function on any $\mathbb{Q}$-linear combination of the roots and their powers. $\endgroup$ – Matt B Aug 26 '15 at 18:49

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