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Given the following system of polynomial equations: $$ \left\{\begin{array}{lclclcr} x & + & y & + & z & = & 1 \\ x^{2} & + & y^{2} & + & z^{2} & = & 14 \\ x^{3} & + & y^{3} & + & z^{3} & = & 36 \end{array}\right. $$ What is $x^{5} + y^{5} + z^{5}\ {\large ?}$ .


How should I approach this? Is there a general formula for this kind of system?

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    $\begingroup$ Try using this approach by blackpenredpen. $\endgroup$ – Andrew Chin Aug 10 at 21:38
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Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. So we have $$\begin{cases} p=1\\p^2-2q=14\\p^3-3pq+3r=36 \end{cases}$$ $$\begin{cases} p = 1\\q = -\frac{13}{2}\\ r = \frac{31}{6} \end{cases}$$ By consecutive eliminating the highest powers terms we get $$x^5+y^5+z^5-(x+y+z)^5+5(xy+yz+xz)(x+y+z)^3-5(xy+yz+xz)^2(x+y+z)-5(x+y+z)^2xyz+5(xy+yz+xz)xyz=0$$ In other words, $x^5+y^5+z^5=p^5-5qp^3+5q^2p+5p^2r-5qr$ $=\frac{877}{2}$

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The requested exponent does not matter that much. Each of the three is a root of the same $$ 6 t^3 - 6 t^2 - 39 t - 31, $$ one real and two complex conjugates, but we don't need them. Each also obeys $$ 6 t^{n+3} - 6 t^{n+2} - 39 t^{n+1} - 31 t^n, $$ so that $a_n=x^n + y^n + z^n$ is a sequence with linear recurrence $$ a_{n+3} = a_{n+2} + \frac{13}{2} a_{n+1} + \frac{31}{6} a_n $$ or $$ a_1 = 1, \; a_2 = 14, \; a_3 = 36, \; a_4 = \frac{793}{6}, \; $$ $$ a_5 = \frac{877}{2}, \; a_6 = \frac{17803}{12}, \; a_7 = \frac{180601}{36}, \; a_8 = \frac{1218641}{72}, \; $$

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    $\begingroup$ This should be the accepted answer. $\endgroup$ – Alexey Burdin Aug 10 at 23:18
  • $\begingroup$ @AlexeyBurdin Thanks. For a while there this type of question was quite frequent on this site, so I had a memory that this sort of thing ought to work; then I answered one within the past few months and the details turned out to be easy enough. $\endgroup$ – Will Jagy Aug 10 at 23:23
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    $\begingroup$ found the previous time I did this, that OP was not enthusiastic math.stackexchange.com/questions/3749238/… $\endgroup$ – Will Jagy Aug 10 at 23:35
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    $\begingroup$ How do you find that they are all zeros of that polynomial? $\endgroup$ – vonbrand Aug 11 at 2:04
  • $\begingroup$ @WillJagy What do you mean by “Each of the three is a root of the same [polynomial]”? The exponents I gave (1, 2, and 3) aren’t roots of that polynomial. $\endgroup$ – paper man Aug 12 at 5:32

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