An algebra problem by ابراهيم فقرا

If $(P_i=x^i+y^i+z^i)$

According to Newton's identities,

$P_4=\frac{4P_3P_1}{3}-P_2{P_1}^2+\frac{{P_1}^4}{6}+\frac{{P_2}^2}{2}$

If $P_1=P_2=P_3=P_4=k$ , then we get the equation : $k^4-6k^3+11k^2-6k=0$

The solutions of this equation are $0,1,2,3$ , so the answer is $6$

Relevant wiki: Newton's Identities

Let $P_n = x^n + y^n+z^n$ , where $n$ is a positive integer, $S_1=x+y+z=k$ , $S_2 = xy + yz+zx$ , and $S_3 = xyz$ . Given are $P_1=P_2=P_3=P_4=k$ . By Newton's sums or Newton's identities, we have:

$\begin{aligned} P_1 & = S_1 = k \\ P_2 & = S_1P_1 - 2S_2 = k^2 - 2S_2 = k & \small \color{#3D99F6} \implies S_2 = \frac {k^2-k}2 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = k^2 - \frac {k^3-k^2}2 + 3S_3 = k & \small \color{#3D99F6} \implies S_3 = \frac {k^3 - 3k^2+2k}6 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_3 = k^2 - \frac {k^3-k^2}2 + \frac {k^4-3k^3+2k^2}6 = k \end{aligned}$

$\begin{aligned} \implies k^2 - \frac {k^3-k^2}2 + \frac {k^4-3k^3+2k^2}6 - k & = 0 & \small \color{#3D99F6} \text{Multiply both sides by }6 \\ 6k^2 - 3k^3 + 3k^2 + k^4 - 3k^3+2k^2 - 6 k & = 0 \\ k^4 - 6k^3 + 11k^2 - 6k & = 0 \\ k(k^3-6k^2+11k-6) & = 0 \\ k(k-1)(k-2)(k-3) & = 0 \end{aligned}$

Therefore, the sum of possible values of $k$ is $0+1+2+3 = \boxed 6$ .