Solving More Equations

The possible values of $k$ are $0$ , $1$ , $2$ , and $3$ , which are obvious. If $x=y=z=0$ , then $k=0$ , If one of $x$ , $y$ , and $z$ is $1$ , and the other two are $0$ , then $k=1$ . If two of $x$ , $y$ , and $z$ is $1$ , and the remaining one is $0$ , then $k=2$ . If $x=y=z=1$ , then $k=3$ . There is no other solution because $x \ne x^2 \ne x^3 \ne x^4$ except for $x=0$ or $x=1$ . Therefore the sum of all possible values of $k$ is $0+1+2+3 = \boxed 6$ .

We can however solve for $k$ using Newton's sums or Newton's identities as follows. Let $P_n = x^n+y^n+z^n$ , where $n$ is a positive integer. Then we have $P_1=P_2=P_3=P_4=k$ . Let also $S_1 = x+y+z$ , $S_2 = xy + yz + zx$ , and $S_3 = xyz$ . Then we have:

$\begin{aligned} P_1 & = S_1 = x+y+z = k \\ P_2 & = S_1P_1 - 2S_2 = k^2 - 2S_2 = k & \small \color{#3D99F6} \implies S_2 = \frac {k(k-1)}2 \\ P_3 & = S_1P_2 - S_2P1 + 3S_3 = k^2 - \frac {k^2(k-1)}2 + 3S_3 = k & \small \color{#3D99F6} \implies S_3 = \frac {k(k-1)(k-2)}6 \\ P_4 & = S_1P_3 - S_2P2 + S_3P_1 = k^2 - \frac {k^2(k-1)}2 + \frac {k^2(k-1)(k-2)}6 = k \end{aligned}$

From the equation of $P_4$ :

$\begin{aligned} k^2 - k - \frac {k^2(k-1)}2 + \frac {k^2(k-1)(k-2)}6 & = 0 \\ 6k(k-1) - 3k^2(k-1) + k^2(k-1)(k-2) & = 0 \\ k(k-1)(6 - 3k + k(k-2)) & = 0 \\ k(k-1)(k-2)(k-3) & = 0 \\ \implies k & = 0, 1, 2, 3 \end{aligned}$