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If suppose $N \neq A$ , then $P$ must satisfy both $P + A = PA$ and $P + N = PN$ . But since both equations share common "roots", namely $P$ , it is impossible to determine the solutions for $N \neq A$ .

So it remains to determine the solutions for $N = A$ . For the third equation, $\begin{array}{rl} AP + NP &= PNAP\\ AP + AP &= P^2A^2\\ 2AP &= P^2A^2\\ 0 &= P^2A^2 - 2AP\\ 0 &= PA(PA - 2) \end{array}$ Either $PA = 0$ or $PA = 2$ . The first case is trivial since the solutions are $P = N = A = 0$ and $PPAP = 0$ . Let's look at $PA = 2$ . Vieta's formula shows that if $P + A = PA = 2$ , where $P$ and $A$ are roots of the quadratic equation, then $\begin{array}{rl} x^2 - 2x + 2 &= 0\\ x^2 - 2x + 1 &= -2 + 1\\ (x - 1)^2 &= -1\\ x &= 1 \pm i \end{array}$ So $A$ and $N$ are both either $1 + i$ and $1 - i$ . Letting $P_1 = 1 + i$ and $P_2 = 1 - i$ and noting that $A_1P_1 = A_2P_2 = 2$ $\begin{array}{rl} A_1P_1 \cdot P_1P_1 + A_2P_2 \cdot P_2P_2 &= A_1P_1\left(P_1P_1 + P_2P_2\right)\\ &= 2\left((1 + i)^2 + (1 - i)^2\right)\\ &= 2\left(1 + 2i - 1 + 1 - 2i - 1\right)\\ &= 0 \end{array}$ Thus, the sum of all coordinates is $\boxed{0}$ .