Strict equal pairs

To solve this problem, we have to take the following three cases:

Case 1: $P = P^2$ only and $P^3 = P^4$ only

$P = P^2 \implies P^2 - P = 0 \implies P(P-1) = 0 \implies P \in \{0,1\}$

$P^3 = P^4 \implies P^4 - P^3 = 0 \implies P^3(P - 1) = 0 \implies P \in \{0,1\}$

For $P=0$

$P = P^2 = P^3 = P^4 = 0$

For $P=1$

$P = P^2 = P^3 = P^4 = 1$

Since all the terms become equal, these are not our solutions.

Case 2: $P = P^3$ only and $P^2 = P^4$ only

$P = P^3 \implies P^3 - P = 0 \implies P(P^2 -1) = 0 \implies P(P+1)(P-1) = 0 \implies P \in \{0,1,-1\}$

$P^2 = P^4 \implies P^4 - P^2 = 0 \implies P^2 (P^2 -1) = 0 \implies P^2 (P+1)(P-1) = 0 \implies P \in \{0,1,-1\}$

We've already seen above that for $P = 0,1$ , all the terms become equal and therefore both of those are not our solutions here as well.

Checking for $P=-1$

$P = P^3 = -1 \qquad \text{ and } \qquad P^2 = P^4 = 1$

So, $P=-1$ is one of our solutions.

Case 3: $P = P^4$ only and $P^2 = P^3$ only

$P = P^4 \implies P^4 - P = 0 \implies P(P^3 - 1) = 0 \implies P(P-1)(P^2+P+1) = 0 \implies P \in \{0,1,\omega,\omega^2\}$

where $\omega = \dfrac{-1 + \sqrt 3 i}{2}$ .

$P^2 = P^3 \implies P^3 - P^2 = 0 \implies P^2 (P-1) = 0 \implies P \in \{0,1\}$

Disregarding the two solutions $P=\omega,\omega^2$ as they are not the common solutions, we're left with $P=0,1$ . We've already seen above that these two values are not our solutions.

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Thus the only value of $P$ which satisfies our criteria is $P=-1$ .

Hence $P^5 = {(-1)}^5 = \boxed{-1}$ .