1. Given relation

We have $P^2=I-P.$ So every higher power of $P$ can be reduced using this relation.

2. Find the pattern of powers of $P$

Using $P^2=I-P$:

$P^3=P\cdot P^2=P(I-P)=P-P^2=P-(I-P)=2P-I.$

$P^4=P\cdot P^3=P(2P-I)=2P^2-P=2(I-P)-P=2-3P,$ so $P^4=2I-3P.$

Continue a little:

$P^5=P(2I-3P)=2P-3P^2=2P-3(I-P)=5P-3I.$

This suggests a Fibonacci-type pattern. Let $P^n=a_n I+b_n P.$ Using the recurrence from $P^{n+2}=P^n-P^{n+1},$ we get coefficients following the same recurrence.

From direct computation:

$P^1=P,$ $P^2=I-P,$ $P^3=-I+2P,$ $P^4=2I-3P,$ $P^5=-3I+5P,$ $P^6=5I-8P,$ $P^7=-8I+13P,$ $P^8=13I-21P.$

Also, $P^6+P^7=(5I-8P)+(-8I+13P)=-3I+5P=P^5,$ so the recurrence is consistent.

3. Use the given equations

We are given $P^{\alpha}+P^{\beta}=\gamma I-29P \quad ...(1)$ and $P^{\alpha}-P^{\beta}=\delta I-13P \quad ...(2)$

Add and subtract these equations.

Adding:

$2P^{\alpha}=(\gamma+\delta)I-42P$ so $P^{\alpha}=\frac{\gamma+\delta}{2}I-21P.$

Thus we need a power of $P$ whose coefficient of $P$ is $-21$.

From the list above, $P^8=13I-21P.$ Hence $\alpha=8$ and $\frac{\gamma+\delta}{2}=13 \Rightarrow \gamma+\delta=26.$

Subtracting:

$2P^{\beta}=(\gamma-\delta)I-16P$ so $P^{\beta}=\frac{\gamma-\delta}{2}I-8P.$

Thus we need a power of $P$ whose coefficient of $P$ is $-8$.

From the list, $P^6=5I-8P.$ Hence $\beta=6$ and $\frac{\gamma-\delta}{2}=5 \Rightarrow \gamma-\delta=10.$

4. Solve for $\gamma$ and $\delta$

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