We are given $\alpha+\beta=1, \qquad \alpha\beta=-1.$

Also, $p_n=\alpha^n+\beta^n.$

Since $\alpha,\beta$ are roots of $x^2-(\alpha+\beta)x+\alpha\beta=0,$ they satisfy $x^2-x-1=0.$

Hence each of $\alpha,\beta$ satisfies $r^2=r+1.$

Therefore the sequence $p_n=\alpha^n+\beta^n$ satisfies the recurrence $p_{n+2}=p_{n+1}+p_n.$

This is because $\alpha^{n+2}=\alpha^{n+1}+\alpha^n, \qquad \beta^{n+2}=\beta^{n+1}+\beta^n,$ and adding gives the result.

The question states $p_{n-1}=11 \quad \text{and} \quad p_{n+1}=29.$

Using the recurrence, $p_{n+1}=p_n+p_{n-1}.$

So, $29=p_n+11,$ which gives $p_n=18.$

Therefore, $p_n^2=18^2=324.$

Comparison with stored answer: Stored correct answer = $324$.

Our derived answer matches it.