1. Given data and basic relations

The roots $\alpha,\beta$ of $x^2-ax-b=0$ satisfy $\alpha+\beta=a, \qquad \alpha\beta=-b.$

Also, $P_n=\alpha^n-\beta^n.$

We are given: $P_3=-5\sqrt7\,i,\quad P_4=-3\sqrt7\,i,\quad P_5=11\sqrt7\,i,\quad P_6=45\sqrt7\,i.$

Let $P_n=\sqrt7\,i\,Q_n.$ Then $Q_3=-5,\quad Q_4=-3,\quad Q_5=11,\quad Q_6=45.$

2. Recurrence satisfied by $P_n$

Since $\alpha,\beta$ are roots of $x^2-ax-b=0$, each root satisfies $r^2=ar+b.$ Hence the sequence $\alpha^n$ and $\beta^n$ each satisfy $r^n=a r^{n-1}+b r^{n-2}.$ Subtracting, we get $P_n=aP_{n-1}+bP_{n-2}.$ So for $n=5,6$, $P_5=aP_4+bP_3,$ $P_6=aP_5+bP_4.$

Substitute the given values: $11\sqrt7 i=a(-3\sqrt7 i)+b(-5\sqrt7 i),$ $45\sqrt7 i=a(11\sqrt7 i)+b(-3\sqrt7 i).$

Cancel $\sqrt7 i$: $11=-3a-5b \quad ...(1)$ $45=11a-3b \quad ...(2)$

3. Solve for $a$ and $b$

From (1): $3a+5b=-11.$ From (2): $11a-3b=45.$

Solve:

Multiply the first by $11$: $33a+55b=-121.$ Multiply the second by $3$: $33a-9b=135.$ Subtract: $64b=-256 \Rightarrow b=-4.$ Then from $3a+5b=-11$: $3a-20=-11 \Rightarrow 3a=9 \Rightarrow a=3.$

Thus, $\alpha+\beta=3,\qquad \alpha\beta=4.$

So $\alpha,\beta$ are roots of $x^2-3x+4=0.$

4. Find $\alpha^4+\beta^4$

Let $S_n=\alpha^n+\beta^n.$ We need $|S_4|$.

Using standard identities: $S_1=\alpha+\beta=3,$ $S_2=(\alpha+\beta)^2-2\alpha\beta=3^2-2\cdot 4=9-8=1.$

Now, $S_4=(S_2)^2-2(\alpha\beta)^2=1^2-2\cdot 4^2=1-32=-31.$ Therefore, $|\alpha^4+\beta^4|=|-31|=31.$

5. Answer check

Derived answer: $31$.

This matches the stored correct answer.