Find $U+V$ given $U\alpha^{-5} + V\beta^{-5} = 22$, when $\alpha, \beta$ are roots of $x^2-x-1$ and $U,V$ are integers.
For simplicity I only show the U term: $$U+V = U\alpha^{-5}\alpha^{5}+ ... \\ = U \alpha^{-5}(\alpha^4 \alpha)\\ = U \alpha^{-5}((\alpha+1)^2 \alpha)+ ...\\ =U \alpha^{-5}((\alpha +1) \alpha + 2(\alpha+1) + \alpha)+ ... \\ = U \alpha^{-5}(\alpha +1+\alpha + 2 \alpha + 2 + \alpha)+ ...\\ = U \alpha^{-5 }(5\alpha + 3) + ... \\ = 5 U \alpha^{-4}+5V\beta^{-4} + 3 \cdot 22 $$
Now How to get rid of $\alpha^{-4}, \beta^{-4}$? Please help me, do give hints only.
Thanks a lot!
Start with $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$ since these satisfy $x^2 = x+1$. Clearly, both $\alpha$ and $\beta$ are non-zero.
Now, $\alpha^{-1} = \alpha - 1$. Square both sides : $\alpha^{-2} = \alpha^2 - 2\alpha + 1 = 2 - \alpha$. Squaring again, $\alpha^{-4} = (2-\alpha)^2 =$ $ 4 - 4 \alpha + \alpha^2 = 5 - 3 \alpha$. Another $\alpha$ multiplication gives $\alpha^{-5} = 5\alpha^{-1} - 3 = 5\alpha - 8$. Can you see the Fibonacci sequence here?
The exact same applies for $\beta$.
Therefore, $U\alpha^{-5} + V \beta^{-5} = U(5 \alpha - 8) + V(5 \beta- 8) = 22$. Hence, $5\alpha U + 5\beta V = 22 + 8(U+V)$.
Hence, $5(1 + \sqrt 5)U + 5(1 - \sqrt 5)V = 44 + 16(U+V)$, rewrite to $5\sqrt{5}(U-V) = 44 + 11(U+V)$. Hence, we get that $U-V = 0$ otherwise the LHS will be irrational and hence $44 = -11(U+V)$ so $U+V = -4$. Of course, we get $U= V = -2$ from here.
Key idea : if $\alpha$ is the root of a quadratic equation, higher powers of $\alpha$ can be expressed as $c \alpha + d$ for some $c,d$.