I am trying this problem:
Find the maximum common divisor (in $ \mathbb{Z} $) of the numbers $ a, \ b $ such that $ (1 + \sqrt{5})^{2020} = a + b \sqrt{5} $.
Taking norms (the context is the ring $ \mathbb{Z}[\sqrt{5}] $), we have: $$ N(1 + \sqrt{5}) = 1^2 - 5 \cdot 1^2 = -4 $$ and then, if $ d = mcd(a, b) \in \mathbb{Z} $, $$ N(d) = d^2 \mid (-4)^{2020} = 2^{4040} $$ So, $ d $ must be $ 1 $ or a power of two. But I am stuck now. I have tried more things unsuccessfully. By inspection: $$ (1 + \sqrt{5})^2 = 6 + 2 \sqrt{5} \longrightarrow d = 2 $$ $$ (1 + \sqrt{5})^3 = 16 + 8 \sqrt{5} \longrightarrow d = 2^3 $$ $$ (1 + \sqrt{5})^4 = 56 + 24 \sqrt{5} \longrightarrow d = 2^3 $$ $$ (1 + \sqrt{5})^5 = 176 + 80 \sqrt{5} \longrightarrow d = 2^4 $$ $$ (1 + \sqrt{5})^6 = 576 + 256 \sqrt{5} \longrightarrow d = 2^6 $$ I can't clear see a pattern here.
$$\frac{\sqrt5+1}{2}=\varphi$$ $$(1+\sqrt5)^{2020}=2^{2020}\varphi^{2020}$$ As is well known, $\varphi^n=f_n\varphi+f_{n-1}$, where $f_n$ is the Fibbonachi sequence. $$2^{2020}\varphi^{2020}=2^{2020}(f_{2020}\varphi+f_{2019})=2^{2020}\frac{f_{2020}\sqrt5+f_{2020}+2f_{2019}}{2}$$ $$\gcd(a,b)=2^{2019}gcd(f_{2020},f_{2020}+2f_{2019})$$ We note the Fibbonachi sequence is 3-periodic modolu 2 where $f_1\equiv1$. thus, $f_{2020}$ is odd and since the $\gcd$ is a power of $2$,
$$gcd(a,b)=2^{2019}$$