Real part of $(1+2i)^n$ Is it true that for all $n\in \mathbb{N}$, $n\ge 2$ we have
$$|\textrm{Re}((1+2i)^n)|>1?$$
I do know  de Moivre's Theorem.
I do not know how to show that $|\sqrt{5}^n\cos(n\arccos\left ( \frac{1}{5} \right ))|>1$ because the value $\cos(n\arccos\left ( \frac{1}{5} \right ))$ can become (theoretically) arbitrarily small.

Recursion for $\boldsymbol{r_n=\mathrm{Re}\!\left((1+2i)^n\right)}$
Both $1+2i$ and $1-2i$ satisfy the equation $z^2-2z+5=0$. Therefore, the real part of $(1+2i)^n$
$$
r_n=\frac{(1+2i)^n+(1-2i)^n}2
$$
satisfies
$$
r_n=2r_{n-1}-5r_{n-2}
$$

mod $\boldsymbol{4}$
Since $r_0=r_1=1$, we have that $r_n\equiv1\pmod{4}$ for all $n$.
Therefore, if $\left|r_n\right|\le1$, we must have $r_n=1$.

mod $\boldsymbol{25}$
Compute the first $24$ terms:
$$
\small r_n=1,1,\underbrace{22,14,18,16,17,4,23,\overset{\normalsize r_9}{1},12,19,3,11,7,9,8,21,2,24,13,6}_{\text{period of $20$}},22,14,\dots\pmod{25}
$$
This means that if $n\ge2$ and $r_n=1$, we must have $n\equiv9\pmod{20}$.

mod $\boldsymbol{11}$
Since $(1+2i)^3=-11-2i\equiv9i\pmod{11}$, Little Fermat says
$$
(1+2i)^{60}\equiv(9i)^{20}=81^{10}\equiv1\pmod{11}
$$
Therefore, $r_n\pmod{11}$ repeats every $60$ terms. However,
$$
\begin{align}
r_9&\equiv0\pmod{11}\\
r_{29}&\equiv2\pmod{11}\\
r_{49}&\equiv9\pmod{11}
\end{align}
$$
Thus, if $n\equiv9\pmod{20}$, $r_n\not\equiv1\pmod{11}$.

Conclusion
$\left|r_n\right|\gt1$ for $n\ge2$.
