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.
 A: 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 $n\ge2$.
A: Because $1+2i$ has integer real and imaginary parts, $\text{Re}((1+2i)^n)$ is always an integer.  It's also easy to see that it is always odd.  So the real question is, can it be $\pm 1$?  
If $\theta = \arctan(2)$, $\text{Re}((1+2i)^n) = 5^{n/2} \cos(n\theta)$.  For this to be $\pm 1$ requires $|n \theta - k \pi/2| < 2\cdot 5^{-n/2}$ for some odd integer $k$.  That means that 
$\pi/\theta$, which we know is an irrational number, is very closely
approximated by a rational.  That's very unlikely to be the case, but I don't know of any way to prove it.  There are lower bounds on how closely $\pi$ can be approximated by rationals, but I'm not aware of anything similar for $\pi/\theta$.
A: $a_{2n} = Re((1+2i)^{2n}) = Re((-3+4i)^n) = Re((1+4(-1+i))^n)$
Applying the Binomial expansion we get $(1+4(-1+i))^n$ as a series that converges $2$-adically :
$(1+4(-1+i))^n = 1 + 4(-1+i)\binom n1 -32i \binom n2 + \ldots + 4^k(-1+i)^k\binom nk + \ldots$
Since $v_2(4^k/k!) \ge k \to \infty$, this gives a $2$-adic power series
$(1+4(-1+i))^n = 1 + a_1 n + a_2 n^2 + \ldots$ where $a_k \in 2^k \Bbb Z_{(2)}[i]$
Taking the real part we get $a_{2n} = 1 + b_1 n + b_2 n^2 + \dots$
Computing the first few coefficients mod $8$, we get $b_1 = 4 \pmod 8$ and $b_n = 0 \pmod 8$ for $n > 1$, which shows that the map $n \mapsto a_{2n}$, when extended continuously to $\Bbb Z_{(2)}$, is a bijection from $\Bbb Z_{(2)}$ to $1+4\Bbb Z_{(2)}$. So it takes the value $1$ exactly once, at $n=0$.
For $a_{2n+1}$ the same thing works : $(1+2i)^{2n+1} = (1+2i)^{2n}(1+2i)$, so we just multiply each $a_k$ by $(1+2i)$ before taking real parts. Again we end up on $a_{2n+1} = 1 + 4n \pmod 8$, so it is again a bijection from $\Bbb Z_{(2)}$ to $1+4\Bbb Z_{(2)}$, it takes the value $1$ exactly once, at $n=0$.
You can be a bit more precise and show that $v_2(a_{2n} - a_{2m}) = v_2(a_{2n+1} - a_{2m+1}) = v_2(4(n-m)) = 2+v_2(n-m)$
This also implies that $(a_n)$ can only take a particular value at most twice, and so $|a_n| \to \infty$.
A: Hint:
All parts are integer.
Let $|\Re(1+2i)^n|=1=5^{n/2}|\cos(n\alpha)|$, with $\alpha=\arctan(2)$.
Then we must have $|\Im(1+2i)^n|=5^{n/2}|\sin(n\alpha)|=5^{n/2}\sqrt{1-5^{-n}}=\sqrt{5^n-1}\in\mathbb N$, i.e. $5^n-1$ is a perfect square.
