<s>[INST] 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.
 [/INST] $a_{2n} = Re((1+2i)^{2n}) = Re((-3+4i)^n) = Re((1+4(-1+i))^n)$
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 $|a_n| \to \infty$.
</s>