TITLE: Real part of $(1+2i)^n$
QUESTION [10 upvotes]: 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.

REPLY [4 votes]: 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.