I came across the following problem, which turned out to be surprisingly hard:

Show that $\underset{n\rightarrow \infty}{\lim} \left| \mathrm{Re}((\frac{1+i\sqrt{7}}{2})^n)\right| = \infty.$

Intuitively, setting $z=\frac{1+i\sqrt{7}}{2}=\sqrt{2}e^{i\theta},$ we see that $|\mathrm{Re}(z^n)|=2^{\frac{n}{2}}|\cos (n\theta)|,$ so that as long as $n\theta \ (\mathrm{mod}\ \pi)$ does not approach $\frac{\pi}{2}$ exponentially fast on subsequences, $|\mathrm{Re}(z^n)|$ should go to $\infty.$ But how to prove that ?

Here is a partial argument:

Set $z^n=\frac{a_n + i b_n\sqrt{7}}{2}$ where $a_n, b_n \in \mathbb{Z}.$ Then, taking square moduli, $a_n,b_n$ and $d=n+2$ are integer solutions of the diophantine equation:

$$2^d=a^2+7b^2.$$

Then the abc conjecture implies that this equation has finitely many solutions with bounded $a.$

I would like to see an elementary solution to this problem.

I came across the following problem, which turned out to be surprisingly hard:

Show that $\underset{n\rightarrow \infty}{\lim} \left| \mathrm{Re}((\frac{1+i\sqrt{7}}{2})^n)\right| = \infty.$

Intuitively, setting $z=\frac{1+i\sqrt{7}}{2}=\sqrt{2}e^{i\theta},$ we see that $|\mathrm{Re}(z^n)|=2^{\frac{n}{2}}|\cos (n\theta)|,$ so that as long as $n\theta \ (\mathrm{mod}\ \pi)$ does not approach $\frac{\pi}{2}$ exponentially fast on subsequences, $|\mathrm{Re}(z^n)|$ should go to $\infty.$ But how to prove that ?

Here is a partial argument:

Set $z^n=\frac{a_n + i b_n\sqrt{7}}{2^n}$ where $a_n, b_n \in \mathbb{Z}.$ Then, taking square moduli, $a_n,b_n$ and $d=n+2$ are integer solutions of the diophantine equation:

$$2^d=a^2+7b^2.$$

Then the abc conjecture implies that this equation has finitely many solutions with bounded $a.$

I would like to see an elementary solution to this problem.