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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.

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  • $\begingroup$ I added a power of $n$ in the denominator of $z^n$. I presume this is what you intended. $\endgroup$
    – Anthony Quas
    Commented Jun 17, 2019 at 16:05
  • $\begingroup$ The command of Mathematica DiscreteMaxLimit[RealAbs[ComplexExpand[Re[((1 + I*Sqrt[7])/2)^n]]], n -> Infinity] outputs $\infty$. $\endgroup$
    – user64494
    Commented Jun 17, 2019 at 16:59
  • $\begingroup$ @user64494 I would not trust Mathematica on this particular kind of problem. Also, the output gives no hint of any kind of proof of the result. $\endgroup$
    – Somos
    Commented Jun 17, 2019 at 18:35
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    $\begingroup$ These type of problems are easy consequences of the finiteness of solutions (in number fields) to the $S$-unit equation, but I don't expect more elementary arguments (for the general version of the problem). You can also use Baker's theorem to get effective bounds. $\endgroup$
    – Pound Sterling
    Commented Jun 17, 2019 at 23:20
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    $\begingroup$ Looks somewhat similar to mathoverflow.net/questions/273112/… $\endgroup$
    – fedja
    Commented Jun 18, 2019 at 3:11

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