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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.

Tzara_T'hong
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    Did you try $n=2$? – Robert Israel Mar 14 '16 at 18:38
  • I corrected my question. – Tzara_T'hong Mar 14 '16 at 18:40
  • $1+2i=\sqrt{5}e^{i\arctan(2)}$, hence $$(1+2i)^n=\sqrt{5}^n\cdot{e^{in\arctan(2)}}=\sqrt{5}\cos(n\arctan(2))+\sqrt{5}i \sin(n\arctan(2))$$ Can you continue? – Galc127 Mar 14 '16 at 18:43
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    Maybe this is the wrong tack but by the binomial theorem wont the real part always be an integer. So suffices other options are the real part can't be 0 or +/- 1. Is that doable? – fleablood Mar 14 '16 at 18:54
  • The real part is always an integer $\equiv1\pmod4$, so, following fleablood's suggestion, it suffices to exclude the possibility that it is equal to $1$. – Jyrki Lahtonen Mar 14 '16 at 19:05
  • @JyrkiLahtonen I know this. – Tzara_T'hong Mar 14 '16 at 19:08
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    This question is related to the one at http://math.stackexchange.com/questions/1697058/on-equations-m21-5n – Barry Cipra Mar 14 '16 at 19:23
  • Furthermore, for the real part to be equal to $1$, it has to be congruent to $1$ w.r.t. any modulus $m$. With $m=5$ we get that $n$ has to be $\equiv1\pmod4$. With $m=13$ we get that $n$ has to be $\equiv0,1\pmod{12}$. Putting those two together leaves the possibility $n\equiv1\pmod{12}$. Doesn't look very useful :-( – Jyrki Lahtonen Mar 14 '16 at 19:31
  • I forgot the name of the method, but there are some $p$-adics methods that can sometimes solve this kind of problems. – mercio Mar 14 '16 at 19:32
  • Yet another useless equivalent formulation. Define the sequence $(G_n)$ by declaring $G_0=G_1=1$, $G_{n+2}=2G_{n+1}-5G_n$ for all $n\ge0$. Prove that $G_n\neq1$ for all $n>1$. – Jyrki Lahtonen Mar 14 '16 at 20:15
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    @JyrkiLahtonen: looking mod $4$, $11$, and $25$ does give a full result. – robjohn Mar 15 '16 at 12:36
  • @BarryCipra: thanks for pointing that out! I have added an answer there using my answer here. – robjohn Mar 16 '16 at 16:29

4 Answers4

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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 $n\ge2$.

robjohn
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    When I first read the question, I thought it said $\ge1$. I didn't get back for quite a while before I could revisit my answer. I modified it with a better partial answer. Now, the answer is complete. Perhaps the downvotes can be removed. – robjohn Mar 15 '16 at 11:11
  • Nice solution! If $|r_n|\le 1$ we must have $r_n=1$ or $r_n=-1$. And we have to check the case $r_n = 24 \pmod {25}$ too, but your method also works well for it. – Tzara_T'hong Mar 16 '16 at 17:59
  • @Andrey: True, but neither $r_n=-1$ nor $r_n=0$ are allowed because $r_n\equiv1\pmod4$. – robjohn Mar 16 '16 at 18:06
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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 $\pi/\theta$.

Robert Israel
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$a_{2n} = Re((1+2i)^{2n}) = Re((-3+4i)^n) = Re((1+4(-1+i))^n)$

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 $|a_n| \to \infty$.

mercio
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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.