Prove that $\not \exists$ $n \in \mathbb N\setminus \{0\}$ such that $(2+i)^n=(2-i)^n$

Question

Prove that $\not \exists$ $n \in \mathbb N\setminus \{0\}$ such that $(2+i)^n=(2-i)^n$.

(IMO)

I found a proof considering that $2+i=2-i+2i$, then expanding the terms in the right of $(2+i)^n=[(2-i)+(2i)]^n$ using Binomial theorem. Then, assuming that there is an $n$ such that $(2+i)^n=(2-i)^n$ leads to an absurd: $2^{2n}=5(a+bi)$, $a,b\in \mathbb Z$.

I'm looking for proofs using other arguments, if possible.

Thanks and sorry if this is a duplicate.

Answer 1

Hint $${2+i\over 2-i} = {(2+i)^2\over 5} = {3\over 5}+{4\over 5}i = \cos \phi+ i\sin \phi$$

Where $\tan \phi = {4\over 3}$. So $$ \cos (n\phi)+i\sin (n\phi) =1$$ So $\sin (n\phi) =0$ and thus $\boxed{n\phi = \pi \cdot k}$ for some integer $k$ and $\cos (n\phi) =1$ and thus $\boxed{n\phi = 2\pi \cdot l}$

Answer 2

Consider Thomas Andrew's answer to this question. We can use the same arguments to prove that

For $p,q\in\mathbb Z$, if $\gcd(p,q)=1$, and if $\vert pq\vert>1$, then $(p+qi)^n$ cannot be real nor purely imaginary.

Imitating Proof:

First notice that no rational integer ($>1$) can divide $z:=p+qi$, otherwise $p$ and $q$ are not co-prime.

Then note that if a Gaussian prime $\mathfrak p$ divides $z$, then $\mathfrak p^n\mid z^n$. If $z^n$ is real or purely imaginary, then $\overline{\mathfrak p}^n\mid z^n$ as well (bar denotes complex conjugation), since $i$ is a unit. This shows that if $\mathfrak p$ and $\overline{\mathfrak p}$ are co-prime, then a rational prime power $N(\mathfrak p)$ divides $z$, a contradiction.

So the only possible prime factors of $z$ are rational primes $\equiv3\pmod4$ and $1\pm i$. Again rational primes are excluded. So the only possible prime factor of $z$ is $1\pm i$. And $1\pm i$ cannot appear with multiplicity $\geq2$, as if $(1+i)^2=2i$ divides $z$, then $2$ divides $z$ as well. Therefore $z\in\left\{\pm1\pm i,\pm 1, \pm i\right\}$ and $\vert pq\vert\leq1$. $\square$

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Now $(2+i)^n$ can never be real, and hence $(2+i)^n\ne(2-i)^n$ for $n\in\mathbb N\setminus\left\{0\right\}$.

Notes:

1. It is of course not necessary to show that $z^n$ cannot be purely imaginary (which I thought was), but I think it could be useful, so I left the proof there.

2. The whole idea comes from the cited answer. If this should be community wiki, please point it out. Thanks in advance.

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Hope this helps.

Answer 3

Let $z_1 = 2 + i \space$ and $z_2 = 2 - i$.
So the equation is now $z_1^n = z_2^n$.

$$|z_1| = \sqrt{2^2 + 1^2} = \sqrt{5} = \sqrt{2^2 + (-1)^2} = |z_2|$$

$z_1$ is of the form $z_1 = a_1 + b_1i, \space a_1, b_1 \in \mathbb{R}$, but it can be written in it's trigonometric form $z_1 = |z_1|(\space cos(\alpha) + sin(\alpha)i\space)$, where $\alpha = arctg(\frac{b_1}{a_1}) = arctg(\frac{1}{2})$.

Similarly $z_2=|z_2|(\space cos(\beta) + sin(\beta)i \space)$, where $\beta = arctg(\frac{-1}{2}) = -arctg(\frac{1}{2}) = -\alpha$.

So we have $$z_1^n = |z_1|^n(\space cos(n\alpha) + sin(n\alpha)i \space) \\ z_2^n = |z_2|^n(\space cos(n\beta) + sin(n\beta)i \space)$$

The equation is now $$|z_1|^n(\space cos(n\alpha) + sin(n\alpha)i \space) = |z_2|^n(\space cos(n\beta) + sin(n\beta)i \space)$$

Since $|z_1|^n = |z_2|^n$, simplifying, the equation becomes $\space cos(n\alpha) + sin(n\alpha)i = cos(n\beta) + sin(n\beta)i$

The real part from the left must be equal to the real part from the right and the imaginary part as well.

$$\bigg\{\begin{align} cos(n\alpha) = cos(n\beta) \\ sin(n\alpha) = sin(n\beta) \end{align}$$

Now the first equation: $\begin{align} cos(n\alpha) = cos(n\beta) \\ \alpha = -\beta \end{align}\bigg\}\Longleftrightarrow cos(n\alpha) = cos(-n\alpha)$, which is true because $cos$ is an even function so $cos(x) = cos(-x), \space \forall \space x \in \mathbb{R}$.

The second equation after substituting $\beta$ with $-\alpha$ becomes $sin(n\alpha) = sin(-n\alpha)$, since $sin$ is an odd function this means that $sin(x) = -sin(x), \space \forall \space x \in \mathbb{R}$, so the equation is now $sin(n\alpha) = -sin(n\alpha)$.

This is split in two cases one where $sin(n\alpha) = 0$ and one where $sin(n\alpha) \ne 0$

It begins with the first case when $sin(n\alpha) = 0$.
This is only true when $n\alpha = k\pi, \space k \in \mathbb{Z}$, so only when $n = \frac{k\pi}{\alpha} = \frac{k\pi}{arctg(\frac{1}{2})}, \space k \in \mathbb{Z}$.

The second case where $sin(n\alpha) \ne 0$ leads to a contradiction because when dividing both sides by $sin(n\alpha)$ the equation is $1 = -1$ which is obviously false.

In conclusion $n = \frac{k\pi}{arctg(\frac{1}{2})}$, where $k \in \mathbb{Z}$.
When $k \ne 0, n \notin \mathbb{N}$ and when $k = 0, n = 0$ so there exists no $n \in \mathbb{N} - \{0\}$ such that the initial equation is satisfied.

Answer 4

Here's a very simple-minded proof using the Gaussian integers. All congruence here are over the Gaussian integers $\Bbb Z[i]$. If $$(2+i)^n=(2-i)^n$$ for an integer $n\ge1$, then $$(2+i)^{n+1}=(2+i)(2-i)^n=5(2-i)^{n-1}$$ so that $$(2+i)^{n+1}\equiv0\pmod 5.$$ But modulo $5$, the powers of $(2+i)$ cycle repeat as follows: $$2+i,\quad3+4i,\quad2+i,\quad3+4i,\ldots$$ and none is divisible by $5$.

Answer 5

It is enough to notice that $\frac{2+i}{2-i}$ is not a root of unity, since its minimal polynomial over $\mathbb{Q}$ is not a monic polynomial in $\mathbb{Z}[x]$, i.e. $\frac{2+i}{2-i}$ is an algebraic number but not an algebraic integer. On the other hand the minimal polynomial of a (non-real) root of unity is a cyclotomic polynomial $\Phi_m(x)\in\mathbb{Z}[x]$, which is a palyndromic polynomial and so a monic polynomial.

This implies that $\arctan(2)\not\in\pi\mathbb{Q}$, among other things.

Answer 6

$a,b = 2\!-\!i,2\!+i\,$ are coprime by $\,(1\!+\!i)a-b = 1,\,$ so $\ \overbrace{a\mid b^{\phantom i}\! b^{n-1}\!\Rightarrow a\mid b^{n-1}\Rightarrow\cdots \Rightarrow \color{#c00}{a\mid 1}}^{\large\rm Euclid's\ Lemma\ \&\ induction}.\,$ Symmetry $\,\Rightarrow\, \color{#0a0}{b\mid 1}\,$ too, $ $ so $\,(\color{#c00}{1/a})(\color{#0a0}{1/b}) = 1/5\in \Bbb Z[i],\,$ contradiction.