Determine whether ${\frac{2+i}{2-i}}$ is a root of unity

Question

I need to determine whether ${\dfrac{2+i}{2-i}}$ is a root of unity.
At first, I expressed this number as ${\dfrac{3}{5}+\dfrac{4}{5}i}$.
Then I tried to use a formula for $\sin{nx}$, where x = ${acos{\dfrac35}}$ and realized that n is even, then, using the properties of divisibility, realized that n divides 4. But I don't know what to do now.

Answer 1

The ring $\mathbb{Z}[i]$ is an Euclidean domain, so this ring is a unique factorisation ring. Show that $2+i$ and $2-i$ are non associated primes, so an equality $(2+i)^n=(2-i)^n$ with $n\geq 1$ is impossible due to the unicity of the factorisation.

Answer 2

Suppose that $$\left(\frac 35+\frac 45i\right)^n=1$$

Then, applying the binomial theorem and taking imaginary parts, $$\sum_{j=0}^{\lfloor (n-1)/2\rfloor}(-1)^j\binom n{2j+1}\left(\frac 35\right)^{n-2j-1}\left(\frac 45\right)^{2j+1}=0$$ or, multiplying by $5^n/4$, $$\sum_{j=0}^{\lfloor (n-1)/2\rfloor}(-1)^j\binom n{2j}3^{n-2j-1}4^{2j}=0$$

Therefore, $4$ is a root of the polynomial $$\sum_{j=0}^{\lfloor (n-1)/2\rfloor}(-1)^j\binom n{2j}3^{n-2j-1}X^{2j}$$

whose costant term is $3^{n-1}$. Since clearly $4$ doesn't divide $3^{n-1}$, this is impossible.