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Is there any integer solution $(a,b,n)$ for $n\geq 2$ such that $$\operatorname{Re}\left((a+bi)^n\right)=\pm1$$ except $(a,b)=(\pm1,0),(0,\pm1)$?

My attempt:

If $a\equiv b\pmod{2}$, then $\operatorname{Re}(a+bi)^{n}$ is even, because $(a+bi)^n=2\left(\frac{a^2-b^2}{2}+abi\right)(a+bi)^{n-2}$. So $a\not\equiv b\pmod{2}$.

If $n$ is even, write $c+di=(a+bi)^{n/2}$, then $\operatorname{Re}(a+bi)^n=\operatorname{Re}(c+di)^2=c^2-d^2=(c+d)(c-d)$.
So, $\operatorname{Re}(a+bi)^n=\pm1\implies (c,d)=(\pm1,0),(0,\pm1)\implies (a+bi)^n=\pm1\implies n|a+bi|=1$
Since $|a+bi|>1$, $n$ is not even.

If $n$ has an odd divisor $p$, write $c+di=(a+bi)^{n/p}$, then $\operatorname{Re}(a+bi)^n=\operatorname{Re}(c+di)^p$.
Since $\operatorname{Re}(c+di)^p$ can be divided by $c$, we have that $\operatorname{Re}(a+bi)^{n}=\pm1 \implies c=\operatorname{Re}(a+bi)^{n/p}=\pm1$.
For $p=n$, we have that $\operatorname{Re}(a+bi)^n=\pm1\implies a=\pm1$

Therefore, we only need to consider it for $\operatorname{Re}(1+2mi)^n=\pm1$ where $n$ is a odd prime, and $m\neq 0$. However, I don't know how to progress from here. Assignment for $n$ shows that it has no solution for some $n$, but not for every $n$.

Any help would be appreciated. Thank you.

dodicta
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1 Answers1

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Let $n\geq 2$. Suppose $$ (a+bi)^n = \pm1+ci $$ for some integer $c$. Then taking the complex norm, which is multiplicative, defined by $$ N(x+yi) := x^2+y^2 $$ gives us $$ (a^2+b^2)^n = 1+c^2, $$ Setting $(x,y)=(a^2+b^2,|c|)$, we may rewrite this as $$ x^n-y^2 = 1 $$ Now if $x,y>0$, then since $n\geq 2$, by Catalan's conjecture we have $$ a^2+b^2 = x=3 $$ This is impossible, therefore it must be either $x=0$ or $y=0$.

The case $x=0$ is trivial since it forces $a=b=0$, which fails.

For the other case $y=0$ we have a necessary condition $$ (a^2+b^2)^n = 1 $$ Then since $a^2+b^2\geq 0$ this means $a^2+b^2=1$. This can only happen for $(a,b) = (\pm 1,0), (0,\pm 1)$.

Yong Hao Ng
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