Can $\operatorname{Re}(a+bi)^{n}$ be overlapped with $a,b\in\mathbb{Z}$ fixed?

Question

Is there any integer solution for $$\operatorname{Re}((a+bi)^{m})=\operatorname{Re}((a+bi)^{n})$$ except $(m,n)=(0,1),(1,3)$, where $0\leq m<n,\ |a|\neq |b|,\ a\neq 0,\ b\neq 0$?

In other words,

Can $\operatorname{Re}(a+bi)^{n}$ be overlapped with $a+bi\in\mathbb{Z}[i]$ fixed except for some trivial cases?

This is a generalization of my earlier question, Is there any integer solution for $\operatorname{Re}(a+bi)^n=\pm1$, where $n\geq 2$, except $(a,b)=(\pm1,0),(0,\pm1)$?. The answer to this question is no. So, we have no solution for $m=0$.

I checked for every $0<|a|,|b|\leq 10000,\ |a|\neq|b|,\ 0\leq m<n\leq 1000$, then only found these:
$(a,b,m,n)=(\pm 2,\pm 1,1,3), (\pm 7,\pm 4,1,3), (\pm 26,\pm 15,1,3), (\pm 97,\pm 56,1,3), (\pm 362,\pm 209,1,3), (\pm 1351,\pm 780,1,3), (\pm 5042,\pm 2911,1,3)$

They are the integer solutions for $\operatorname{Re}(a+bi)^{1}=\operatorname{Re}(a+bi)^{3} \iff a^2-3b^2=1$.
I couldn't find any solutions for $(m,n)\neq (1,3)$.

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PS
Just for your information, I also checked for $\operatorname{Im}(a+bi)^n$, then I found these solutions:
$(a,b,m,n)=(-2,\pm4,2,3),(8,\pm24,4,5),(9,\pm15,2,3),(-32,\pm56,2,3),(121,\pm209,2,3),(-450,\pm780,2,3),(1681,\pm2911,2,3)$

Except for $(a,b,m,n)=(8,\pm24,4,5)$, they are the integer solutions for $\operatorname{Im}(a+bi)^{2}=\operatorname{Im}(a+bi)^{3} \iff 3a^2b-b^3=2ab \iff 3a^2-2a=b^2$.

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I would appreciate any help. Thank you for your cooperation.

Answer 1

This answer is to prove that there are no solutions for some particular values of $(m, n)$.

$m=1$, $n=4$:

In that case, the equation becomes $a = a^4 - 6a^2b^2 + b^4$, that is, $a(1+4ab^2) = (a^2-b^2)^2 $. It is easy to rule out $a = \pm b$. So suppose $a \neq \pm b$. Because $a$ and $1+4ab^2$ are coprime, $a$ and $1+4ab^2$ are $\pm$ squares. Suppose $a > 0$. Then $4ab^2$ and $1+4ab^2$ are squares. Thus $4ab^2=0$, a contradiction. Suppose $a < 0$. Then $-4ab^2$ and $-4ab^2-1$ are squares, so that $-4ab^2=1$, a contradiction.

$m=2$, $n = 3$:

In this case, $a^2-b^2 = a^3-3ab^2$. Rewrite it as $b^2(3a-1) = a^2(a-1) $. Because $a$ and $3a-1$ are coprime, $a^2 \mid b^2$, so that $a \mid b$. Then $3a-1 \mid a-1$. But $|3a-1| > |a-1|$ when $a \neq 0$, a contradiction.

$m = 2$, $n = 4$:

In this case, $a^2-b^2 = a^4 - 6a^2b^2 + b^4$. That is, $(2(a^2-b^2)-1)^2 = 16 a^2b^2+1$. Necessarily, $16a^2b^2 = 0$.

Answer 2

We prove the following claim:

If $n$ is odd and $m$ is even, then

• $z$ cannot be an integer multiple of $1+i$;
• $\displaystyle\frac{n+1}2\le m\le n-1$.

So for example, this solves cases like $(m,n)=(18,95)$.

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Case: $n$ odd and $m$ even

Writing $z=a+bi=re^{i\theta}$, the condition $\Re(z^n-z^m)=0$ gives $\cos m\theta=r^{n-m}\cos n\theta$.

When $r=\sqrt{a^2+b^2}$ is an integer, this forces $\cos\theta=p/q$ with $(p,q)=1$ so the rational root test can be used on the equation $$r^{n-m}T_n(\cos\theta)-T_m(\cos\theta)=0\tag1$$ where $T_\bullet$ denotes the Chebyshev polynomial of the first kind.

If $n$ is odd and $m$ is even, the roots must be of the form $\cos\theta=\pm1/q$ with $q\mid r$ as $a=r\cos\theta$ is an integer. Thus we have $a=\pm r/q$ and $b=\pm a\sqrt{q^2-1}\in\Bbb Z$. However, there is no integer $q>1$ such that $q^2-1$ is a square so no solutions can exist.

When $r$ is a non-integer, squaring $(1)$ and using the product identity for the Chebyshev polynomial yields $$r^{2(n-m)}T_{2n}(\cos\theta)-T_{2m}(\cos\theta)+r^{2(n-m)}-1=0\tag2$$ after rearrangement of terms. Notice that $r^2$ is necessarily an integer so $\cos^2\theta$ must be rational; the same equation can be used for the the rational root test as all powers of $\cos\theta$ are even.

If $n$ is odd and $m$ is even, the roots must be of the form $\cos^2\theta=p/q$ where $p=1,2$ and $q\mid r^2$. Simultaneously from the definition, we have $\cos^2\theta=a^2/(a^2+b^2)$ which automatically excludes $p=2$ since $(p,q)=1$. Thus $q=1+b^2/a^2\in\Bbb Z$ so $a\mid b$.

Let $b=ca$ where $c\in\Bbb Z^+$. Then $\Re(1+ci)^m=a^{n-m}\Re(1+ci)^n$ and taking the complex norm yields \begin{align}(1+c^2)^m&=a^{2(n-m)}w^2+y^2\tag3\\(1+c^2)^n&=w^2+z^2\tag4\end{align} where $w,y,z$ are positive integers. This means that $(1+c^2)^m-y^2\ge(1+c^2)^n-z^2$ which is true if $(1+c^2)^m\ge(1+c^2)^n-\lfloor(1+c^2)^{n/2}\rfloor^2$; that is, by minimising $y$ and maximising $z$. By using $\lfloor x\rfloor\ge x-1$, we have the necessary condition $(1+c^2)^m\ge2(1+c^2)^{n/2}-1$, which is met if $(1+c^2)^{m-n/2}\ge2$.

Therefore, if $c>1$ then we obtain the inequality $(n+1)/2\le m\le n-1$, and if $c=1$ (that is, in the special case of $z$ being an integer multiple of $1+i$), we have $(n+3)/2\le m\le n-1$.

We can solve the case of $c=1$ completely. Since $\Re(1+i)^n=2^{n/2}\cos(n\pi/4)$, we have the equality $$2^{(n-m)/2}\cos(n\pi/4)=a^{n-m}\cos(m\pi/4).\tag5$$ As the LHS is nonzero, $m$ must be an integer multiple of $4$. As $n$ is odd, taking absolute values on both sides gives $2^{(n-m-1)/2}=a^{n-m}$ but this has no results over the integers.

Answer 3

Just for completeness sake, the case $m=1$, $n=2$ is impossible:

If $a$ and $b$ are nonzero integers with $|a|\neq|b|$ and
$$\operatorname{Re}\left(a+bi\right)=\operatorname{Re}\left((a+bi)^2\right),$$ then $a=a^2-b^2$, or equivalently $$a^2-a-b^2=0.$$ Then the quadratic $X^2-X-b^2$ has two integral roots, so its discriminant $$\Delta=(-1)^2-4(-b^2)=4b^2+1,$$ is a perfect square. Of course $4b^2=(2b)^2$ is itself a perfect square, and the only two consecutive perfect squares are $0$ and $1$, so $b=0$, contradicting the assumption that $b\neq0$.

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The case $m=1$, $n=5$ is impossible too, by a similar but more involved approach:

Let $a$ and $b$ be integers such that $$\operatorname{Re}\left(a+bi\right)=\operatorname{Re}\left((a+bi)^5\right).$$ Then expanding both sides shows that $$a=a^5-10a^3b^2+5ab^4.$$ Because $a\neq0$ it follows that $$a^4-10a^2b^2+5b^4=1,$$ and this Thue equation has only finitely many integral solutions $(a,b)$. With the help of a computer we find that they are $(\pm1,0)$, so there are no nontrivial solutions for $(m,n)=(1,5)$.

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In general if $m=1$ and $n$ is odd then the diophantine equation $$\operatorname{Re}\left(a+bi\right)=\operatorname{Re}\left((a+bi)^n\right),$$ boils down to a Thue equation $$f(a,b)=1,$$ where $f$ is a homogeneous polynomial of degree $n-1$. By Thue's theorem there are only finitely many integral solutions if $n-1\geq3$, i.e. if $n>3$. Moreover, there exists an effective algorithm for finding all integral solutions. A quick check with PARI/GP shows that there are no integral solution other than $(a,b)=(\pm1,0)$ for odd $n\leq27$. Perhaps someone more familiar with PARI/GP or similar software can check for higher values of $n$ as well.

One simple observation to make from the Thue equation is that $a$ and $b$ must be coprime.

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Some other simple observations:

0. Without loss of generality $b>0$.
1. It suffices to consider $m$ and $n$ with $\gcd(m,n)=1$.
2. If $n$ is even and $m$ is odd then $a\mid b^n$.