Finding three perfect complex powers with equal norm and cancelling imaginary parts 
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*The Problem
I've been curious about finding 3 distinct Gaussian integers with equal norm whose imaginary parts cancel out. So $\alpha, \beta, \gamma\in\mathbb{Z}[i]$ with $|\alpha|=|\beta|=|\gamma|$ and $\alpha+\beta+\gamma\in\mathbb{Z}$. To rule out some trivial solutions, we should require $\alpha, \beta, \gamma$ to be strictly complex as well. Otherwise we get the infinite families of solutions satisfying
$$\text{Im}(\alpha)=0\quad \text{Im}(\beta)=-\text{Im}(\gamma)$$
This isn't too hard by itself but I'm looking for solutions where $\alpha,\beta,\gamma$ are also perfect powers. So in other terms,
$$|\alpha^2|=|\beta^2|=|\gamma^2|\quad\alpha^2+\beta^2+\gamma^2\in\mathbb{Z}$$
$$|\alpha^3|=|\beta^3|=|\gamma^3|\quad\alpha^3+\beta^3+\gamma^3\in\mathbb{Z}$$
$$|\alpha^4|=|\beta^4|=|\gamma^4|\quad\alpha^4+\beta^4+\gamma^4\in\mathbb{Z}$$
$$...$$

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*Computational Approach
Trying my best to recall from an old absract algebra college class, for Gaussians, the norm of a product is the product of the norm. So for any positive integer $k$ the statement $|\alpha^k|=|\beta^k|=|\gamma^k|$ implies more simply that $|\alpha|=|\beta|=|\gamma|$ which itself means that $\alpha,\beta,\gamma$ have the same factorization into Gaussian primes up to the conjugation of their prime factors. With this in mind, I took random sets of Gaussian primes, $\{\pi_1,...,\pi_r\}$, made a list of all the values which $\text{Im}(\prod\pi_i^k)$ could take on for each choice of conjugation for the $\pi_i$'s, and then checked if any 3 non-zero values from that list could add to zero. I did my best to reduce by any symmetries along the way.
With this I found non-trivial solutions for when $\alpha, \beta, \gamma$ are NOT perfect powers. Example:
$$\alpha = 249+88i\quad \beta = 241+108i\quad \gamma =177-196i$$
constructed with norm $5\cdot13\cdot29\cdot37$. There actually seem to be lots of these non-power non-trivial solutions. Here's a bigger one:
$$\alpha=7042031+2529728i\quad\beta= 6747407+3234536i\quad\gamma= 4771057-5764264i$$
with norm $5\cdot 13\cdot 17\cdot 37\cdot 41\cdot 51\cdot 73\cdot 89\cdot 97$.
I also found some solutions with $k=2$:
$$\alpha^2=156191+19800i\quad\beta^2= -80609+135240i\quad\gamma^2= 27391-155040i$$
$$\alpha=-396-25i\quad\beta= 196+345i\quad\gamma= 304-255i$$
But for $k\ge 3$ my program turned up no solutions. And it looks like from a probability perspective, the odds of encountering a solution become exponentially worse as $k$ grows.

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*Algebraic Approach
So after hitting a dead end with the computer, my next attempt was to try and build up a solution. I've seen a few problems recently where the non-trivial solutions to an equation were found by first finding the trivial solutions, then finding transformations that apply to all solutions, then using those transformations to build up non-trivial solutions from the trivial ones.
For instance here, any $\alpha,\beta,\gamma,k$ that meet our criteria, trivial or not, can be transformed into a new solution by either scaling or conjugation.
$$(\alpha,\beta,\gamma;k)\rightarrow (\bar{\alpha}, \bar\beta, \bar\gamma;k) \quad \quad\quad(\alpha,\beta,\gamma;k)\rightarrow (c\alpha, c\beta, c\gamma;k)$$
But neither of these transormations are helpful. My next thought was to just look for polynomials $f,g,h\in\mathbb{Z}[x_1,x_2,x_3]$ such that for some $k\ge 2$
$$f\bar f=g \bar g = h\bar h \quad\text{and}\quad f^k+g^k+h^k\in\mathbb{Z}[x_1^k+x_2^k+x_3^k]$$
since any such polynomials act as a valid transformation :
$$(\alpha,\beta,\gamma;k)\rightarrow(f(\alpha,\beta,\gamma),g(\alpha,\beta,\gamma),h(\alpha,\beta,\gamma);k)$$
This is where I'm out too deep. I feel like the elementary symmetric polynomials could be useful here. But I'm not sure how. And I'm not sure what problems are created by applying conjugation to terms of a polynomial.
I also had the thought to use the divisibility of $k$. For instance, any solution to $k=4$ is necesarily also a solution for $k=2$. So maybe solutions to $k=4$ could be built up from $k=2$ solutions? But I didn't get anywhere with this.

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*Question
So I guess I'm wondering the usual, has this been studied anywhere? Has anyone enumerated solutions? Or shown there are no solutions for some sufficiently large $k$? Have any similar questions been answered? Any help is appreciated.
 A: So if $\alpha,\beta,\gamma$ have the same prime factorization up to conjugation of the prime factors, then we can place any prime factor into one of four buckets depending on whether it 1) appears with the same conjugation in $\alpha,\beta,$ and $\gamma$, 2)appears with the same conjugation in $\alpha$ and $\beta$ but the opposite conjugation in $\gamma$ , 3) same thing but for $\beta$, or 4) same thing but for $\gamma$.
This isn't as rigorous as I'd like, but I think this means for any $\alpha,\beta,\gamma\in\mathbb{Z}[i]$ with the same norm, there exist $w,x,y,z\in\mathbb{Z}[i]$ such that
$$\alpha=w\bar{x}yz\quad\beta=wx\bar{y}z\quad\gamma=wxy\bar{z}$$
And we are now looking for $w,x,y,z$ such that
$$w^k\bar x^ky^kz^k+w^kx^k\bar y^kz^k+w^kx^ky^k\bar z^k\in\mathbb{R}$$
or equivalently such that
$$\text{Im}(w^k\bar x^ky^kz^k+w^kx^k\bar y^kz^k+w^kx^ky^k\bar z^k)=0\quad\quad(1)$$
Now equation (1) actually turns out to be equivalent to a more useful equation:
$$\frac{\text{Re}(w^k)}{\text{Im}(w^k)}=\frac{\text{Re}(x^ky^kz^k)-4\text{Re}(x^k)\text{Re}(y^k)\text{Re}(z^k)}{\text{Im}(x^ky^kz^k)+4\text{Im}(x^k)\text{Im}(y^k)\text{Im}(z^k)}\quad\quad(2)$$
Showing equations (1) and (2) are equivalent was tedious for me but comes down to expanding out each Re$()$ and Im$()$ with the following identities
$$\text{Im}(a+b)=\text{Im}(a)+\text{Im}(b)\quad\text{Re}(a+b)=\text{Re}(a)+\text{Re}(b)$$
$$\text{Im}(ab)=\text{Im}(a)\text{Re}(b)+\text{Re}(a)\text{Im}(b)\quad\text{Re}(ab)=\text{Re}(a)\text{Re}(b)-\text{Im}(a)\text{Im}(b)$$
and then cancelling/collecting terms. These identities are just the definition of addition and multiplication for $a,b\in\mathbb{C}$.
Looking at equation (2) now, this means $w^k$ is a rational multiple of $u^k-4v$ where $u=xyz$ and $$v=\text{Re}(x^k)\text{Re}(y^k)\text{Re}(z^k)-\text{Im}(x^k)\text{Im}(y^k)\text{Im}(z^k)i$$
So all in all, we're looking for solutions to
$$cw^k=u^k-4v$$
where $c\in\mathbb{Q}$ is a rational number, the terms $u,v,w\in\mathbb{Z}[i]$ are Gaussian integers, and $v$ is defined as above for some factorization of $u$ into 3 factors $x,y,z$. There are a few other restrictions one would have to place on $x,y,z$ to ensure that the complex numbers $\alpha,\beta,\gamma$ of our original problem are distinct and strictly complex.
But I think this is as far as I'll go. The equation $cw^k=u^k-4v$ looks like the sort of thing that's at the very threshold of modern algebra and I'd rather not touch it.
