Find all Gaussian integers $α, β, γ$ such that $αβγ = α + β + γ = 1$ I tried to solve for this by assuming $α=a+bi$, $β=c+di$, and $γ=e+fi$, and explicitly solving this by equal $a+c+e=1$, $b+d+f=0$, and similarly for $αβγ=1$. Is there any other easier approach for this problem? I know $(1, i, -i)$  is a pair of solution. But is there any other? 
 A: Since $\alpha,
\beta,\gamma$ divide $1$, they are all units. The group of units is just $\{\pm 1,\pm i\}$. If any of $\alpha,\beta,\gamma$ are equal from that list they cannot add to $1$ while simultaneously multiplying to $1$, and with three distinct $\alpha,\beta,\gamma$, there will always be a pair which are additive inverses, hence the third must be $1$, and the other two must be $\pm i$.
A: Ignoring order, you have found the only possible solution. If $\alpha \beta \gamma = 1$, then it is a necessary but not sufficient condition that the norm function give $N(\alpha \beta \gamma) = 1$. This narrows the choices down to the units, 1, $-1$, $i$, $-i$ for each of the three variables.
But $\alpha + \beta + \gamma = 1$ means that $\alpha + \beta = 0$ (you could choose to take out a different variable but it won't change the end result). Therefore $\alpha = -\beta$ and $\beta = -\alpha$. And of course $\gamma = 1$.
So essentially this boils down to two possible solutions. $\alpha = 1$ and $\beta = -1$ works for the addition but not the multiplication: $\alpha \beta \gamma = -1$, not 1. This leaves the solution you have already found: $\alpha = i$, $\beta = -i$. This works because $\alpha + \beta = 0$ and $\alpha \beta = 1$, hence $\alpha + \beta + \gamma = \alpha \beta \gamma = 1$ as desired.
EDIT: This is how I got from $\alpha + \beta + \gamma = 1$ to $\alpha + \beta = 0$: I subtracted 1 from both sides to get $\alpha + \beta + \gamma - 1 = 0$. If $\gamma = 1$, then $\gamma - 1 = 0$ so "$\gamma - 1$" can be removed from the equation.
2ND EDIT: It seems perfectly obvious to me why at least one of $\alpha$, $\beta$ or $\gamma$ has to be 1. But since I don't want anyone to say I just copied off their answer, I'm going to prove this by brute force:


*

*$\alpha = \beta = \gamma \neq 1$ can't work because $u^n = 1$ with $u$ a unit other than 1 requires an even exponent or a doubly even exponent.

*$i, -1, -1$ doesn't work for the multiplication.

*$i, i, -1$ works for the multiplication but not the addition.

*$i, i, -i$ doesn't work for the multiplication.

*$i, -i, -i$ doesn't work for the multiplication either.

*$-i, -i, -1$ doesn't work for the multiplication either.

*$i, -i, -1$ doesn't work for the addition.


I don't need to hear criticism for this being inefficient but I do need to hear if I neglected to cover one or two cases.
A: Taking norms we have $(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) = 1$ in $\mathbb{Z}$ then the only solutions here are $\pm 1$ and $0$ for each factor. On the other hand, $a + c + e = 1$ and $b + d + f = 0$ put restrictions to be considered for the solutions $x + yi = (\pm 1, 0), (0, \pm 1)$ with $x = a, c, e$ and    $y = b, d, f$. The only solutions for the system are $(\alpha, \beta, \gamma) = (1, \pm i, \mp i)$, $(\pm 1, \mp 1, 1)$ (don’t forget $\mathbb{Z} \subset \mathbb{Z}[i]$) and all the corresponding permutations that could be taken as equivalent. 
