Solve an equation over the Gaussian integers

Find all the Gaussian integers \pi and \rho such that 5-i = \pi(1+2i) + \rho and N(\rho) < N(1+2i).


First, let \pi = a_1 + a_2i and \rho = b_1 + b_2i. Expanding and comparing real and imaginary parts, we have the following linear system of equations.

\left[ \begin{array}{cc} 1 & \text{-}2 \\ 2 & 1 \end{array} \right] \left[ \begin{array}{c} a_1 \\ a_2 \end{array} \right] = \left[ \begin{array}{c} 5-b_1 \\ -1-b_2 \end{array} \right].

Evidently, \left[ \begin{array}{cc} 1 & \text{-}2 \\ 2 & 1 \end{array} \right] \left[ \begin{array}{cc} 1 & 2 \\ \text{-}2 & 1 \end{array} \right] = 5I. This yields the following equality.

\left[ \begin{array}{c} 5a_1 \\ 5a_2 \end{array} \right] = \left[ \begin{array}{c} 3-b_1-2b_2 \\ -11 + 2b_1 - b_2 \end{array} \right].

Reducing mod 5 yields the following two equations: 3 \equiv b_1 + 2b_2 mod 5 and 1 \equiv 2b_1 - b_2 mod 5. A simple but tedious case analysis reveals that this system has two solutions: (0,-1) and (1,1).

Now recall the constraint that N(\rho) < N(1+2i); that is, b_1^2 + b_2^2 < 5. Thus (|b_1|, |b_2|) must be one of (0,0), (0,\pm 1), (0, \pm 2), (\pm 1, 0), (\pm 1, \pm 1), and (\pm 2, 0). Thus we have \rho \in \{ 1+i, -i \}.

Thus the solutions (\pi,\rho) of our original equation are (1-2i,-i) and (-2i,1+i).

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