Is Chinese Remainder Theorem relevant here?

{ Z = ( q 1 ) ( 1 i ) Z = ( q 2 ) ( 1 2 i ) i Z = ( q 3 ) ( 1 4 i ) + 2 Z = ( q 4 ) ( 3 2 i ) + 2 i \large{\begin{cases} Z &=& (q_{1})(1 - i) \\ Z &=& (q_{2})(1 - 2i) - i \\ Z &=& (q_{3})(1 - 4i) + 2 \\ Z &=& (q_{4})(3 - 2i) + 2i \\ \end{cases} }

Let Z Z be a Gaussian integer and q 1 q_{1} , q 2 q_{2} , q 3 q_{3} , & q 4 q_{4} be the Gaussian integral quotients of the system above, what is the least value of Z 2 | Z | ^{2} ?


The answer is 178.

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1 solution

Mark Hennings
Jan 9, 2020

Since the Gaussian integers form a Euclidean domain, the Chinese Remainder Theorem is applicable, and we note that 1 i 1-i , 1 2 i 1-2i , 1 4 i 1-4i and 3 2 i 3-2i are coprime in the Gaussian integers. We obtain that Z = ( 37 29 i ) p 195 11 i p Z [ i ] Z \; = \; (37 - 29i)p - 195 - 11i \hspace{2cm} p \in \mathbb{Z}[i] and it is easy to show that Z 2 |Z|^2 is minimized when p = 3 + 3 i p = 3 + 3i , giving the minimum value of Z 2 |Z|^2 as 3 + 13 i 2 = 178 |3 + 13i|^2 = \boxed{178} .

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