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Number theory Difficulty 6.4 National olympiad Prove it

24. (Division with remainder in Z[1]\boldsymbol{Z}[\sqrt{-1}]) Let αj=aj+ibjZ[1],aj,bjZ\alpha_{j}=a_{j}+\mathrm{i} b_{j} \in \boldsymbol{Z}[\sqrt{-1}], a_{j}, b_{j} \in \boldsymbol{Z} (j=0,1)(j=0,1).
(i) Prove: There must exist η1,α2Z[1]\eta_{1}, \alpha_{2} \in \boldsymbol{Z}[\sqrt{-1}], satisfying
α0=η1α1+α2,0N(α2)<N(α1)\alpha_{0}=\eta_{1} \alpha_{1}+\alpha_{2}, \quad 0 \leqslant N\left(\alpha_{2}\right)<N\left(\alpha_{1}\right)

where N[α]=r2+s2,α=r+s1,r,sQN[\alpha]=r^{2}+s^{2}, \alpha=r+s \sqrt{-1}, r, s \in Q, here N(α)N(\alpha) is called the norm of α\alpha;
(ii) Are η1,α2\eta_{1}, \alpha_{2} in (i) unique? What is the maximum number of solutions?

Solution

None

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.