Biquadratischer Zahlbereich/1 Modulo 4/Aufgabe/Lösung

Wegen ${}{\sqrt {a}}=2x-1$ und ${}{\sqrt {b}}=2y-1$ ist direkt

{}{\begin{aligned}z&={\frac {1+{\sqrt {ab}}}{2}}\\&={\frac {1+{\sqrt {a}}{\sqrt {b}}}{2}}\\&={\frac {1+(2x-1)(2y-1)}{2}}\\&={\frac {4xy-2y-2x+2}{2}}\\&=2xy-y-x+1\\&\in \mathbb {Z} [x,y].\end{aligned}}

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