Wegen a = 2 x − 1 {\displaystyle {}{\sqrt {a}}=2x-1} und b = 2 y − 1 {\displaystyle {}{\sqrt {b}}=2y-1} ist direkt
![{\displaystyle {}{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/760a9dd2f8072495e1a1d3bd3c281d31800f5263)