求$x,y,z,w$ 使$x+y\sqrt2+z\sqrt3+w\sqrt6$的极小多项式的常数项为1

hbghlyj

求$x,y,z,w\inZ$ 使$x+y\sqrt2+z\sqrt3+w\sqrt6$的极小多项式的常数项为1？

即
$(x-y \sqrt{2}+z \sqrt{3}-w \sqrt{6})(x+y \sqrt{2}-z \sqrt{3}-w \sqrt{6})(x-y \sqrt{2}-z \sqrt{3}+w \sqrt{6})(x+y \sqrt{2}+z \sqrt{3}+w \sqrt{6})=1$
展开得
\[ x (x (-12 w^2 + x^2 - 4 y^2 - 6 z^2) + 48 w y z) + w^2 (36 w^2 - 24 y^2 - 36 z^2) + y^2 (4 y^2 - 12 z^2) + 9 z^4 - 1=0 \]
但这好像没有用。

hbghlyj

WolframAlpha算出$(1+\sqrt2)^m(2+\sqrt3)^n,\forall m,n\inZ$ 都满足条件。
例如$(1+\sqrt2)^1(2+\sqrt3)^{-1}$的极小多项式$=x^4 - 8 x^3 - 10 x^2 + 8 x + 1$

hbghlyj

$x+y\sqrt2+z\sqrt3+w\sqrt6$的极小多项式的常数项为1，则$\exists m,n\inZ:x+y\sqrt2+z\sqrt3+w\sqrt6=(1+\sqrt2)^m(2+\sqrt3)^n$
对吗？

hbghlyj

hbghlyj 发表于 2024-4-5 14:27
$x+y\sqrt2+z\sqrt3+w\sqrt6$的极小多项式的常数项为1，则$\exists m,n\inZ:x+y\sqrt2+z\sqrt3+w\sqrt6=(1+\sqrt2)^m(2+\sqrt3)^n$

设$X+Y\sqrt2:=(x+y \sqrt{2}-z \sqrt{3}-w \sqrt{6})(x+y \sqrt{2}+z \sqrt{3}+w \sqrt{6})$
则$X-Y\sqrt2=(x-y \sqrt{2}-z \sqrt{3}+w \sqrt{6})(x-y \sqrt{2}+z \sqrt{3}-w \sqrt{6})$
相乘得$X^2-2Y^2=1$，所以$\exists m,n\inZ:X+Y\sqrt2=(1+\sqrt2)^m$，
代回得$(1+\sqrt2)^m=(x+y \sqrt{2}-z \sqrt{3}-w \sqrt{6})(x+y \sqrt{2}+z \sqrt{3}+w \sqrt{6})$
即$(1+\sqrt2)^m=(x+y \sqrt{2})^2-3(z+w \sqrt{2})^2$
然后怎么办呢