Întrebare numere rationale

Determinati numerele rationale a si b pentru care a radical2 + b totul supra radical2 -1 apartine Q.

Multumesc!

\[\frac{a\sqrt{2}+b}{\sqrt{2}-1}\in \mathbb{Q}\]
\[{}^{\sqrt{2}+1)}\frac{a\sqrt{2}+b}{\sqrt{2}-1}=\frac{2a+b\sqrt{2}+a\sqrt{2}+b}{1}=\]
\[\underbrace{(2a+b)}_{\in \mathbb{Q}}+\underbrace{(a+b)}_{\in \mathbb{Q}}\sqrt{2}\]
\[2a+b\in \mathbb{Q}\,\forall a,b\in \mathbb{Q}\]
\[a+b\in \mathbb{Q}\,\forall a,b\in \mathbb{Q}\]
\[\underbrace{(2a+b)}_{\in \mathbb{Q}}+\underbrace{(a+b)}_{\in \mathbb{Q}}\sqrt{2}\in \mathbb{Q}\]\[\Rightarrow a+b=0\Rightarrow a=-b,\,a,b\in \mathbb{Q}\] .