<s>[INST] Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$ Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.
I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.
 [/INST] Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then
$$
1=a+6b+(3a+b)\sqrt{2},
$$
i.e.
$$
3a+b=0,\ a+6b=1.
$$
It follows that 
$$
a=-\frac{1}{17},\ b=\frac{3}{17}.
$$
Now
$$
(1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}),
$$
i.e.
$$
x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}.
$$
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