I am working on showing that $\{a+b\sqrt{2}$ | $a,b \in \mathbb Z \} $ is an euclidean domain.
There was a similar problem showing that $\{a+b\sqrt{-2}$ | $a,b \in \mathbb Z \} $ was an euclidean domain, but that was easier because for complex numbers one can use the norm $\phi (z) = \bar z \cdot z$.
Does there exist a similar function $\phi$ which is natural to consider for the ring $\{a+b\sqrt{2}$ | $a,b \in \mathbb Z \} $ ?
Yes, it is $$ a + b \sqrt{2} \mapsto N(a + b \sqrt{2}) = \lvert a^{2} - 2 b^{2} \rvert. $$ The analogy with the other case you mention is provided by the fact that the extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ has a single non-trivial automorphism, which is $$ \sigma : a + b \sqrt{2} \mapsto a - b \sqrt{2}, $$ so that $$ N(x) = \lvert x \cdot \sigma(x) \rvert, $$ where you have to take the absolute value, as $a^{2} - 2 b^{2}$ might well be negative.
In the other case you mention, the automorphism is just complex conjugation.