 Solution (source code)

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= Solution

An <integral basis> of a degree-$d$ <number field> is a $\mathbb Z$-basis of its <ring of integers>: each <algebraic integer> has a unique integer linear combination of those $d$ elements. For $k=\mathbb Q(\sqrt2)$, the standard quadratic-integer criterion gives $\mathcal O_k=\mathbb Z[\sqrt2]$, since $2$ is squarefree and is not congruent to one modulo four. Thus
$$
\boxed{1,\sqrt2\text{ is an integral basis of }k.}
$$

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