 Solution

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$, andThe proposed integral basis for coprime biquadratic radicands congruent to two and one isEach entry is an algebraic integer, by the ring of integers of a quadratic field formula and closure under products. Their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/4$, so they are rationally linearly independent. It remains to show that they span every algebraic integer.

Write $\gamma =(a+b\alpha +c\beta +d\alpha \beta )/2$ initially with rational $a,b,c,d$. The field traces to $\mathbb{Q}(\alpha )$ and $\mathbb{Q}(\alpha \beta )$ are $a+b\alpha$ and $a+d\alpha \beta$. Since $m_1\equiv m_1{m}_2\equiv 2(\mathrm{mod}4)$, the corresponding integral basis in each subfield gives $a,b,d\in \mathbb{Z}$. The field trace to $\mathbb{Q}(\beta )$ is $a+c\beta$. Membership in $\mathbb{Z}[\omega ]$ says that $2a,2c$ are integers of equal parity. As $a$ is already an integer, this forces $c\in \mathbb{Z}$ too.

Now the field norm to $\mathbb{Q}(\beta )$ isBy the ring of integers of a quadratic field formula, $A/2$ is an integer and has the same parity as $ac-m_{1}bd$. Thus $A$ is even, giving $a\equiv c(\mathrm{mod}2)$, andUsing $m_1\equiv 2$, $m_2\equiv 1$ and $a\equiv c(\mathrm{mod}2)$, this becomesHence $b\equiv d(\mathrm{mod}2)$. Finally,an expression with integer coefficients. This proves the asserted integral basis without assuming a general theorem about products of elements of an integral basis.