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So far I have that $B = \{1,\sqrt{3},\sqrt{5},\sqrt{3}\sqrt{5}\}$ is a $\mathbb{Q}$-basis for $\mathbb{Q}(\sqrt{3},\sqrt{5})$.

I think the discriminant of $B$ is $2^83^25^2$, which implies that, if $B$ is not an integral basis, then there is an algebraic integer of the form $\frac{1}{p}(a + b\sqrt{3} + c\sqrt{5} + d\sqrt{3}\sqrt{5})$ for integers $0 \leq a,b,c,d \leq p-1$ and $p$ one of $2,3,5$.

Beyond this, I'm not sure how to proceed.

Any help would be much appreciated.

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  • $\begingroup$ See this post for the discriminant of $\Bbb Q(\sqrt{m},\sqrt{n})$. $\endgroup$ – Dietrich Burde Jul 3 '19 at 15:11
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The formula for the discriminant of $\Bbb Q(\sqrt{m},\sqrt{n})$ is given in Theorem $3$ of the paper Integers of biquadratic fields, page $525$. Theorem $2$ lists the integral bases for each case. Since $\Bbb Q(\sqrt{3},\sqrt{5})=\Bbb Q(\sqrt{3},\sqrt{15})$ we are in the case $(3,3)\bmod 4$.

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