Is $7 + 5 \sqrt{2}$ a sum of squares in $\mathbb{Q}(\sqrt{2})$?
I believe it is probably not.
Suppose $7 + 5 \sqrt{2} = (a_1 + b_1 \sqrt{2})^2 + \ldots + (a_n + b_n \sqrt{2})^2 $. By factoring out the common denominator $d$ we can assume that all $a_i, b_i$ are integers and replace $7 + 5 \sqrt{2} $ with $7d + 5d \sqrt{2} $. Then I notice that
$a_1^2 + \ldots + a_n^2 + 2(b_1^2 + \ldots b_n^2) $ is divisible by $7$
and
$2a_1b_1 + \ldots + 2a_nb_n$ is divisible by $5$.
I don't see any contradictions yet, though.