Global approximation of finitely many square classes

Choose representatives $x_v\in Q_v^{\times }$ of the classes ${\xi }_v$. By [L1], there exists $r\in {\mathbb{Q}}^{\times }$ sufficiently close to every $x_v$ that $r\mathrm{/}{x}_v$ is a square in $Q_v^{\times }$ for each $v\in S$. Thus $r$ already has the required local square classes on $S$.

Only finitely many primes outside $S$ occur to odd valuation in $r$; let $P$ be that set and put $M:={\prod }_{p\in P}p$, with $M:=1$ when $P=\mathrm{\varnothing }$. Choose an odd prime $\mathrm{\ell }\mathrm{\notin }S\cup P$ such that $M\mathrm{/}\mathrm{\ell }$ is a square in every completion $Q_v$ with $v\in S$. This is a finite list of sign and congruence conditions, so [L2] supplies such a prime. Now set $t:=r\frac{M}{\mathrm{\ell }}\mathrm{.}$ For each $v\in S$, the factor $M\mathrm{/}\mathrm{\ell }$ is a local square, so $t$ and $r$ define the same square class in $Q_v^{\times }\mathrm{/}(Q_v^{\times }{)}^2$. For $p\in P$, the extra factor $M$ changes the odd valuation of $r$ to an even one; for $p\mathrm{\notin }S\cup P\cup \{\mathrm{\ell }\}$ the valuation was already even and stays even; and at $\mathrm{\ell }$ the new valuation is $v_{\mathrm{\ell }}(r)-1$, which is odd because $\mathrm{\ell }\mathrm{\notin }P$. Thus $t$ has the prescribed local square classes on $S$ and has even valuation at every prime outside $S$ except possibly the one extra odd prime $\mathrm{\ell }$.