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$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

Can someone prove the following lemma: Let $G$ be a finite group and $z$ a central element of order $2$. Then either

(1) $G \cong H \times (\ZZ/2 \ZZ)$, with $z$ generating the $\ZZ/2 \ZZ$ or

(2) There is some $w \in G$ with $w^2=z$?

Be warned that this is false if you replace $2$ by some other prime: In the group $\left( \begin{smallmatrix} 1 & \FF_p & \FF_p \\ 0 & 1 & \FF_p \\ 0 & 0 & 1 \end{smallmatrix} \right)$, the element $\left( \begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right)$ is central and $p$-torsion, but is not a $p$-th power when $p$ is odd.

Given this, I believe I can prove the Hasse principle whenever the Galois closure of $\KK$ has even degree over $\QQ$.

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

Can someone prove the following lemma: Let $G$ be a finite group and $z$ a central element of order $2$. Then either

(1) $G \cong H \times (\ZZ/2 \ZZ)$, with $z$ generating the $\ZZ/2 \ZZ$ or

(2) There is some $w \in G$ with $w^2=z$?

Be warned that this is false if you replace $2$ by some other prime: In the group $\left( \begin{smallmatrix} 1 & \FF_p & \FF_p \\ 0 & 1 & \FF_p \\ 0 & 0 & 1 \end{smallmatrix} \right)$, the element $\left( \begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right)$ is central and $p$-torsion, but is not a $p$-th power when $p$ is odd.

Given this, I believe I can prove the Hasse principle whenever the Galois closure of $\KK$ has even degree over $\QQ$.

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.