Given a polynomial $P(X)$ with coefficients in $\mathbb{Q}(\sqrt{2},\sqrt{91})$ how do I find values of $X$ in $\mathbb{Q}(\sqrt{2},\sqrt{91})$ such that $P(X)$ is a perfect square in $\mathbb{Q}(\sqrt{2},\sqrt{91})$? Or how do I decide whether no such values of $X$ exist?
In particular, consider the polynomial
$P(X)=(6845983737169536\sqrt{91}\sqrt{2}+144586598507324711)X^4++(648643816782394524\sqrt{2}+74356950326415948\sqrt{91})X^3++(53552805056233956\sqrt{91}\sqrt{2}+964911661738142866)X^2--(594454520900196684\sqrt{2}+63627500651668308\sqrt{91})X++(12210708574543356\sqrt{91}\sqrt{2}+198775894389522551)$
Is there an $X_0$ in $\mathbb{Q}(\sqrt{2},\sqrt{91})$ such that $P(X_0)$ is a perfect square? can one find such an $X_0$ explicitly?
I am trying to find a an embedding of a certain graph in $\mathbb{Q}(\sqrt{2},\sqrt{91})\times \mathbb{Q}(\sqrt{2},\sqrt{91})$, and everything reduces to answering the question above.
