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Let $p_1$ etand $p_2$ two prime primes numbers $\equiv 1\pmod 4$. If we note by $\varepsilon_m$ the fundamental unit of real quadratic field $\mathbb{Q}(\sqrt m)$, then how can be solved in $\mathbb{Q}(\sqrt{2},\sqrt{p_1p_2})$ the following equation: \begin{equation*} \varepsilon_{2}\varepsilon_{p_1p_2}\varepsilon_{2p_1p_2}=X^2 \end{equation*}

Let $p_1$ et $p_2$ two prime numbers $\equiv 1\pmod 4$. If we note by $\varepsilon_m$ the fundamental unit of real quadratic field $\mathbb{Q}(\sqrt m)$, then how can be solved in $\mathbb{Q}(\sqrt{2},\sqrt{p_1p_2})$ the following equation: \begin{equation*} \varepsilon_{2}\varepsilon_{p_1p_2}\varepsilon_{2p_1p_2}=X^2 \end{equation*}

Let $p_1$ and $p_2$ two primes numbers $\equiv 1\pmod 4$. If we note by $\varepsilon_m$ the fundamental unit of real quadratic field $\mathbb{Q}(\sqrt m)$, then how can be solved in $\mathbb{Q}(\sqrt{2},\sqrt{p_1p_2})$ the following equation: \begin{equation*} \varepsilon_{2}\varepsilon_{p_1p_2}\varepsilon_{2p_1p_2}=X^2 \end{equation*}

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root of a unit in a real biquadratic field

Let $p_1$ et $p_2$ two prime numbers $\equiv 1\pmod 4$. If we note by $\varepsilon_m$ the fundamental unit of real quadratic field $\mathbb{Q}(\sqrt m)$, then how can be solved in $\mathbb{Q}(\sqrt{2},\sqrt{p_1p_2})$ the following equation: \begin{equation*} \varepsilon_{2}\varepsilon_{p_1p_2}\varepsilon_{2p_1p_2}=X^2 \end{equation*}