Consider the system $$ D_1 X^2 + D_4 W^2 = D_2 Y^2 \quad \quad D_1 X^2 - D_4 W^2 = D_3 Z^2 $$ where $D_i$ are strictly positive squarefree and pairwise coprime integers. How can I show that there is a solution $(W, X, Y, Z)$ in $\mathbb{Q}_p$ for all $p \nmid 2 D_1 D_2 D_3 D_4$. The solution must also satisfy $W, X,Y,Z \neq 0$.
I started by fixing some $p$ and trying to find a solution in $\mathbb{F}_p$ first. The $D_i$ are all units, so I divide by $D_2$ and $D_3$ respectively to get $$ (D_1 / D_2)X^2 + (D_4 / D_2) W^2 = Y^2 \quad \quad (D_1 / D_3) X^2 - (D_4 / D_3)W^2 = Z^2 $$ and I tried to use some pigeonhole principle kind of argument to show that there must be some values of $X$ and $W$ such that $(D_1 / D_2)X^2 + (D_4 / D_2) W^2$ and $(D_1 / D_3)X^2 - (D_4 / D_3) W^2$ are both squares.
I know that there are $(p+1)/2$ squares in $\mathbb{F}_p$. If I can show that an expression like $(D_1 / D_2)X^2 + (D_4 / D_2) W^2$ takes on at least $((p+1)/4)$ or $((p+1)/4) + 1$ values that are squares (depending on whether $p \equiv 1$ or $3$ mod 4), and same for the other equation, then by pigeonhole principle, the two equations must have a common square. But I have difficulty showing this (because two $(X_1, W_1)$ and $(X_2, W_2)$ could 'collide').
Thank you for the help.
Edit If there is a solution in $\mathbb{F}_p$, is there a version of Hensel's lemma that allows to lift this solution to $\mathbb{Q}_p$? I've only ever seen Hensel's for the roots of a single polynomial in one variable, and not for simultaneous equations.