Let $p$ be a prime, and let $u_1, u_2, u_3$ be squarefree integers not divisible by $p$, with the product $u_1 u_2 u_3$ a square. Let $c_1, c_2, c_3$ be integers not divisible by $p$.
I want to show that the system $$ u_1 y_1^2 = x - pc_1 $$ $$ u_2 y_2^2 = x - pc_2 $$ $$ u_3 y_3^2 = x - pc_3 $$ has a solution $(x, y_1, y_2, y_3)$ in $\mathbb{Q}_p$ if the other system $$ u_1 a_1^2 + pb_1^2 = 1 $$ $$ u_2 a_2^2 + pb_2^2 = 1 $$ has solutions $(a_1, b_1)$ and $(a_2, b_2)$ in $\mathbb{Q}_p$.
Context: I'm reading this paper by Sir Swinnerton-Dyer and he makes this statement and I'm trying to fill in the details. Paper: here
On page 521: "Let $\mathbf{u} = (u_1, u_2, u_3)$ be in $U_\mathcal{B}$ and let $p$ be a prime dividing $b$. If the $u_i$ are all units at $p$, the condition that the $2$-covering associated with $\mathbf{u}$ should be locally soluble at $p$ is $$ \chi_p(u_1) = \chi_p(u_2) = 0 $$."
The relevant $2$-covering associated with a triple $(m_1, m_2, m_3)$ (which WLOG can assume $m_i$ squarefree) is the system of equations $$ m_1 y_1^2 = x - bc_1 $$ $$ m_2 y_2^2 = x - bc_2 $$ $$ m_3 y_3^2 = x - bc_3 $$ I think WLOG, one can assume that $b$ and $c_i$ have no common factors.
Also, the author defined $$ \chi_p(u) = (u, p)_p $$ where $(u, v)_p$ is the additive Hilbert symbol $$ (u, v)_p = \begin{cases} 0 & \text{ if } ux^2 + vy^2 = 1 \text{ is soluble in } \mathbb{Q}_p \\ 1 & \text{otherwise} \end{cases} $$
Counterexample: $p=2, u_1:=u_2:=-1, u_3:=1, c_1:=1, c_2:=c_3:=5$.
Namely, the "second system" is now solved by $a_i=b_i=1$, but if there existed $x, y_i$ which solve the first system, the second and third equation of that first system would imply that $y_2=y_3=0$ (because $-1$ is not a square in $\mathbb Q_2$), a fortiori $x=5p=10$. But then the first equation of the first system would read $-y_1^2=8$ which is not true for any $y_1 \in \mathbb Q_2$.
I am pretty sure that for all $p\neq 2$, the second system and $p \nmid u_i$ force all $u_i$ to be squares in $\mathbb Q_p$, relying on the crucial fact that there, all elements of $1+p\mathbb Z_p$ are squares. This in turn makes the first system easily solvable, but then the whole approach seems like overkill. Also, in the case $p=2$, the conditions almost still force all $u_i$ to be squares, but there is this accidental extra solution of $u=-1$ which led me to the above counterexample.