Let $\alpha, \beta , \gamma$ are non-zero elements of $Q_p$, show that $$(\alpha\gamma,\beta\gamma)=(\alpha,\beta)(\gamma,-\alpha\beta)$$, where $(\alpha,\beta)=1 $ or $-1$ whether $X^2-\alpha Y^2-\beta Z^2$ represents $0$ or not, i.e. Hilbert symbol.
I was thinking but Could not find any hint. Somebody please tell me how to show it.
If I let $\alpha=p^iu , \beta=p^jv , \gamma=p^kt$ and use $(\alpha,\beta)=(w/p)$ where $w=(-1)^{ij}u^jv^i$ formula for odd primes, it is not helping.
(note: for simplicity of writing I used $a,b,c$ instead of $\alpha, \beta, \gamma$ - my Greek letters are on my other keyboard).
Split both sides down using multiplicativity: $(ac, bc) = (a,b) (a,c) (b,c) (c,c)$, $(a,b)(c,-ab) = (a,b)(c,-1)(a,c)(b,c)$. All that remains to prove is $(c,c) = (c,-1)$, which is a consequence of $(c,c)(c,-1) = (c,-c) = 1$.
However, this is non-constructive, i.e. given two solutions of $a x^2 + b y^2 - z^2 = 0$ and $c x^2 - ab y^2 - z^2 = 0$, it does not give a solution of $ac x^2 + bc y^2 -z^2 = 0$ (and whatever happens in the non-square case).