I wish to prove the following lemma.
Lemma. Let $(a,b)_p$ be the Hilbert symbol evaluated at the prime $p$. Then, it satisfies the following property $$(a,b)_p=(ac^2,b)_p=(a,bc^2)_p.$$ I guess I see the rough idea: given the equation $ax^2+by^2=z^2$, we wish to see that the solutions to the equation $ac^2x^2+by^2=z^2$. Assume $(x,y,z)\neq (0,0,0)$ as the result is clear if not. Then, it follows that $(cx, y, z)$ is a solution to this equation which is non-zero and hence, is equal to $(a,b)_p$.
Do you think this is formal enough to prove the lemma?
I would write: For $\color{red}{c\neq 0}$ we have that:
$(x,y,z) \neq(0,0,0)$ is a solution to $ac^2X^2+bY^2=Z^2$
if and only if
$(cx,y,z) \neq(0,0,0)$ is a solution to $aX^2+bY^2=Z^2$.