quadratic equation modulo product of coprime squarefree integers Let $a, b, c$, pairwise coprime squarefree integers.
Suppose $au^2 + bv^2 + cw^2 ≡ 0 (mod\ |abc|)$ with $au^2 , bv^2 , cw^2$  pairwise coprime.
Prove that if $(x,y,z) \in Λ_0 := \{(x, y, z) ⊂ \Bbb Z^3 : aux + bvy + cwz ≡ 0 (mod\ |abc|)\}$ then $ax^2 + by^2 + cz^2 ≡ 0 (mod\ |abc|)$.
I don't know where to start. I tried to prove that $ax^2 + by^2 + cz^2 ≡ 0 (mod\ |a|)$ and do similarly modulo $b$ and $c$ but didn't succeed.
Thank you for your help.
 A: You have that $a, b, c$ are pairwise coprime squarefree integers, where
$$au^2 + bv^2 + cw^2 \equiv 0 \pmod{|abc|} \tag{1}\label{eq1A}$$
with $au^2, bv^2, cw^2$ being pairwise coprime. Note this means that $u$ and $v$ are coprime with $c$. This can be seen quite easily from assuming otherwise, e.g., $\gcd(u,c) = d \gt 1$, then $d \mid bv^2$, but since $\gcd(b,c) = 1$, then $d \mid v^2$, so $au^2$ and $bv^2$ can't then be pairwise coprime.
Next, the question says that if $(x,y,z)$ is such that
$$aux + bvy + cwz \equiv 0 \pmod{|abc|} \tag{2}\label{eq2A}$$
then to prove that
$$ax^2 + by^2 + cz^2 \equiv 0 \pmod{|abc|} \tag{3}\label{eq3A}$$
I'll first show \eqref{eq3A} is true modulo $|c|$. From \eqref{eq1A}, we have
$$au^2 + bv^2 \equiv 0 \pmod{|c|} \implies au^2 \equiv -bv^2 \pmod{|c|} \tag{4}\label{eq4A}$$
From \eqref{eq2A}, we have
$$aux + bvy \equiv 0 \pmod{|c|} \implies aux \equiv -bvy \pmod{|c|} \tag{5}\label{eq5A}$$
Multiplying both sides of \eqref{eq5A} by $u$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
(au^2)x & \equiv -buvy \pmod{|c|} \\
(-bv^2)x & \equiv -buvy \pmod{|c|} \\
vx & \equiv uy \pmod{|c|}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
In the last line, I used that $b$ and $v$ are relatively prime to $c$. Next, multiply both sides of \eqref{eq6A} by $bvy$, plus use \eqref{eq4A} and \eqref{eq5A}, to get
$$\begin{equation}\begin{aligned}
(bv^2)xy & \equiv buvy^2 \pmod{|c|} \\
(-au^2)xy & \equiv buvy^2 \pmod{|c|} \\
(-au)xy & \equiv bvy^2 \pmod{|c|} \\
-a(uy)x & \equiv bvy^2 \pmod{|c|} \\
-a(vx)x & \equiv bvy^2 \pmod{|c|} \\
-ax^2 & \equiv by^2 \pmod{|c|} \\
ax^2 + by^2 & \equiv 0 \pmod{|c|}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Similarly, you can show that $by^2 + cz^2 \equiv 0 \pmod{|a|}$ and $ax^2 + cz^2 \equiv 0 \pmod{|b|}$. As $a,b,c$ are relatively prime to each other, you can put these together to get that \eqref{eq3A} holds.
Note I did not use anywhere that $a,b$ and $c$ are square-free integers, so I'm note sure why this was stated as the question's result holds without that requirement.
