# 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.

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{aligned} (au^2)x & \equiv -buvy \pmod{|c|} \\ (-bv^2)x & \equiv -buvy \pmod{|c|} \\ vx & \equiv uy \pmod{|c|} \end{aligned}\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{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}\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.

• Excellent, thank you for your thourough answer – PerelMan Oct 1 at 2:00