<s>[INST] The equations $ax^2+by^2+cz^2=0$ and $ax^2+by^2+cz^2=1$ 
Let $a,b,c -$ a non-zero integers. It is known that the equation $ax^2+by^2+cz^2=0$ it has a non-zero integer solution. Prove that the equation $ax^2+by^2+cz^2=1$ has solution in rational numbers.

My work so far:
Let $a>0,b>0,c<0$ and $z<0$.
Let $(x,y,z) -$ solution of $ax^2+by^2+cz^2=0$. Then:
If $c=-n^2$ that
$$ax^2+by^2+cz^2=0 \Leftrightarrow$$
$$ax^2+by^2-n^2z^2=0 \Leftrightarrow$$
$$25ax^2+25by^2-25n^2z^2=0 \Leftrightarrow$$
$$a(5x)^2+b(5y)^2-16n^2z^2=9n^2z^2 \Leftrightarrow$$
$$a\left(\frac{5x}{3nz}\right)^2+b\left(\frac{5y}{3nz}\right)^2+c \left(\frac{4}{3n}\right)^2=1$$
Hence, $\left(\frac{5x}{3nz};\frac{5y}{3nz};\frac{4}{3n}\right)$ solution of $ax^2+by^2+cz^2=0$.
Let $c\not=-n^2$.
I need help here...
 [/INST] By the hypothesis, there exists rational $u,v$ such that $au^2+b v^2+c=0$. This means that $a(-\frac{u^2}{c})+b(-\frac{v^2}{c})=1$. Now we want a solution of the equation $ax^2+b y^2+c z^2=1$ in rational numbers $x,y,z$. Replacing $c$ by $-au^2-bv^2$ Give $a(x^2-u^2z^2)+b(y^2-v^2z^2)=1$. Hence, if we can find rational $x,y,z$ such that $x^2-u^2z^2=-\frac{u^2}{c}$ and $y^2-v^2z^2=-\frac{v^2}{c}$, we are done. This imply that $(\frac{x}{u})^2=z^2-\frac{1}{c}= (\frac{y}{v})^2$. So we put $x=ku$ and $y=kv$, with the property that $k^2=z^2-1/c$. There is a lot of solutions, we can take $2k=1-1/c$ and $2z=(1+1/c)$, this gives $2x=(1-1/c)u$ and $2y=(1-1/c)v$.
It remain to show that these values gives a solution, that is easy. 
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