How do we conclude that (a,b)=1

[evinda]

Hi! (Nerd)

Suppose that we have the equation $ax^2+by^2+cz^2=0, a,b,c \in \mathbb{Q}$.

Without loss of generality, we suppose that $gcd(a,b,c)=1$.

Also, we can consider that $a,b,c$ are square-free.

We can suppose that $(a,b)=(b,c)=(a,c)=1$.

Proof of the last sentence:

Let $d=(a,b)>1$

Since $a,b$ are square-free $\Rightarrow d$ is also square-free.
$$a=da' \\ b=db'$$
$$\Rightarrow (a',b')=1$$

It stands that $(d,c)=1$.

Proof:

Let $d'=(d,c)>1$

$$\Rightarrow \begin{Bmatrix}
d' \mid d\\ \\

d' \mid c
\end{Bmatrix} \Rightarrow \begin{Bmatrix}
d' \mid a\\ \\
d' \mid c
\end{Bmatrix} \Rightarrow d' \mid (a,b,c)=1, \text{ Contradiction} \Rightarrow (d,c)=1$$

So, when $(x,y,z)$ an integer solution of $ax^2+by^2+cz^2=0 \\ \Rightarrow d(a'x^2+b'y^2)+cz^2=0 \\ \Rightarrow d \mid cz^2, (d,c)=1 \Rightarrow d \mid z^2$

Since $d$ is square-free, $d \mid z^2 \Rightarrow d \mid z \Rightarrow z=dz'$

The initial equation:
$$a'x^2+b'y^2+(cd)z^2=0$$

We concluded to an equivalent equation, where $(a',b')=1$

With the same procedure, we prove that $(a,b)=(b,c)=(c,d)=1$
Could you explain me why, having the equivalent equation, we conclude that $(a,b)=1$ ?

 

[jvicens]

Hi there! It seems like you are trying to prove that $gcd(a,b,c)=1$ in the given equation. Let me explain it in a simpler way:

1. Suppose we have the equation $ax^2+by^2+cz^2=0$, where $a,b,c$ are integers and $gcd(a,b,c)=1$.
2. Without loss of generality, we can assume that $a,b,c$ are square-free, meaning they do not have any repeated prime factors.
3. We can also assume that $(a,b)=(b,c)=(a,c)=1$, which means that $a,b,c$ do not have any common prime factors.
4. Now, let's consider the equation $ax^2+by^2+cz^2=0$ and suppose that $(x,y,z)$ is an integer solution.
5. From the given information, we know that $gcd(a,b,c)=1$, which means that $a,b,c$ do not have any common prime factors.
6. Since $a,b,c$ are square-free, it follows that $gcd(a,b)=gcd(b,c)=gcd(a,c)=1$, which means that $a,b,c$ do not have any common prime factors with each other.
7. Therefore, we can conclude that $(a,b)=(b,c)=(a,c)=1$, which proves the last sentence in the given forum post.

I hope this helps clarify the reasoning behind the proof. Let me know if you have any other questions. Keep up the good work!