- #1
evinda
Gold Member
MHB
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Hello! (Wave)
I have to check if the equation $3x^2+5y^2-7z^2=0$ has a non-trivial solution in $\mathbb{Q}$. If it has, I have to find at least one. If it doesn't, I have to find at which p-adic fields it has no solution.
I used the following theorem:
We suppose that $a,b,c \in \mathbb{Z}, (a,b)=(b,c)=(a,c)=1$.
$abc$ is square-free. Then, the equation $ax^2+by^2+cz^2=0$ has a non-trivial solution in $\mathbb{Q} \Leftrightarrow$
1. $a,b,c$ do not have the same sign.
2. $\forall p \in \mathbb{P} \setminus \{ 2 \}, p \mid a$, $\exists r \in \mathbb{Z}$ such that $b+r^2c \equiv 0 \pmod p$ and similar congruence for the primes $p \in \mathbb{P} \setminus \{ 2 \}$, for which $p \mid b$ or $p \mid c$.
3. If $a,b,c$ are all odd, then there are two of $a,b,c$, so that their sum is divided by $4$.
4. If $a$ even, then $b+c$ or $a+b+c$ is divisible by $8$.
Similar, if $b$ or $c$ even.The first sentence is satisfied.
For the second one:
$$p=3:$$
$$5+x^2(-7) \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 2 \mod 3$$
$$\left ( \frac{2}{3} \right)=-1$$So, we see that the equation hasn't non-trivial solutions in $\mathbb{Q}$.
To check if there is a solution in $\mathbb{Q}_2$, I used the following lemma:
If $2 \nmid abc$ and $a+b \equiv 0 \pmod 4$, then the equation $ax^2+by^2+cz^2=0$ has at least one non-trivial solution in $\mathbb{Q}_2$.
In our case, $a+b=8 \equiv 0 \pmod 4$, so there is no solution in $\mathbb{Q}_2$, right?
For $p=3,5,7$, I used the following lemma:
Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p \mid a$, and $Q: ax^2+by^2+cz^2=0$ a quadratic form.
Then there is a solution to $\mathbb{Q}$ over $\mathbb{Q}_p$ iff $-\frac{b}{c}$ is a square $\mod p$.
$$\left( -\frac{5}{-7}\right)=\left( \frac{5}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_3$.
$$\left( \frac{-3}{-7} \right)=\left( \frac{3}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_5$.
$$\left( -\frac{3}{5}\right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_7$.
It remains to check if the equation has non-trivial solutions in $\mathbb{Q}_p, p \neq 2,3,5,7$.
Can we do this, by only using the pigeonhole principle?
Or do we have to apply Hensel's Lemma? If so, how could we do this? (Thinking)
I have to check if the equation $3x^2+5y^2-7z^2=0$ has a non-trivial solution in $\mathbb{Q}$. If it has, I have to find at least one. If it doesn't, I have to find at which p-adic fields it has no solution.
I used the following theorem:
We suppose that $a,b,c \in \mathbb{Z}, (a,b)=(b,c)=(a,c)=1$.
$abc$ is square-free. Then, the equation $ax^2+by^2+cz^2=0$ has a non-trivial solution in $\mathbb{Q} \Leftrightarrow$
1. $a,b,c$ do not have the same sign.
2. $\forall p \in \mathbb{P} \setminus \{ 2 \}, p \mid a$, $\exists r \in \mathbb{Z}$ such that $b+r^2c \equiv 0 \pmod p$ and similar congruence for the primes $p \in \mathbb{P} \setminus \{ 2 \}$, for which $p \mid b$ or $p \mid c$.
3. If $a,b,c$ are all odd, then there are two of $a,b,c$, so that their sum is divided by $4$.
4. If $a$ even, then $b+c$ or $a+b+c$ is divisible by $8$.
Similar, if $b$ or $c$ even.The first sentence is satisfied.
For the second one:
$$p=3:$$
$$5+x^2(-7) \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 2 \mod 3$$
$$\left ( \frac{2}{3} \right)=-1$$So, we see that the equation hasn't non-trivial solutions in $\mathbb{Q}$.
To check if there is a solution in $\mathbb{Q}_2$, I used the following lemma:
If $2 \nmid abc$ and $a+b \equiv 0 \pmod 4$, then the equation $ax^2+by^2+cz^2=0$ has at least one non-trivial solution in $\mathbb{Q}_2$.
In our case, $a+b=8 \equiv 0 \pmod 4$, so there is no solution in $\mathbb{Q}_2$, right?
For $p=3,5,7$, I used the following lemma:
Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p \mid a$, and $Q: ax^2+by^2+cz^2=0$ a quadratic form.
Then there is a solution to $\mathbb{Q}$ over $\mathbb{Q}_p$ iff $-\frac{b}{c}$ is a square $\mod p$.
$$\left( -\frac{5}{-7}\right)=\left( \frac{5}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_3$.
$$\left( \frac{-3}{-7} \right)=\left( \frac{3}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_5$.
$$\left( -\frac{3}{5}\right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_7$.
It remains to check if the equation has non-trivial solutions in $\mathbb{Q}_p, p \neq 2,3,5,7$.
Can we do this, by only using the pigeonhole principle?
Or do we have to apply Hensel's Lemma? If so, how could we do this? (Thinking)