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Could you give me some hints how I can solve the following exercise?

Check if the equation $3x^2+7y^2-5z^2=0$ has a non-trivial solution in $\mathbb{Q}$ . If it has a solution, find at least one. If it does not have, find at which p-adic fields it has no rational solution.

EDIT:

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.
    1. $\forall p \in \mathbb{P} \setminus \{ 2 \}, p \mid a$, $\exists r \in \mathbb{Z}$ 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 $p \mid c$.
    2. If $a,b,c$ are all odd, then there are two of $a,b,c$, so that their sum is divided by $4$.
    3. If $a$ even, then $b+c$ or $a+b+c$ is divisible by $8$. Similar, if $b$ or $c$ even.
Andrés E. Caicedo
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evinda
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2 Answers2

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After playing around, the suspicion is that there is no solution. In the absence of a single prime that jumps to your eye (three is just too many) start checking $p=2$, which is always good for a surprise. Any rational solution in $\mathbb Q_2$ may be scaled to that $x,y,z$ are in $\mathbb Z_2$ and at least one of them is odd. Then consider this in $\mathbb Z/4\mathbb Z$, where squares are either $0$ or $1$: We have $3x^2+7y^2-5z^2\equiv 3(x^2+y^2+z^2)\not\equiv 0\pmod 4$ because $x^2+y^2+z^2$ is $1$ or $2$ or $3$.

Hagen von Eitzen
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  • Hagen von Eitzen I edited my post and added a theorem.

    The first sentence is satisfied. But, the second sentence isn't for $p=3$. So, the equation has no non-trivial solutions in $\mathbb{Q}$, right? Do I have to check the second sentence also for $p=7,5$ to see if there is a solution in $\mathbb{Q}_3$ and $\mathbb{Q}_7$ ?

     – evinda Nov 05 '14 at 21:43
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    @evinda, that theorem is strictly for $\Bbb{Q}$, and does note say anything about non-existence of solutions in the $p$-adics. Do observe that because $\Bbb{Q}$ is a subset of all the $p$-adic fields, the existence of a rational solution implies the existence of a $p$-adic solution for all $p$. – Jyrki Lahtonen Nov 05 '14 at 21:54
  • @JyrkiLahtonen A ok.. So, to check if the equation $3x^2+7y^2-5z^2=0$ has a solution in $\mathbb{Q}_5$ and in $\mathbb{Q}_7$, do I have to write it modulo $5$ and $7$ ?

    If so, it is like that:

    $$ 3x^2+7y^2 \equiv 0 \pmod 5 \Rightarrow 3x^2+2y^2 \equiv 0 \pmod 5 $$

    and

    $$ 3x^2-5z^2 \equiv 0 \pmod 7 \Rightarrow 3x^2+2z^2 \equiv 0 \pmod 7 $$

    How can I check if the above congruences have solutions?

     – evinda Nov 05 '14 at 22:00
  • The brute force way is to try all the possibilities. $x,y\in{0,1,2,3,4}$ in the first, $x,z\in{0,1,2,3,4,5,6}$ in the second. What else? Ok, you list the possible values of $3x^2,2y^2$ and $2z^2$ w.r.t. the relevant modulus to speed up things. – Jyrki Lahtonen Nov 06 '14 at 08:53
  • @JyrkiLahtonen I understand..So, in this way we can check, if there are solutions in $\mathbb{Q}_2,\mathbb{Q}_3, \mathbb{Q}_5, \mathbb{Q}_7$, right? :) $$$$ – evinda Nov 07 '14 at 08:25
  • @JyrkiLahtonen But, I still havent understood what we do, in order to check, if there are solutions in $\mathbb{Q}_p$.. Could you maybe explain it again to me? We want to apply Hensels lemma.. How can we find an $\alpha$, such that $F(\alpha) \equiv 0 \mod p\mathbb{Z}_p$ and $F`(\alpha)\not\equiv 0 \mod p\mathbb{Z}_p$? I am confused right now... – evinda Nov 07 '14 at 08:31
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    If you find $x_0,y_0\in{0,1,2,\ldots,p-1}$ such that $3x_0^2\equiv 5-7y_0^2\pmod p$ (so $z=1$), and $x_0\neq0$, then you use $\alpha=x_0$ and $F(x)=3x^2+(7y_0^2-5)$. – Jyrki Lahtonen Nov 07 '14 at 10:18
  • @JyrkiLahtonen Can we set $z=1$, no matter which equation we have? – evinda Nov 11 '14 at 23:03
  • @evinda: If the equation is homogeneous (all the terms have the same degree), then $(x,y,z)$ is a solution, if and only if $(\lambda x,\lambda y, \lambda z)$ is a solution for $\lambda\neq0$. So if there exists a solution $(x,y,z)$ with $z\neq0$, then by selecting $\lambda=1/z$ we get ... – Jyrki Lahtonen Nov 12 '14 at 05:17
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Just ordinary integers: what happens, exactly, when $$ 3 x^2 + 7 y^2 - 5 z^2 \equiv 0 \pmod 9? $$

More formally, see my answers at Isotropy over $p$-adic numbers

Community
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Will Jagy
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  • Will Jagy I edited my post and added a theorem.

    The first sentence is satisfied. But, the second sentence isn't for $p=3$. So, the equation has no non-trivial solutions in $\mathbb{Q}$, right? Do I have to check the second sentence also for $p=7,5$ to see if there is a solution in $\mathbb{Q}_3$ and $\mathbb{Q}_7$ ?

     – evinda Nov 05 '14 at 21:44