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 have, I have to find at which p-adic fields it has no rational solution.


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.

The first sentence is satisfied.

For the second one:


$$5+x^2(-7) \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 2 \mod 3$$ $$\left ( \frac{2}{3} \right)=-1$$

So, we see that the equation hasn't non-trivial solutions in $\mathbb{Q}$.


To check if there is a solution in $\mathbb{Q}_2$, we use 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 $p=3,5,7$, we use 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? I haven't understood it..

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    $\begingroup$ Rational means rational means rational. You need to say "in which $p$-adic fields does the equation have no solution?" Don't preface "solution" with rational, because if there is no solution in $\Bbb Q$ then there is no "rational" solution! $\endgroup$
    – anon
    Nov 4, 2014 at 17:45
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    $\begingroup$ For all $p$ you can simply set $z=1$ and use the same technique I used in my answer to your previous question. In other words, so that you can then pick $y$ from the set $\{0,1,2,\ldots,p-1\}$ in such a way that Hensel's Lemma proves the existence of an $x\in\Bbb{Z}_p$. The primes 2,3,5,7 need to be dealt with separately. I don't know the answer yet. $\endgroup$ Nov 4, 2014 at 22:34
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    $\begingroup$ I'll look at your theorem later and see how it's relevant, I have to leave in a moment. For your second question - um, what is $x$? What is it? What number, or residue, is it mod $p$? Lastly, for your third question, why would you even think that? I mean, $x^2+y^2+z+1=0$ has a solution, but that doesn't mean it has a solution with $z=1$. The converse of a true implication is not necessarily true. $\endgroup$
    – anon
    Nov 5, 2014 at 18:50
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    $\begingroup$ You still need to prove that you can choose $y$ from the set $\{0,1,2,\ldots,p-1\}$ in such a way that there exist an $x\in\{1,\ldots,p-1\}$ such that $(x,y,1)$ is a solution modulo $p$. That part of my argument goes through with the obvious changes here. Some care is needed to make sure that $x\neq0$. $\endgroup$ Nov 5, 2014 at 19:49
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    $\begingroup$ The reason we suppose $z=1$ is because it's convenient and works. You understand that if there's a solution with $z=1$ then there exists a solution, right? Anyway, your theorem is for $\Bbb Q$, so I don't see how you're applying it for $\Bbb Q_p$. $\endgroup$
    – anon
    Nov 5, 2014 at 23:54

1 Answer 1


Your theorem tells you straight away that the equation has no solutions in $\mathbb{Q}$ (that is, no rational solution). As was also pointed out in the earlier comments, a solution in $\mathbb{Q}_p$ is NOT a 'rational solution', as the values of $x, y$ and $z$ need not be rationals! Without seeing the proof for your theorem, you shouldn't really invoke certain parts of it as you've done afterwards; for example, you've taken a local condition at 3, shown that it's not satisfied, and concluded that there is no solution in $\mathbb{Q}_3$. It's true that no such solution exists - but as your theorem stands, you've not given a proof. The theorem is proved using the Hasse-Minkowski theorem (which says that such equations have a rational root if and only if they have one everywhere locally), and during the proof you will end up showing the following, which is what you need:

Lemma: Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p|a$, and $Q: ax^2 + by^2 + cz^2 = 0$ a quadratic form. Then there is a solution to $Q$ over $\mathbb{Q}_p$ if and only if $-b/c$ is a square mod $p$.

Proof: Suppose a solution exists. Then by scaling, we may assume that $y$ and $z$ lie in $\mathbb{Z}_p$, and indeed that they lie in $\mathbb{Z}_p^\times$. (Suppose without loss of generality that $y \in p\mathbb{Z}_p;$ then as $p|a$, $p|y$ and $p\nmid c$, we must have $p|z,$ and hence $p|x$ by considering the parity of the exponent of $p$ in the sum. So we can divide our trio $(x,y,z)$ by $p$.)

Now consider the equation mod $p$. This becomes equivalent to $(z/y)^2 \equiv -b/c$ mod $p,$ where we can divide by $c$ and $y^2$ as they have invertible image in $\mathbb{F}_p$. So $z/y$ gives the corresponding element of $\mathbb{F}_p$ that squares to $-b/c$.

Conversely, suppose that we have $Z^2 \equiv -b/c$ mod $p$. Then in particular, $p\nmid Z$ and the trio ($x,y,z) = (0,1,Z)$ gives a solution mod $p$. Using Hensel's lemma (and using that $p\neq 2$) we see that this lifts to a solution in $\mathbb{Q}_p$, as required.

This Lemma applies directly to your case, with the obvious symmetry for $p|b$ or $p|c$. In particular, you've shown that $5/7 \equiv 2$ is not a square mod 3, hence there is no solution in $\mathbb{Q}_3$, that $3/7 \equiv 4$ is a square mod 5, so there is a solution in $\mathbb{Q}_5$, and $-5/3\equiv 3$ is not a square mod 7, so there isn't a solution in $\mathbb{Q}_7$.

For primes not dividing $abc$ and not equal to 2, as has already been pointed out, you can use Hensel easily to show that solutions exist via a counting argument on quadratic residues mod $p$.

The case of $\mathbb{Q}_2$ is trickier, as we can't use Hensel or the Lemma directly. Can you prove a similar Lemma to the one above to show that the local conditions at 2 in your theorem are precisely the ones that determine when there is a solution in $\mathbb{Q}_2$? Once you've done that, your work above shows that there is a solution in $\mathbb{Q}_2$, and furthermore you've done all of the work in proving your theorem (once you invoke Hasse-Minkowski, of course!).

  • $\begingroup$ The formula in General there. math.stackexchange.com/questions/738446/… Using the condition for the existence of solutions you can tell when decisions will be. This is to ascertain whether there is an equivalent quadratic form when the root is intact. Usually all boils down to the representation of a number as a sum of squares. But it is not satisfied. And continues to try to find condition as the solution module. $\endgroup$
    – individ
    Nov 7, 2014 at 10:16
  • $\begingroup$ Chris Williams Did you conclude that $\frac{-b}{c}$ is, for example, not a square mod $7$, since the Legendre symbol is equal to $\left ( \frac{\frac{-b}{c}}{7} \right )=-1$ ? $$$$ In my notes, there is the following sentence: If $2 \nmid abc$ and $a+b \equiv 0 \pmod 4$, then the equation $ax^2+by^2+cz^2=0$ has at least one non-trivial solution in $\mathbb{Q}_2$. Can I use this? If so, is it like that? $$$$ $a+b=3+5 \equiv 8 \equiv 0 \pmod 4$ So, there is no solution in $\mathbb{Q}_2$. Or am I wrong? $\endgroup$
    – evinda
    Nov 7, 2014 at 16:28
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    $\begingroup$ For your other question about counting arguments: there are precisely $(p+1)/2$ quadratic residues mod $p$ for any odd prime $p$ (as for any non-zero class $x$ mod $p$, we have $x^2 \equiv (-x)^2$ mod $p$ and $x \neq -x$ mod $p$, whilst the equation $X^2 \equiv a$ mod $p$ has at most 2 solutions in $\mathbb{F}_p \cong \mathbb{Z}/p\mathbb{Z}$. Thus the non-zero elements mod $p$ pair up and we get $(p-1)/2$ non-zero residues). Then fix $z=1$. There are $(p+1)/2$ possible values of $3x^2$ and $-7 + 5y^2$ for $x,y$ mod $p$. Thus, by the pigeonhole principle, $\exists x,y$ s.t. we have a solution! $\endgroup$ Nov 7, 2014 at 18:23
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    $\begingroup$ Suppose $a$ is a non-zero residue. Then $a \equiv x^2$ mod $p$ for some $x$ mod $p$. Then we also have $a \equiv (-x)^2$ mod $p$, and $x\neq -x$ mod $p$ (as then $2x \equiv 0$ mod $p$, so $x \equiv 0$ mod $p$ contradiction). So as there are at MOST 2 solutions to $X^2 - a \equiv 0$ mod $p$, $x$ and $-x$ are precisely the elements that square to $a$ mod $p$. Thus all of the $(p-1)$ non-zero elements in $\mathbb{Z}/p\mathbb{Z}$ pair off, with each pair giving a unique residue. So there are $1 + (p-1)/2 = (p+1)/2$ residues (where we add 1 as 0 is a residue). $\endgroup$ Nov 7, 2014 at 18:33
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    $\begingroup$ Sorry, I've realised I made a sign error in one of the comments above; it should read 'there are $(p+1)/2$ possible values of $3x^2$ and $(p+1)/2$ possible values of $7 - 5y^2$ for $x,y$ mod $p$.' Then by the PP you have that for some $x,y$, we have $3x^2 = 7 -5y^2$ mod $p$, which then gives a solution. (I didn't have space to write that last sentence in the above comment, either). $\endgroup$ Nov 7, 2014 at 18:43

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