Proof that a diophantine equation does not have non-trivial solutions

Question

Consider integer variables $x,y,z$ and the diophantine equation $$z^2=6x^2+2y^2$$ I have the following proof that the above equation does not have a non-trivial integer solution i.e. the only solution is $(x,y,z)=(0,0,0)$

If $z^2 = 2(3x^2 +y^2)$ has an integral solution, then $x$ and $y$ must be odd. That is, $x^2 \equiv 1(\mathrm{mod}\ 8)$ and $y^2 \equiv 1(\mathrm{mod}\ 8)$. This implies that $3x^2 +y^2 \equiv4(\mathrm{mod}\ 8)$. This means that $z^2$ must be divisible by 8 but not by 16. Clearly there is no such integer $z$.

I understand the proof except two things:

• What is an "integral" solution? Does the author mean integer solution?
• Why $x$ and $y$ must be odd?

Related to that is the following diophantine equation $$z^2=nx^2+y^2$$ for integer $n>6$ and I want to find the smallest possible $n$ s.t. the equation does not have a non-trivial integer solution. I tried $n=7$ i.e. $$z^2=7x^2+y^2$$ but I don't know how to start.

Answer 1

For the second part, if $z^2=nx^2+y^2$ then $$z^2-y^2=nx^2=(z+y)(z-y)$$

The two factors differ by $2y$ hence are both odd or both even. You can choose $x=2$ and set $z+y=2n, z-y=2$ so that $z=n+1, y=n-1$ to get a solution for any $n$.

There are many other solutions - I chose $x=2$ to guarantee I could get two factors of the same parity whatever $n$ would be. You are only at a loss for values of $y$ and $z$ if $n\equiv 2 \bmod 4$ and $x$ is odd.

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So there is a solution

$$(n+1)^2=n\times 2^2+(n-1)^2$$ for every $n$

My other observation was in relation to how free we might be to choose $x$ once we are given a value of $n$ - for most values of $n$ there are lots of solutions.

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The point here is, to follow what I think is being said in comments, that in $n$ is odd - say $11$, we can choose $x$ to be any number we like - odd or even, and there will be values of $y$ and $z$ that we can find.

Say $x=3$, then $(z+y)(z-y)=99$. Let $z+y=11, z-y=9$ so that $z=10, y=1$ we have $$10^2=11\times 9 +1$$

If $x$ were $4$, we'd have $(z+y)(z-y)=176$ and we could choose $z+y=22, z-y=8$ with $z=15, y=7$ and $$225=11\times 16+49$$

This works for any odd $n$.

Suppose $n=2$ so that $n$ is even and not divisible by $4$, then choose $x$ even, say $x=2$ so that $(y+z)(y-z)=8$ and we can have $z+y=4, z-y=2$ with $z=3, y=1$ and $$9=2\times 4 +1$$

However if we try $x$ odd in this case, say $x=2m+1$ we have $$(z+y)(z-y)=2x^2$$ Now $x^2=4m(m+1)+1$ is one more than a multiple of $8$ and $2x^2$ is $2$ more than a multiple of $16$, and is not divisible by $4$. However $z+y$ and $z-y$ have the same parity so their product is either odd or a multiple of $4$.

Answer 2

If $$6x^2+2y^2-z^2=0$$ has an integral solution (one with integer values), then it has one with $\gcd(x,y,z)=1$. If $x$ and $y$ have opposite parities then $3x^2+y^2$ is odd, so that $z^2=2(3x^2+y^2)$ is twice an odd number, which is impossible for a square. So $x$ and $y$ must have the same parity. But $z^2$ is even, so $z$ is even, so if $x$ and $y$ are even, then $2\mid\gcd(x,y,z)$. So we may reduce to the case where $x$ and $y$ are both odd.

$n=7$ doesn't work: $$4^2=7+3^2.$$

Answer 3

For any even positive integer $n,$ you get nontrivial $x^2 + n y^2 = z^2$ with $$ (n-1)^2 + n ( 2)^2 = (n+1)^2 $$

When $n$ is odd, switch to $$ \left( \frac{n-1}{2} \right)^2 + n ( 1)^2 = \left( \frac{n+1}{2} \right)^2 $$

Answer 4

Concerning the diophantine equation (1) $nx^2+y^2=z^2$ for $n\in \mathbf N$, allow a late comer to bring some complements to the answers given by @Will Jagy and by @Mark Bennett.

1) The homogeneity of equation (1) suggests to first determine the rational solutions, and actually one can parametrize all of them. The case $z=0$ being equivalent to the rationality of $\sqrt n$, we may suppose $z\neq 0$ and divide by $z$. Writing $x'=x/z, y'=y/z$, one then gets the rational equation (2) $y'^2+nx'^2=1$ , or $N(\alpha)=1$, where $\alpha=y'+x'\sqrt {-n} \in \mathbf Q(\sqrt {-n})$ and $N$ denotes the norm map of $\mathbf Q(\sqrt {-n})/\mathbf Q$. This amounts to determine all the elements of norm $1$ of the quadratic field $\mathbf Q(\sqrt {-n})$. A quick solution is given by Hilbert's thm. 90, which states that any such $\alpha$ is of the form $\sigma (v)/v$, where $v\in \mathbf Q(\sqrt {-n})^*$ and $\sigma$ is the automorphism sending $\sqrt {-n}$ to $-\sqrt {-n}$, i.e. $\alpha=(a-b\sqrt {-n})/(a+b\sqrt {-n})$, with $a,b \in \mathbf Q$, or equivalently Re$(\alpha)=(a^2-nb^2)/(a^2+nb^2)$, Im$(\alpha)=2ab/(a^2+nb^2)$. This gives the parametrization of all the rational solutions of (2).

2) Conversely,if $(x',y')$ is a solution of (2), writing $x'=x/z,y'=y/z$ (after reduction to a common denominator if necessary), one gets the integral solutions of (1). This is analogous to the determination of the pythagorean triples.

NB: The key simplification lies in the square in the RHS . In general the study of the diophantine equation $nx^2+y^2=m$ belongs to CFT, see e.g. the book of D. Cox, "Primes of the form $x^2+x^2+ny^2$" ./.