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.
Best Answer
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: $$4^2=7+3^2.$$