Show that $x^2+5y^2=2z^2$ has only trivial solution $(x,y,z)=(0,0,0)$

Question

I have no way to start with this problem. Can anyone give me a hint?

Show that $x^2+5y^2=2z^2$ has only trivial solution $(x,y,z)=(0,0,0)$

Answer 1

Assume that there is no prime dividing all of them (by them I mean $x,y,z$). Otherwise, factor out by it.

Looking at it modulo $5$ it says: $x^2=2z^2$. The squares modulo $5$ are $0,1,4$. This means that if $x\neq 0$ then $z\neq 0$, the possibilities remaining do not give a result.

Then, $x$ is a multiple of $5$, and so is $z$. Then $25\mid 5y^2$, so $y$ is a multiple of $5$. Contradiction.

Answer 2

$2$ is not a quadratic residue modulo $5$, so if you consider the equation modulo $5$, you get

$$x^2 \equiv 2z^2 \pmod{5},$$

which has only the trivial solution $x \equiv z \equiv 0 \pmod{5}$.

Inserting that into the original, it follows that $5^2 \mid 5y^2 \Rightarrow 5 \mid y$. Dividing out the common factor $5$ gives a solution $\xi^2 + 5\eta^2 = 2 \zeta^2$, which again has all three multiples of $5$. etc. ad infinitum.