Showing that there's no solution to the congruence $x^{2}+3y^{2}\equiv2\mod3$

Question

I was given the following question:

Let $x,y\in \mathbb{Z}$, show that the congruence $x^{2}+3y^{2}\equiv2\mod3$ has no solution.

Here's my attempt so far:

$x^{2}+3y^{2}\equiv2\mod3$

$\Rightarrow x^{2}+3y^{2}-2\equiv0\mod3$

Thus: $3\mid x^{2}+3y^{2}-2$

Therefore the summation have to be divisible by 3, so we can exclude $3y^{2}$.

Now, let's assume by contradiction that there's a $k\in\mathbb{Z}$ such that $x^{2}-2=3k$.

$\Rightarrow x^{2}-2-3k=0$

And I don't know how to continue.

Am I doing something wrong? What is the right direction?

Thank you!

Answer 1

hint

You just need to prove that there is no $ x\in \Bbb N$ such that

$$x^2\equiv 2 \pmod 3$$.

Let $ x \in \Bbb N$.

$$x=0\implies x^2\equiv 0\mod 3$$ $$x=1\implies x^2\equiv 1\mod 3$$ $$x=2\implies x^2\equiv 1\mod 3$$ assume that $x\ge 3$.

If $x \equiv 0\mod 3 $ then $x^2\equiv 0 \mod 3$.

If $ x$ is not a multiple of $3$, then $gcd(x,3)=1$ and by Fermat's Little theorem $$x^{\phi(3)}=x^2\equiv 1 \mod 3$$

Done.