$x^2 + 2y^2=3^k$

Question

Let $x,y \in \mathbb Z$ such that both $x$ and $y$ are not factor of $3$. Prove that $\forall k \in \mathbb N$ there exists $(x , y)$ that $x^2 + 2y^2 = 3^k$.

I know that $x^2 + 2y^2$ is divisible by $3$ always. But how can we prove that there exists a solution $\forall k \in \mathbb N$? Thank you!

Answer 1

Use these:

$1^2+2\cdot1^2=3$

$1^2+2\cdot2^2=3^2$

$(3x)^2+2(3y)^2=3^2(x^2+2 y^2)$