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

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!

• Have you tried induction? – Eleven-Eleven Jun 9 '20 at 12:42
• Yes , but how can we do that? – dark.nes_s Jun 9 '20 at 12:43
• Hint: Think about expressions like $\left(1+\sqrt {-2}\right)^k=\left(a_k+b_k\sqrt {-2}\right)$. – lulu Jun 9 '20 at 12:46
• Well, I think the base case is obvious (can you think of an (x, y) for $x^2+2y^2=3$?). Note you also need the fact that $x$ and $y$ are not multiples of $3$. If you work now in mod $3$, and you square $x$ and $y$ what does that mean? – Eleven-Eleven Jun 9 '20 at 12:51
• Thank you very much! – dark.nes_s Jun 9 '20 at 13:19

## 1 Answer

Use these:

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

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

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

• See also en.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity – lhf Jun 9 '20 at 13:16
• That's pretty much the whole solution! – Anas A. Ibrahim Jun 9 '20 at 13:24
• I took out the word Hint in deference to the comments – J. W. Tanner Jun 9 '20 at 13:50