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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!

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    $\begingroup$ Have you tried induction? $\endgroup$ – Eleven-Eleven Jun 9 '20 at 12:42
  • $\begingroup$ Yes , but how can we do that? $\endgroup$ – dark.nes_s Jun 9 '20 at 12:43
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    $\begingroup$ Hint: Think about expressions like $\left(1+\sqrt {-2}\right)^k=\left(a_k+b_k\sqrt {-2}\right)$. $\endgroup$ – lulu Jun 9 '20 at 12:46
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    $\begingroup$ 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? $\endgroup$ – Eleven-Eleven Jun 9 '20 at 12:51
  • $\begingroup$ Thank you very much! $\endgroup$ – dark.nes_s Jun 9 '20 at 13:19
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Use these:

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

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

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

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

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