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I wanted to solve the equation $x^2-5y^2 = -4$ with $x$ and $y$ integers.

Let $\omega=\frac{1+\sqrt5}{2}$ and $A = \mathbb{Z}[\omega]$. One can reduce the Pell equation to finding the elements of $A$ that have a norm equal to $-4$.

I know that $N(2) = 4$, $N(\omega) = -1$, so that for all integer $n$, $N(\pm2\omega^{2n+1}) = -4$.

How can I prove that there are no other solution?

I am familiar with Dirichlet's unit theorem, and I have managed (using ad hoc inequalities) to show that $\omega$ is a fundamental unit.

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And, given the fundamental solutions, all solutions can be found by taking any solution $(x,y)$ and (repeatedly) finding $$ (9x+20 y, 4 x+9y) $$

Note: complete proof is given by working out the Conway topograph for the form $x^2 - 5 y^2$ and noting all occurrences of $x^2 - 5 y^2 = -1$ (to double) and primitive $x^2 - 5 y^2 = -4.$ I believe i have done exactly that in an earlier question. For example, see The quadratic diophantine $ k^2 - 1 = 5(m^2 - 1)$

Here is the picture; note that you can see the matrix corresponding to the $ (9x+20 y, 4 x+9y) $ formula explicitly. Meanwhile, the river is periodic, one section taken to the next by the map $ (9x+20 y, 4 x+9y). $ Next, all occurrences of $-1$ or $-4$ occur along the river itself, not further away. Finally, it suffices to take (up to $PSL_2 \mathbb Z$) $(-1,1), (1,1), (4,2)$ as the "fundamental" solutions, as $-1$ occurs just once per cycle. Or we could name $ (1,1), (4,2), (11,5)$ and just forget about minus signs entirely.

For $+4,$ we can take $ (2,0), (3,1), (7,3).$

Books with theorems used include Buell, Binary Quadratic Forms; Buchmann and Vollmer, Binary Quadratic Forms; Conway, The Sensual Quadratic Form.

enter image description here

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The Alpertron gives the following fundamental solutions:

$$(x,y)=(\pm 1, \pm 1), (\pm 4, \pm 2)$$

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  • $\begingroup$ Sure, and this is consistent with my conjecture, as $2\omega=1+\sqrt5$ and $2\omega^3=4+2\sqrt5$. But how do you prove it? $\endgroup$ – Yann Oct 2 '13 at 21:44
  • $\begingroup$ The website gives step-by-step justification if you ask for it. $\endgroup$ – vadim123 Oct 2 '13 at 22:42

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