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In the course of solving some other tasks I came across the following two equations where I need to show that there are only finitely many solutions to it respectively none at all. However, I have some trouble to actually show that there are no (more) solutions to it.

It would be great if one of you could give me some argument.

The first one: we have $a, b \in \mathbb{Z}$ and $d \equiv 1 \mod 4$ with $d \in \mathbb{Z}_{< 0}$ but $d \neq -3$. The equation is $$a^2+ab+b^2\frac{1-d}{4}= \pm 1$$ with the only solutions being $(\pm 1, 0)$.

I computed that of course those are solutions and the fact that $d \neq -3$ excludes the solutions $(1,-1), (-1,1), (0,1), (0,-1)$. How do I show that there are no more solutions to it?

The second one: Have $x,y \in \mathbb{Z}$ and show that there are no solutions $(x,y)$ in $\mathbb{Z}^2$ for $$ \vert x^2-10y^2 \vert = \{2,3\}.$$

I assume this could somehow work by considering two cases and using some modulo argument but I'm not sure.

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Hint for your first equation: it seeks for units in the ring of integers of $\mathbb{Q}[\sqrt{d}]$ (why?). There is a standard way of rewriting your equation $(a+b/2)^2-d(b/2)^2=\pm1$. Now the left hand side is a positive quadratic form so there aren't much possibilities for $a,b$...

Hint for your second equation: try a reduction mod $5$. Usually reducing mod $d$ gives useful information, where $d$ is the discriminant of your quadratic form. Another approach to this pell equation would be using the fact that if a pell equation has solution, it has a small solution (see every resource on pell equations), so that you only need to check a finite number of points. However it is in fact always possible to find a local obstruction mod some prime number, by Hasse-Minkowski.

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  • $\begingroup$ Thank you for your answer, and in particular the hints to equation 1. By this rewriting it's so much easier to solve it :) Thank you! $\endgroup$ – SallyOwens Feb 1 '17 at 10:08
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The Diophantine equation $x^2-dy^2=n$ is also known as generalized Pell's equation, and the first equation is known as "Representing an integer by a positive definite binary quadratic form". For solving such equations see here, and here. In the case of $x^2-10y^2=\pm2,\pm 3$ there is no integral solution because of congruence conditions. This works as in this question.

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  • $\begingroup$ Both of your answers were very helpful! Sorry that I can only mark one of them as my favourite answer :( $\endgroup$ – SallyOwens Feb 1 '17 at 10:11

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