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I'm trying to find all integer solutions of an equation $x^2 + dy^2 = c$ with $d,c \in \mathbb{Z}_{>0}$. I'm well aware of the methods that exists when $d \in \mathbb{Z}_{<0}$ or when $c$ is a square. But what if both are not the case? I'm especially interested in the equation $x^2 + 7y^2 = 116$.

I've tried splitting it up in $\mod 29$ and $\mod 4$, but couldn't find all the solutions in $\mod 29$. Any thoughts?

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    $\begingroup$ there is an algorithm for this by Hardy, Muskat, and Williams , Math. Comp (1990). Also, this new book looks interesting, these authors mostly deal with contests books.google.com/… $\endgroup$ – Will Jagy Aug 24 '15 at 17:49
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    $\begingroup$ meanwhile, $x^2 + 7 y^2 \equiv 4 \pmod 8$ means both $x,y$ are even, divide both by 2 and you just need to solve $u^2 + 7 v^2 = 29,$ which gives $(\pm 1, \pm 2).$ Then go back and double $\endgroup$ – Will Jagy Aug 24 '15 at 18:20
  • $\begingroup$ if both are odd you get $0 \pmod 8,$ if one is odd and one even you get odd. You should confirm that odd squares are always $1 \pmod 8,$ whatever method will allow you to believe it and remember $\endgroup$ – Will Jagy Aug 24 '15 at 19:17
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    $\begingroup$ Looked at your earlier questions, definitely recommend Cassels for the Hilbert symbol and related, it is also inexpensive. I have pretty much every book in English on quadratic forms, take my word for it: store.doverpublications.com/0486466701.html $\endgroup$ – Will Jagy Aug 24 '15 at 19:20
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    $\begingroup$ math.stackexchange.com/questions/268402/… $\endgroup$ – Will Jagy Aug 24 '15 at 19:29
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For this particular equation if $|y|>4$ then $7y^2>116$,also if $|x|>10$ then $x^2>116$(for $x,y$ integers).Now by trial you have that $$(x,y)=(2,4),(-2,4),(2,-4),(-2,-4)$$

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I might be a little under-qualified for this question, but here's my take on it:

The function you are looking at is an ellipse elongated along the x-axis, so solutions have a bounded domain in the integers, which is c. We can focus on just one quadrant since an ellipse has 4-fold symmetry, so it makes sense to take x and y to also be both positive, and then extrapolate the other solutions by changing the signs.

If we want to show that we have all integer solutions, we can reduce in a modulus or modulo d and just try all the congruence classes of that form in the integers, as you already have done. Since the domain is bounded, the answer can't be too big. That means we have finitely many integers to try.

This should work for any case just to find the solutions, I don't yet have a more elegant approach (I will edit this post soon). There may be some help in thinking geometrically a bit more.

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