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Suppose we have a conic $ax^2 + bxy + cy^2 + dx + ey + f = 0$ where $a,b,c,d,e,f \in \mathbb{Q}$.

Is there a way of computing the integer points on this curve. Since it is affine an not projective we can't just find the rational points and clear denominators.

Thanks

nigel
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  • This is at least as hard as solving Pell's equations (http://en.wikipedia.org/wiki/Pell's_equation). Note that it's equivalent to require $a, b, c, d, e, f \in \mathbb{Z}$ after clearing denominators. – Qiaochu Yuan Jul 19 '14 at 17:52
  • Constantly this question appears. Look there or in my answers. http://math.stackexchange.com/questions/526164/how-to-solve-an-equation-of-the-form-ax2-by2-cx-dy-e-0/829148#829148 If formulas little will look in the blog. http://www.artofproblemsolving.com/Forum/blog.php?u=206450 – individ Jul 19 '14 at 17:53

2 Answers2

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April 2017. There has been some interest for the case with some coefficients $0.$ I decided to write it with $c=0.$

IF $c=0,$ so that $$a x^2 + b x y + d x + e y + f = 0, $$

THEN $$ ( a b x + b^2 y + b d - a e ) ( b x + e) = - b^2 f - a e^2 + bde. $$ The right hand side is an integer constant, all its divisors, positive and negative, can be found. For each divisor $$ t | (- b^2 f - a e^2 + bde), $$ solve $$ bx + e = t, $$ $$ a b x + b^2 y + b d - a e = \frac{- b^2 f - a e^2 + bde}{t}, $$ and make sure that both $x,y$ come out as integers.

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IF we take instead $a=0,$ so that $$ b x y + c y^2+ d x + e y + f = 0, $$

THEN $$ ( b^2 x + b c y + b e - cd ) ( b y + d) = - b^2 f - c d^2 + bde. $$

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IF we have both $a=c=0,$ $$ b x y + d x + e y + f = 0, $$

THEN $$ ( bx+e ) ( b y + d) = - b f + de, $$ where we see we got to erase one factor of $b$ throughout.

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Will Jagy
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April 2017: if, in the letters named in the question, we have $$ a=c=0, $$ see, among many such questions, How to solve Diophantine equations of the form $Axy + Bx + Cy + D = N$?

Oh, forgot, as Qiaochu says, integers.

Took me a while; never bothered to write out the general case before, but it came out alright. In any case, you can check what I wrote against a multiple of what you wrote, see if i got it all correct.

Define $$ \Delta = b^2 - 4 a c $$

Then you are solving $$ \left( \Delta y + bd -2ae\right)^2 - \Delta \left(2ax+by+d \right)^2 = \left(bd -2ae \right)^2 - \Delta \left(d^2 - 4 a f \right) $$ where the final quantity, $ \left(d^2 - 4 a f \right),$ is not squared. As you can see, there is a solution when $f=0$ with $x,y = 0.$

If $ \Delta = b^2 - 4 a c $ is negative, there are, at most, finitely many solutions. If $ \Delta$ is zero or a positive square, the left hand side factors and there are finitely many solutions, if any. If $ \Delta$ is positive and not a square, there is a Pell type equation, if there are any solutions there are infinitely many. Finding all of them is a mess unless the right hand side has very small absolute value. Even then you need that subset of the Pell-like solutions that allow integer values of $x,y.$

Will Jagy
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  • Actually, unless I am very confused, this does not work. Consider the hyperbola $$ y = \frac{a x + b}{c x + d}.$$ The good news is that $\Delta$ is a perfect square. The bad news is that the equation seemingly educes to $0=0.$ – Igor Rivin Apr 30 '17 at 02:41
  • @Igor I did that one a few minutes ago, https://math.stackexchange.com/questions/2258280/whats-the-best-strategy-for-finding-the-integers-solutions-of-an-hyperbola/2258313#2258313 I am not 100 percent sure, it is certainly possible that the answer above fails for square $\Delta.$ I do not think I even considered that at the time. – Will Jagy Apr 30 '17 at 02:48
  • @Igor I guess I will look at it tomorrow with $\Delta = \delta^2$ and a third case with $\Delta = 0,$ if I get anything I like edit it in. – Will Jagy Apr 30 '17 at 03:00
  • I might be making an arithmetic error (the OP of the question you mention had also posted it on the mathematica board, obnoxiously, and that was why I was looking at it). I found your old answer, but it seems to reduce to $0=0$ using your method here. – Igor Rivin Apr 30 '17 at 03:09
  • @Igor done what I intended. Taking either $a=0$ or $c=0$ or both. The missing case is a parabola, when $\Delta = 0$ but at least one of $a,b,c$ is nonzero. – Will Jagy Apr 30 '17 at 23:52