integral point on conics 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
 A: 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 $x,y.$
A: 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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