Ellipses and lattice points
For any integer N , does there exist an ellipse (but not a circle) with integer coefficients, such that it passes through exactly N lattice points?
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2 solutions
That is great!
How did you come across the idea of finding rational points on the circle which you used?
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I took a chance. First, I started with the idea of Pythagorean triples, and decided that wasn't going to go anywhere. So I tried something else.
It turns out they're related anyway. You can see how the formula for generating Pythagorean triples can come from this.
@Deeparaj Bhat suggested me this solution:
Let N be given. Then, choose N pythagorean triples: ( a 1 , b 1 , h 1 ) , ( a 2 , b 2 , h 2 ) ⋯ ( a N , b N , h N ) such that the h i are the hypotenuses and all h i are distinct. Pick a circle of radius of the lowest common multiple of these values.
@Pi Han Goh Must it pass through exactly N points? If so, it is not obvious that this solutions work (because the scaling might give us additional integer points).
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That's a good point, that's an interesting twist to this problem . For example, what if an odd N is specified? Say, N=2017, no more no less?
" a total of N lattice points" <<<< This does mean "exactly N lattice points" , no?
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My point is that Michael and Agnishom-Deeparaj's solutions currently demonstrate "There are at least N lattice points on the circumference".
It is not clear to me yet why exactly N points can be achieved.
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@Calvin Lin – Ah my bad. I was wondering why I didn't understand Michael's solution.
Do you think that, for clarity, it's better to rephrase my question to "... passes through precisely N lattice points... " ?
If we allowed the ellipses to be circles, the truth of the statement is simply the consequence of Schinzel's Theorem .
For any rational λ > 1 and positive integer n , it is possible to find an ellipse with the ratio of the semimajor and semiminor axes being equal to λ which passes through exactly n lattice points. A proof can be found here .
Well, this was an odd observation to make, I hadn't even thought of it until you brought it up. Okay, easy proof: Consider an unit circle centered at ( 0 , 0 ) and a line passing through ( − 1 , 0 ) , with rational slope s . The coordinates of where it intersects the circle is
( 1 + s 2 1 − s 2 , 1 + s 2 2 s )
which is rational also. Draw as many lines as you like with rational slopes through ( − 1 , 0 ) . We have a bunch of points on the circle which are rational. Multiply everything by the least common multiple of all the denominators, and then scale up either the x or y axis of the circle by any integer. You have an ellipse that passes through N integer lattice points, for any N one pleases.
Thumbs up for great problem!
Edit: Now that "a total of N " is taken to mean "exactly N points, no more no less", a proof is still needs to be provided, and the explanation given above is not it.