# Conics passing through integer lattice points

1. Find parabola that passes through exactly 3 lattice points.
2. Find parabola that passes through exactly 4 lattice points.
3. Prove that for any $n\geq 0$, there is a circle that passes through exactly $n$ lattice points and make an example.
4. Prove that for any $n\geq 0$, there is an ellipse that passes through exactly $n$ lattice points and make an example.
5. Prove that for any $n\geq 0$, there is a hyperbola that passes through exactly $n$ lattice points and make an example.
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The close vote with reason "not a real question" seems a bit excessive, but @Kim, I'd take that as a warning shot that this kind of contextless list of imperatives isn't particularly well received here. Please add some explanations (which lattice are you talking about? $\mathbb Z^2$?), give the motivation for or source of this problem, add the homework tag if it's homework, and formulate the question as a question and not as a list of orders. Thanks. (Also the title shouldn't be a copy of the tags but a more specific summary of the question.) –  joriki Jul 19 '12 at 1:26
I feel that as of late, maybe in the past 2 or 3 weeks, the MSE users have been very downvote-happy. I wonder wether they really are or if I'm just imagining things... –  Olivier Bégassat Jul 19 '12 at 3:25
The lattice points I talking in Z^2. I cannot find the parabola which passes through exactly 3 and 4 lattice points. –  Kim Sokun Aug 3 '12 at 10:14

We solve the hyperbola question. First note that the hyperbola $2x^2-2y^2=1$ has no points on the integer lattice.

We first deal with the case where $n \gt$ is even, even though the proof we give later for $n$ odd actually works for all non-zero $n$.

Look at the hyperbola with equation $x^2-y^2=3^k$, or equivalently $(x-y)(x+y)=3^k$. We obtain all the solutions by putting $x-y=u$, $x+y=v$, where $u$ and $v$ are integers and $uv=3^{k}$. The number of possibilities where $u$ and $v$ are positive is $k+1$, since $u$ can take on all values from $3^0$ to $3^k$. Double this to include the negative possibilities. So the number of lattice points is $2(k+1)$. Now let $k=\frac{n}{2}-1$.

Next we deal with the case where $n$ is odd. Actually, the oddness of $n$ will be irrelevant.

Consider the hyperbola $(4x+1)^2-y^2=5^k$, or equivalently $(4x+1 -y)(4x+1+y)=5^{k}$. The total number of ordered pairs $(u,v)$ of integers (possibly both negative) such that $uv=5^k$ is $2(k+1)$.

So exactly as earlier, the equation $z^2-y^2=5^k$ has precisely $2(k+1)$ integer solutions. In all of these solutions, $z$ is odd.

Note that exactly one of the odd integers $z$ and $-z$ is congruent to $1$ modulo $4$. So the number of solutions of $(4x+1)^2-y^2=5^k$ is $\frac{2(k+1)}{2}$, that is, $k+1$. Now let $k=n-1$.

Remark: The circle case is slightly more delicate. The easiest thing to deal with is $n$ a multiple of $4$, we can use a nice circle centered at the origin, like in our first hyperbola example.

When $n$ is a multiple of $2$ but not of $4$, a trick like the second idea used for the hyperbola works, in the same way. For $n$ odd one has to use a somewhat fancier trick.

The "ellipse" case is taken care of by circles, since every circle is an ellipse. If we want to make a non-circular ellipse, we can use a trick. One idea that works (for circles centered at the origin) is to replace $x^2+y^2=a$ by the ellipse $x^2+9y^2=9a$.

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