Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  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.
share|improve this question
8  
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
2  
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
add comment

1 Answer 1

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 $x^2+9y^2=9a$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.