Sign up ×
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.

Prove that for any $n\geq 0$, there is a hyperbola that passes through exactly $n$ lattice points (= points with integer coordinates) and find an example.

For example it is easy to see that the hyperbola $xy=1$ passes throught exactly two lattice points. It is also easy to see that the hyperbola $$ x^2-2y^2=1 $$ passes through infinitely many lattice points, because this is a Pell equation and its integer solutions are related to the units of the ring of algebraic integers. The intermediate case of a prescribed number of lattice points requires another idea.

share|cite|improve this question
I removed some good comments as obsolete because they referred to an earlier version of the question. –  Jyrki Lahtonen Dec 29 '14 at 9:30
Associated meta post. –  Bill Dubuque Dec 29 '14 at 15:08

1 Answer 1

Note that the hyperbola $2x^2-2y^2=1$ has no points on the integer lattice.
We next deal with the case where $n \gt 0$ 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 $k=\frac{n}{2}-1$.

Finally 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 $k=n-1$.

share|cite|improve this answer

Your Answer


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.