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.

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    $\begingroup$ I removed some good comments as obsolete because they referred to an earlier version of the question. $\endgroup$ Commented Dec 29, 2014 at 9:30
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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 $k=n-1$.


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