Although it's being removed from the curriculum nowadays, there is something that often happens in high school mathematics or first-year college calculus where you learn how to transform one conic section to another one of the same type with a matrix. In particular, we can get from the circle \(x^2+y^2=9\) to \(x^2+2y^2=9\) by multiplying the vector \((x,y)\) by the matrix \(\begin{pmatrix}1& 0\\ 0& 1/\sqrt{2}\end{pmatrix}\); that would not stretch the \(x\)-axis, but shrinks in the \(y\) axis by the appropriate amount.

However, one can also think of it as both conics coming from matrices. \[\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=x^2+y^2\] while \[\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1& 0\\ 0& 2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=x^2+2y^2\; .\] Gauss was interested in extending Fermat's question; namely, what numbers are representable in these ways, as opposed to just a sum of squares? It turns out that many such quadratic forms represent the same sets of integers.

The Sage reference manual even uses our example to demonstrate this: \[\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1& 0\\ 0& 2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=x^2+2y^2\text{ and }\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1& 1\\ 1& 3\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=x^2+2xy+3y^2\] both would fulfill Fermat's result about primes modulo \(8\) we mentioned earlier. So both should represent 11. Clearly \(11=3^2+2\cdot 1^2\) works, but what about the other version?

Looks like \(x=2,y=1\) will do it. (That's because \[x^2+2xy+3y^2=(x+y)^2+2y^2\] and this is a coordinate transformation.)

So there is some very deep theory there, which is another place where lie the beginnings of algebraic number theory, just like with the Gaussian integers. But we'll let it rest there.

Instead, we will continue looking for integer points on a given specific curve. Assuming that ellipses are doable by simply counting, what is next?

The parabola comes to mind. A general parabola would look like \(ny=mx^2\); this can be thought of in your usual terms as \(y=ax^2\) and \(a=m/n\).

Then I can just check all \(x\in\mathbb{Z}\) such that \(n\mid mx^2\). Since \(gcd(m,n)=1\) for this (lowest terms), we would just need in fact that \(n\mid x^2\) (so if \(n\) is prime, \(n\mid x\) suffices)!

So if \(y=mx^2\) for integer \(m\), any \(x\) will do. That makes sense; integer input better give integer output, which would be a lattice point!

If \(2y=x^2\), we just look at it as \(2\mid x\), so that requiring \(x\) even will give lattice points. And so on.