Schinzel's theorem

Schinzel proved this theorem by the following construction. If ${\displaystyle n}$ is an even number, with ${\displaystyle n=2k}$, then the circle given by the following equation passes through exactly ${\displaystyle n}$ points:^[1][2] $${\displaystyle \left(x-{\frac {1}{2}}\right)^{2}+y^{2}={\frac {1}{4}}5^{k-1}.}$$ This circle has radius ${\displaystyle 5^{(k-1)/2}/2}$, and is centered at the point ${\displaystyle ({\tfrac {1}{2}},0)}$. For instance, the figure shows a circle with radius ${\displaystyle {\sqrt {5}}/2}$ through four integer points.

Multiplying both sides of Schinzel's equation by four produces an equivalent equation in integers, $${\displaystyle \left(2x-1\right)^{2}+(2y)^{2}=5^{k-1}.}$$ This writes ${\displaystyle 5^{k-1}}$ as a sum of two squares, where the first is odd and the second is even. There are exactly ${\displaystyle 4k}$ ways to write ${\displaystyle 5^{k-1}}$ as a sum of two squares, and half are in the order (odd, even) by symmetry. For example, ${\displaystyle 5^{1}=(\pm 1)^{2}+(\pm 2)^{2}}$, so we have ${\displaystyle 2x-1=1}$ or ${\displaystyle 2x-1=-1}$, and ${\displaystyle 2y=2}$ or ${\displaystyle 2y=-2}$, which produces the four points pictured.

On the other hand, if ${\displaystyle n}$ is odd, with ${\displaystyle n=2k+1}$, then the circle given by the following equation passes through exactly ${\displaystyle n}$ points:^[1][2] $${\displaystyle \left(x-{\frac {1}{3}}\right)^{2}+y^{2}={\frac {1}{9}}5^{2k}.}$$ This circle has radius ${\displaystyle 5^{k}/3}$, and is centered at the point ${\displaystyle ({\tfrac {1}{3}},0)}$.

The circles generated by Schinzel's construction are not the smallest possible circles passing through the given number of integer points,^[3] but they have the advantage that they are described by an explicit equation.^[2]