TITLE: A circle in the plane contains at most four lattice points?
QUESTION [37 upvotes]: Let $\cal C$ be a circle in ${\mathbb R}^2$ : 
$\cal C=\lbrace (x,y)\in{\mathbb R}^2 | 
(x-x_0)^2+(y-y_0)^2=r^2\rbrace$ for some
constants $x_0,y_0,r$.
What is the maximal number of points 
that can be contained in ${\cal C}\cap {\mathbb Z}^2$ ?
I conjecture it is $4$, attained for the "trivial" case
$x_0=y_0=0,r=1$.

REPLY [77 votes]: The unit circle centred on the origin can be parametrised by $$x=\frac {1-t^2}{1+t^2}; y=\frac {2t}{1+t^2}$$
Any rational value of $t$ gives rational values of $x$ and $y$. This can be scaled by a factor $r$ to give a circle of radius $r$.
Choose $n$ such points, and then choose a radius which clears all the denominators - the resulting circle will have at least $n$ integer points.

REPLY [29 votes]: There is no upper bound. On the circle:
$$ x^2+y^2 = 5^k $$
there are exactly $4k+4$ lattice points. That follows from the fact that the number of representations of $n$ as $x^2+y^2$ is given by four times a multiplicative function that depends on the number of divisors of the form $4k+1$ and the number of divisors of the form $4k+3$.

REPLY [13 votes]: There is no upper bound. Take the product of, for example, several consecutive primes that are $1 \pmod 4,$ and let $r = 5 \cdot 13 \cdot 17 \cdot 29.$ Find all Pythagorean triples with that hypoteneuse $r.$ Those all occur as points on the circle $x^2 + y^2 = r^2$