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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 $x_0=y_0=0,r=1$.

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    $\begingroup$ It can have more - e.g. a circle of radius $\sqrt{5}$ contains 8 points. I believe this number can be arbitrarily large. $\endgroup$
    – Wojowu
    Jul 30, 2015 at 20:59
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    $\begingroup$ There are integers which are sums of squares in several ways. $50$, for example, is $7^2+1^2$ or $5^2+5^2$. $\endgroup$
    – lulu
    Jul 30, 2015 at 20:59
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    $\begingroup$ I note that if a circle passes through three points in $\mathbb{Q}^2$, then the center of the circle is also in $\mathbb{Q}^2$. So we do not lose much generality by saying that the $x_0$ and $y_0$ are rational. Or in fact, by taking $(x_0,y_0)=(0,0)$. $\endgroup$
    – Jeppe Stig Nielsen
    Jul 31, 2015 at 23:43
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    $\begingroup$ @JeppeStigNielsen: If you are interested in whether individual numbers of lattice points can be attained (rather than whether there is an upper bound), then you do loose generality by assuming $(x_0,y_0)=(0,0)$: for that point as centre, the number of lattice points must be a multiple of $4$ for symmetry reasons (but for other centres it does not; see also this question). $\endgroup$
    – Marc van Leeuwen
    Aug 24, 2015 at 9:35
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    $\begingroup$ @MarcvanLeeuwen Interesting. By lattice points I suppose you mean points in $\mathbb{Z}\times\mathbb{Z}$? Do you know what natural numbers $n$ are attainable as lattice points (in this sense) on a circle (whose center can have irrational coordinate(s))? If instead we count the number of points in $\mathbb{Q}\times\mathbb{Q}$, my argument shows (I think) that if that number is finite but exceeds $2$, then it is a multiple of $4$. But do you know what numbers $n$ are attainable as the number of $\mathbb{Q}\times\mathbb{Q}$ points on a circle? $\endgroup$
    – Jeppe Stig Nielsen
    Aug 24, 2015 at 14:18

3 Answers 3

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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.

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    $\begingroup$ This is probably the most elementary solution. Very nice $\endgroup$
    – Ewan Delanoy
    Jul 30, 2015 at 21:09
  • $\begingroup$ Interesting, what about a unit sphere? $\endgroup$
    – John M
    Jul 31, 2015 at 4:01
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    $\begingroup$ A unit sphere is a sphere with a radius of 1. The unit sphere is a unit sphere with it's center at the origin. Not entirely logical but who ever said mathematicians are always logical? $\endgroup$
    – Steven Alexis Gregory
    Jul 31, 2015 at 10:59
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    $\begingroup$ @JohnM The unit sphere centred at the origin includes a unit circle in the $x,y$ plane, and therefore all the rational points on the unit circle - and these can be scaled to integer points. $\endgroup$
    – Mark Bennet
    Jul 31, 2015 at 11:19
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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 $4k+3$.

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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 $x^2 + y^2 = r^2$

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