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Statement

Find the condition for a center of a circle with exactly one lattice point on its circumference (this lattice point must not be the only one lattice point of the disk)

What I have tried:

Part I:

Let $$\mathcal C: (x-x_0)^2 + (y-y_0)^2 = r^2$$ be a circle positioned at $K(x_0,y_0) \in \mathbb R^2$ and $\varepsilon$ be a tangent line to this circle of the form $$\varepsilon : (x-x_0)x_1 + (y-y_0)y_1 = r^2$$ where $P(x_1,y_1)$ is the intersection point.

Our circle $C$ will contain exactly one lattice point $W \in \mathbb Z^2$ if and only if there is only one pair of integers such that $P \equiv W$. In other words (geometrically) the circle has only one tangent passing from a lattice point.

Our problem reduces to finding the condition that $K$ must satisfy in order for $\varepsilon$ to have exactly one integer solution $W'(x',y')$. Therefore let $A = x - x_0, B = y-y_0, r^2 = C > 0$ which leads to a much simpler form: $$Ax_1 + By_1 = C$$

If $B=0$ then the line $x_1 = C/A$ has either no or infinite number of integer solutions. Therefore $B \neq 0 $.

Note that $Ax' + By' = C$ so $$A(x-x') = -B(y-y') \iff y-y' = - \frac A B (x- x')$$

If $ (-A/B) \in \mathbb Q $ then there exist infinitely many sets of integers $(x_1, y_1)$ where $x_1 \neq x'$ which implies that $- A / B $ is irrational (1).

Part II:

Now let $\mathcal C$ be parametrized as follows:

$$\mathcal C: A = r \cos t, \quad B = r \sin t \; \text{such that} \; t \in [0, 2 \pi)$$ which means that $\cot (-t)$ must be irrational according to (1) for all $t \in (0, 2 \pi) - \{ \pi \} = S$. Hence for a random $t \in S$ where $\cot (-t) \in \mathbb Q^c$ we know that there would exist a circle with exactly one lattice point on its circumference.

My questions:

  1. Is the above right and complete (I am not sure for Part II of the proof)?
  2. Are there other solutions to the problem?
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    $\begingroup$ It seems to me that given any point with neither coordinate a half-integer there's a circle centered at that point with exactly one lattice point on its circumference, namely, the circle through the (unique) nearest lattice point. $\endgroup$ Mar 12, 2015 at 11:39
  • $\begingroup$ The point must not be unique $\endgroup$
    – bolzano
    Mar 12, 2015 at 17:10
  • $\begingroup$ Huh? You are asking for "exactly one lattice point on its circumference". If that doesn't make the point unique, I don't know what does. Maybe you should rewrite your question, so I can understand it. $\endgroup$ Mar 12, 2015 at 23:06
  • $\begingroup$ @GerryMyerson Corrected it. $\endgroup$
    – bolzano
    Mar 13, 2015 at 13:45
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    $\begingroup$ By the way, I think the answer is, let $P=(a,b)$ be any point with neither $a$ nor $b$ half an odd integer, and with $a$ and $b$ not both integers. Then there are circles centered at $P$ with arbitrarily large numbers of interior lattice points and exactly one lattice point on the circumference. $\endgroup$ Mar 17, 2015 at 5:45

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A partial answer. Call a point $(x,y)$ ordinary if $2x$ is not an odd integer, $2y$ is not an odd integer, $x+y$ is not an integer, and $x-y$ is not an integer. It's not hard to see that if $(x,y)$ is ordinary then it has a unique nearest lattice point and a unique second-nearest lattice point. So if $(x,y)$ is ordinary then the circle centered at $(x,y)$ with radius equal to the distance to the second-nearest lattice point will enclose the nearest lattice point and will have exactly one lattice point on its circumference.

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