Lattice points inside $x^2+y^2=r^2$

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Asked 12 days ago

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In the interior of the square of side length $N$, whose two corners are lattice points $(m,m)\in\Bbb Z^2$, there are $(N-1)^2$ lattice points.

So, in the intetior of the circle $$x^2+y^2=r^2,\,\, r\in\Bbb R,$$ there must be at least $$\lfloor(\sqrt{\pi}r-1)^2\rfloor$$ lattice points.

Is this true? What is the best lower bound?

Edit: I found this related post.

edited Jun 24 at 5:33

asked Jun 23 at 15:45

Bob Dobbs

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$\begingroup$ Here is the relevant Wiki page: en.wikipedia.org/wiki/Gauss_circle_problem. The best known-bound on the difference between the number of lattice points and $\pi r^{2}$ is $O(r^{0.63})$. $\endgroup$
– Joshua Wang
Commented Jun 23 at 16:14
1

$\begingroup$ It reminds me of this famous video. It doesn't really adress your question though. youtube.com/watch?v=NaL_Cb42WyY $\endgroup$
– Keplerto
Commented Jun 23 at 16:17
2

$\begingroup$ Well, the interior of the square of side length $N=5$ with vertices at $(0,0)$, $(4,3)$ [and two more] has $24$ interior points rather than $(N-1)^2=16$. $\endgroup$
– colt_browning
Commented Jun 24 at 0:14
$\begingroup$ Couldn't this be done by Pick's Theorem? $\endgroup$
– Kamal Saleh
Commented Jun 24 at 5:37

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