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Does there exist a formula for counting the number of lattice points not outside of a square , with the at most information available concerning the square are the position coordinates of the four corners of the square ?

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Have you tried looking at some specific cases (values of n)? –  BobaFret Oct 19 '12 at 4:46
    
@ BobaFret:- I have looked over integer (of-course positive) values of n , there the formula seems to hold –  Souvik Dey Oct 19 '12 at 4:54

2 Answers 2

Consider the square with vertices $(2,0), (4,2), (2,4), $ and $(0,2)$, then I think there are $13 $ points not lying outside the square. But here $n=2\sqrt {2}$ and $(1+[n])^2=9.$

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Did you just calculated by hand those 13 points or , is there a formula ? –  Souvik Dey Oct 19 '12 at 5:16
    
@SouvikDey: Yes I calculated by hand, but I think the no. of lattice points can be calculated depending on the angle of rotation of the square about it's centre. –  pritam Oct 19 '12 at 6:14

What about the square with corners $(\pm1/2,\pm1/2)$? Here $n=1$ is an integer, but there's only one lattice point, the origin, not outside the square, not 4.

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Well okay , so my guess was wrong , I have now fixed the question suitably –  Souvik Dey Oct 19 '12 at 6:27
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By the way, are you familiar with Pick's Theorem? –  Gerry Myerson Oct 19 '12 at 6:33
    
@ Gerry Myerson:- Yes , but that in case of a square with side n , it givesgives 1 + n^2 = i + b/2 , but basically I would like to know i+b –  Souvik Dey Oct 19 '12 at 6:48
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Well, if you know the corners, you might be able to find $b$. For example, suppose $n=5$, and there's a corner at $(0,0)$. If another corner is at $(5,0)$, then $b=20$, but if another corner is $(3,4)$, then $b=4$. –  Gerry Myerson Oct 19 '12 at 12:15

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