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Given a circle about origin with exactly $100$ integral points(points with both coordinates as integers),prove that its radius is either an integer or $\sqrt{2}$ times an integer.

What my solution is: Since circle is about origin, hence, integral points would be symmetric about the $x$-axis and $y$-axis as well as line $x=y$ and line $x+y=0$ ,i.e. if $(x,y)$ is an integral point, so are $(x,-y),(-x,-y),(-x,y),(y,x),(y,-x),(-y,x)$ and $(-y,-x)$.Therefore, we need to consider only a single octant. Since there are a total of $100$ integral points, two cases are possible:
1) radius of the circle is integer.
2) radius of the circle is not an integer.
case 1: If radius is an integer, then $4$ points on the $x$-axis and $y$-axis of the circle would be integral points and hence each octant must have $12$ points(as $100-4=96$ is a multiple of $8$). therefore, this case is consistent.
case2: If radius is not an integer, then $100$ integral points can't be divided into $8$ parts(octants),and points on $x$-axis and $y$-axis of circle are not integral points, therefore points on line $x=y$ and $x+y=0$ must be integral points so as to divide $100-4=96$ points in $8$ parts. But since point on line $x=y$ and circle is of the form $(r\cdot\cos(45^\circ),r\cdot\sin(45^\circ))$,therefore, $r/\sqrt{2}$ is an integer and hence $r=\sqrt{2}\cdot$integer. other points of circle on these lines are consistent with it.

So, i proved that either radius is an integer and if not then it has to be $\sqrt{2}\cdot$integer.

Is there any flaw in my arguments?? I couldn't find the proof to check whether mine is correct. Thanks in advance!!

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Looks good to me. –  Gerry Myerson Jun 20 '12 at 12:52

2 Answers 2

up vote 3 down vote accepted

All the ingredients are here, but the flow of the argument is not optimal. A smooth proof of the claim would begin with "Assume the set $S:=\gamma\cap{\mathbb Z}^2$ contains $100$ elements. Then $\ldots$", or it should begin with "Assume the radius of $\gamma$ is neither an integer nor $\sqrt{2}$ times an integer. Then $\ldots$".

The essential point (which does not come out clearly in your argument) is the following: The group of symmetries of $S$ is the dihedral group $D_4$, which is of order $8$. Since $100$ is not divisible by $8$ this action has nontrivial fixed points, i.e. points on the lines $x=0$, $y=0$, $y=\pm x$.

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For the case of the radius being an integer, you don't have to make any argument at all. What you are trying to prove is that a circle having exactly 100 integral points implies the radius is either integral or $\sqrt 2$integral. This would still be true if none of these circles were integral. It would even be true if there were no circles with exactly 100 integral points.

That said, your argument works fine. It extends to any number of integral points equal to $4 \pmod 8$

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i know that.Thanx anyway. –  Aang Jun 20 '12 at 16:20

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