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!!