$b.)$ Moving the rectangle to the origin we get $A(0,0)$ ,$B(a,b)$,$C(a+c,b-\frac{ac}{b})$,$D(c,-\frac{ac}{b})$ ($a,b,c$ are integers). The sides of the rectangle are : $\sqrt {a^2+b^2}$ and $\frac {c\sqrt {a^2+b^2}}{b}$. So $\sqrt {a^2+b^2}$*$\frac {c\sqrt {a^2+b^2}}{b}=2011$. Which is just $(a^2+b^2)c=2011b$. $2011$ is prime so note : $2011 | a^2+b^2$ so $2011|a$ and $2011|b$. Set $a=2011t$ and $b=2011k$. Putting this into the equation we get $c(t^2+k^2)=k$. This is true only for $(a,b,c)=(0,\pm 2011,\pm 1)$ which means that the sides are parallel to the axes.

One solution to the question a) A= (0,1) B= (2,0) C= (671, 1343) D= (673, 1342)

How we can prove, that the quadrilateral ABCD is rectangle if A (0,0), B (165,198), C(159,203) , D(-6,5) (the solution to a) from the autors)?

Cessio wrote: How we can prove, that the quadrilateral ABCD is rectangle if A (0,0), B (165,198), C(159,203) , D(-6,5) (the solution to a) from the autors)? midpoint of $AB$ coincides with midpoint of $BD$ , it has coordinates $(165/2, 198/2)$ therefore parallelogram $\overrightarrow{AB}\cdot \overrightarrow{AD}=(165,198)(-6,5)=-6 \cdot 165+ 5 \cdot 198=0$ therefore angle $A$ is right parallelogram with a right angle = rectangle

Example for a) A(0,55) B=(15,73) C(81,18) D(66,0)