11.3. Can a square be constructed on a coordinate plane with vertices at integer points and with an area, a) equal to $2000;$ b) equal to $2015?$
Answer: a) can; b) cannot. Solution. a) Since $2000=20^{2} \cdot 5=20^{2}\left(2^{2}+1^{2}\right)$, it is possible to construct a square. Indeed, consider the vertices $A(0 ; 0), B(-20 ; 40)$, $C(20 ; 60), D(40 ; 20)$. The lengths of all sides of the quadrilateral $A B C D$ are $\sqrt{20^{2}+40^{2}}=\sqrt{2000}$, and adjacent sides are perpendicular (which follows from the scalar product of adjacent sides being zero: for example, $(\overrightarrow{A B}, \overrightarrow{B C})=(-20) \cdot 40+40 \cdot 20=0$; or this follows from considering the slopes of adjacent sides: their product is -1). b) We will show that 2015 cannot be represented as the sum of two squares of integers. Indeed, since the squares of integers give a remainder of 0 or 1 when divided by 4, the sum of two squares gives a remainder of 0, 1, or 2. But 2015 gives a remainder of 3 when divided by 4, which contradicts the possibility of such a representation, and thus, it is impossible to construct a square.