2. For two positive numbers $a \neq b$, it is known that

$$
a^{2}-2015 a=b^{2}-2015 b
$$

What is the smallest value that $a^{2}+b^{2}$ can take?
Answer: $2015^{2} / 2=2030112.5$. Solution. By moving and reducing by $a-b \neq$ 0, we get $a+b=2015$. Then $a^{2}+b^{2}=a^{2}+(2015-a)^{2}=2 a^{2}-$ $2030 a+2015^{2}$ and the minimum value is achieved at the vertex of the parabola $2 x^{2}+4030 x+2015^{2}$, i.e., when $a=b=2015 / 2$.