1. Can the parameters $a, b, c$ be chosen such that for all $x$ the equality

$$
(x+a)^{2}+(2 x+b)^{2}+(2 x+c)^{2}=(3 x+1)^{2} ?
$$
Solution. Expanding the brackets: $x^{2}+2 a x+a^{2}+4 x^{2}+4 b x+b^{2}+4 x^{2}+4 c x+c^{2}=9 x^{2}+6 x+1$, from which $(2 a+4 b+4 c) x+a^{2}+b^{2}+c^{2}=6 x+1$. It is sufficient to choose numbers $a, b, c$ such that $a+2 b+2 c=3$, $a^{2}+b^{2}+c^{2}=1$. The numbers $a=1 / 3, b=c=2 / 3$ fit.