1. Find all polynomials $f(x)$ of degree not higher than two such that for any real $x$ and $y$, the difference of which is rational, the difference $f(x)-f(y)$ is also rational.
Answer: linear with a rational leading coefficient or constants.

Solution. Let $f(x)=a x^{2}+b x+c$. Then for any $x$ the difference

$$
f(x+1)-f(x)=\left(a(x+1)^{2}+b(x+1)+c\right)-\left(a x^{2}+b x+c\right)=2 a x+a+b
$$

is rational. If $a \neq 0$ this is impossible, since the equation $2 a x+a+b=\sqrt{2}$ has a solution with respect to $x$. Therefore, $a=0$ and $f(x)=b x+c$. Then the difference

$$
f(2)-f(1)=(2 b+c)-(b+c)=b
$$

is rational. Thus, $b$ is a rational number.

It remains to note that all functions $f(x)=b x+c$ with a rational $b$ fit, since the number

$$
f(x)-f(y)=(b x+c)-(b y+c)=b(x-y)
$$

is rational for any rational difference $x-y$.