1. The coefficients $a, b, c$ of the quadratic trinomial $a x^{2}+b x+c$ are natural numbers not exceeding 2019. Can such a trinomial have two distinct roots that differ by less than 0.01?
Solution. Answer: For example, $1001 x^{2}+2001 x+1000$. The roots of the quadratic equation $a x^{2}+b x+c=0$ differ by the value $\frac{\sqrt{b^{2}-4 a c}}{a}$. For example, let's make $\sqrt{b^{2}-4 a c}=1$. Notice that $(2 n+1)^{2}=4 n(n+1)+1$. Then, for $a=n+1$, $b=2 n+1$, and $c=n$, we have $\frac{b^{2}-4 a c}{a}=\frac{1}{n+1}$, and for example, $n=1000$ (or $n=100$) will work.