Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number?
Answer: 1.

Solution. We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals

$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$