Brazilian Mathematical Olympiad

Let us suppose that all digits between the 1,000,000th and the 3,000,000th decimal digits of $\sqrt{2}$ are zeros. Then
$$\begin{aligned} \sqrt{2} &= \frac{n}{10^{10^6}} + \epsilon, && n \in \mathbb{Z},\ n < 2 \cdot 10^{10^6} \text{ and } 0 < \epsilon < 10^{-3 \cdot 10^6} \\ \Leftrightarrow 2 \cdot 10^{2 \cdot 10^6} - n^2 &= 2n\epsilon 10^{10^6} + \epsilon^2 \cdot 10^{2 \cdot 10^6} \end{aligned}$$
The left hand side is a positive integer. The right hand side is smaller than 1 nevertheless. The result thus follows.