XXI OBM

Let us suppose that all digits between the 1,000,000th and the 3,000,000th decimal digits of $\sqrt{2}$ are zeros. Then $$\begin{array}{rlrl}\sqrt{2}& =\frac{n}{10^{10^6}}+ϵ,& & n\in \mathbb{Z},n<2\cdot 10^{10^6}\text{ and }0<ϵ<10^{-3\cdot 10^6}\\ \iff 2\cdot 10^{2\cdot 10^6}-n^2& =2nϵ10^{10^6}+{ϵ}^2\cdot 10^{2\cdot 10^6}& & \end{array}$$ The left hand side is a positive integer. The right hand side is smaller than 1 nevertheless. The result thus follows.