Saudi Arabia Mathematical Competitions

Assume by contradiction that all integers are odd. Then, we have $$2a-1<2^n\sqrt{2}<2a$$ for some positive integer $a$. Multiplying inequalities (1) by $2$, we get $4a-2<2^{n+1}\sqrt{2}<4a$. Since the integer $\left[2^{n+1}\sqrt{2}\right]$ is odd, it follows $$4a-1<2^{n+1}\sqrt{2}<4a.$$ In analogous way, after $n$ steps, we get $$2^{n+1}a-1<2^{2n}\sqrt{2}<2^{n+1}a$$ From the left inequality above it follows $$a-2^{n-1}\sqrt{2}<\frac{1}{2^{n+1}}$$ hence $$\frac{a^2-2^{2n-1}}{a+2^{n-1}\sqrt{2}}<\frac{1}{2^{n+1}}$$ From the right inequality in the first step we have $2a>2^n\sqrt{2}$ hence $a^2>2^{2n-1}$, that is $a^2-2^{2n-1}\ge 1$. Using this inequality, from above it follows $$\frac{1}{a+2^{n-1}\sqrt{2}}<\frac{1}{2^{n+1}}$$ hence $$2^{n+1}<2^{n-1}\sqrt{2}+a<2^{n-1}\sqrt{2}+\frac{1}{2}(2^n\sqrt{2}+1)$$ From this we get $2^{n+2}<2^{n+1}\sqrt{2}+1$, that is $$2<\sqrt{2}+\frac{1}{2^{n+1}}<\frac{3}{2}+\frac{1}{2}=2$$ contradiction.