Saudi Arabia Mathematical Competitions

Suppose that we have $$\frac{1+\sqrt{8n+1}}{2}<x<\frac{1+\sqrt{8n+9}}{2}$$ for some positive integer $x$. Then $$\sqrt{8n+1}<2x-1<\sqrt{8n+9}$$ hence $8n+1<4x^2-4x+1<8n+9$. It follows $2n<x^2-x<2n+2$, that is $x^2-x=2n+1$, not possible since $x^2-x$ is an even integer.