Number of odd squares

Question

To show that the number of odd squares less than or equal to x is precisely $\lceil \frac{1+\sqrt{x}}{2} \rceil$. I tried a bit for $x$ being an integer by the idea that we are going to pick up the odd squares from $1^2, 2^2,3^2,...,\lceil \sqrt{x}\rceil^2$. If $\lceil \sqrt{x}\rceil$ is even we are going to pick up $\frac{\lceil \sqrt{x}\rceil}{2}$ many odd squares or else $\frac{\lceil \sqrt{x}\rceil + 1}{2}$ many odd squares. How to generalise it?

Answer 1

Hint:

For $(2n-1)^2\le r<(2n+1)^2,$

there are $n$ odd squares