How to simplify this equation? I'm trying to simplify equation
$$n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil$$
into
$$n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor$$
where $x$ is an integer. Indeed, both equations seem to output the same result when testing for some values. Thus far, I've done:
\begin{align*}
&n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil \\
\implies &\sqrt{2x +\frac14} - \frac12 \leq n < \sqrt{2x +\frac14} + \frac12
\end{align*}
The left side easily reduces as follows:
\begin{align*}
&\sqrt{2x +\frac14} - \frac12 \leq n \\
\implies &\sqrt{2x} - \frac12 < n
\end{align*}
but I'm having trouble with the right side which I would wish to reduce as follows:
\begin{align*}
&n < \sqrt{2x +\frac14} + \frac12 \\
\implies &\texttt{???} \\
\implies &n \leq \sqrt{2x} + \frac12
\end{align*}
so that I'm able to obtain my end result.
\begin{align*}
&\sqrt{2x} - \frac12 < n \leq \sqrt{2x} + \frac12\\
\implies &n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor
\end{align*}

Is it possible that that there is an integer $k$ so that $2x < k^2 < 2x +\frac 14$?
As $x$ and $k^2$ are integers  the answer to that is obviously no.
Let's get a little tighter.  Is it possible there is "half-integer" $\frac k2$ where $k\in \mathbb Z$ and $2x < (\frac k2)^2 < 2x+\frac 14$.
That is $8x < k^2 < 8x + 1$.   Again as $8x, k^2, 8x+1$ are all integers the answer is no.
So we either have $(k-\frac 12)^2 < 2x < 2x + \frac 14 < k^2$  (strict inequalitys as $2x+\frac 14$ and $(k-\frac 12)^2$ are not integers) for some integer $k$ or $k^2 \le 2x < 2x+\frac 14\le (k+\frac 12)^2$ for some integer $k$.
Case 1:
$(k-\frac 12)^2 < 2x < 2x +\frac 14 < k^2$.
Then $k-\frac 12 < \sqrt{2x + \frac 14} < k$
$k-1 < \sqrt{2x + \frac 14} < k-\frac 12$
And $\lceil \sqrt{2x +\frac 14} \rceil = k$
And $k-\frac 12 < \sqrt{2x} < k$
$k < \sqrt{2x} +\frac 12 < k + \frac 12$
And so $\lfloor  \sqrt{2x} +\frac 12 \rfloor = k =\lceil \sqrt{2x +\frac 14} \rceil$
Case 2:
$k^2 \le 2x < 2x+\frac 14\le (k+\frac 12)^2$
$k < \sqrt{2x +\frac 14} \le k+\frac 12$
$k-\frac 12 < \sqrt{2x +\frac 14}-\frac 12 \le k$ so
$\lceil \sqrt{2x +\frac 14} \rceil = k$
And $k \le \sqrt {2x} < k+\frac 12$ so
$k+\frac 12 \le \sqrt{2x}+\frac 12 < k+1$ so
$\lfloor  \sqrt{2x} +\frac 12 \rfloor = k =\lceil \sqrt{2x +\frac 14} \rceil$
