Prove that $S_n=5^{4n+3}-{\lfloor\sqrt{5^{4n+3}}\rfloor}^2$ is monotonously increasing The sequence $S_n=5^{4n+3}-{\lfloor\sqrt{5^{4n+3}}\rfloor}^2$ appears to be monotonously increasing:


*

*$S_1=284$

*$S_2=9956$

*$S_3=283261$

*$\dots$


How can I prove (or refute) this?
 A: Let $a_n = \sqrt{5^{4n+3}} = 25^n \sqrt{125}$.
Write $a_n = b_n + e_n$ with $b_n = \lfloor a_n \rfloor$ and $e_n \in [0;1)$.
We have $S_n = a_n^2 - b_n^2 = (2b_n+e_n)e_n$.
The first factor $2b_n+e_n$ is equivalent to $2a_n$ when $n \to \infty$.
Then, since $e_n$ and $e_{n+1}$ are nonzero, we have $S_{n+1}/S_n \sim a_{n+1}e_{n+1}/a_ne_n = 25 e_{n+1}/e_n$
Therefore $S_{n+1}$ has a chance to be less than $S_n$ whenever $e_{n+1} < e_n/25$. Since $e_{n+1} = 25e_n \pmod 1$, this condition only depends on $e_n$ and describes, if we add the condition $0 < e_{n+1}$, an open subset $U$ of $[0;1)$.
If for example $e_n = 2401/2500$ then we do have that $e_{n+1} = 1/100 < e_n/25$. 
If $25e_n = a + e_{n+1}$ and $e_{n+1} < e_n/25$ then $(25e_n - a) < e_n/25$, so $624e_n/25 < a$ and finally $a/25 = 25a/625 < e_n < 25a/624$, which leaves a tiny interval of length $a/15600$.
Considering all those intervals from $a=1$ to $a=24$ we obtain that $U$ has measure $300/15600 = 1/52$.
The last thing to do is to write $\sqrt {125}$ in base $25$ and check that there is an $e_n$ (which you can read immediately from the $25$-ary expansion of $\sqrt {125}$) that falls in $U$.
The intervals are, in base $25$, of the form $(0.a, 0.a0a0a0 \ldots)$ where $a$ is a digit between $1$ and $24$. This makes it easy to recognize at a glance when an $e_n$ is in such an interval.
After checking in wolfram alpha, this does happen (quite late) for $n$ around $185$ where $e_n = 0.17\; 0\; 9\; 18\; 22 \ldots$
$\sqrt 5$ is believed to be normal in all bases, so this would even mean that this happens pseudo-randomly with probability $1/52$. 
