elementary number theory - On primes which have 5 as a quadratic residue

Well, for which primes $p$ is $5$ a quadratic residue?

I went about it this way:

Let $p' = \frac{p-1}{2}$.

Now, from Gauss' Lemma for Quadratic Congruences, $\displaystyle\left(\frac{5}{p}\right) = (-1)^S$, where $S= \displaystyle
\sum_{i=1}^{p'}\lfloor{\frac{10i}{p}} \rfloor$, and where the $\left(\frac{x}{y}\right)$ is the legendre symbol.

Then, we note that $\lfloor \frac{10i}{p} \rfloor = 0$, from $1 \le i \le \lfloor \frac{1}{5} p' \rfloor$. Further,

$\lfloor \frac{10i}{p} \rfloor = 1$, from $ \lfloor \frac{1}{5} p' \rfloor < i \le \lfloor \frac{2}{5} p' \rfloor$,

$\lfloor \frac{10i}{p} \rfloor = 2$, from $ \lfloor \frac{2}{5} p' \rfloor < i \le \lfloor \frac{3}{5} p' \rfloor$, and so forth.

Now, we're primarily concerned with,

$S= \displaystyle
\sum_{i=1}^{p'}\lfloor{\frac{10i}{p}} \rfloor\ mod \ 2 \equiv \lfloor \frac{4}{5} p' \rfloor- \lfloor \frac{3}{5} p' \rfloor + \lfloor \frac{2}{5} p' \rfloor - \lfloor \frac{1}{5} p' \rfloor$.

We get this last expression, by noting that we only get odd values in the intervals $\lfloor \frac{3}{5} p' \rfloor < i \le \lfloor \frac{4}{5} p' \rfloor$ and $\lfloor \frac{1}{5} p' \rfloor < i \le \lfloor \frac{2}{5} p' \rfloor$.

Okay, now my question essentially lies in how to simplify this final expression of floor functions.
How do I do this? Because after that, I can simply look at the conditions required on p for that final expression to be even, in which case $\displaystyle \left(\frac{5}{p}\right) =1$, and 5 to be a quadratic residue.

A further question, if possible, how do I develop some intuition for how to add and subtract floor functions? Is there anything I can read for this?

Thank You!