Motivation: Let $k$ be a positive integer and $m=4k+1$. I want to find the necessary conditions for the following identity to hold： $$\displaystyle\sum_{i=1}^{k}\lfloor\sqrt{im}\rfloor=\frac{m^2-1}{12}$$
Since $\lfloor\sqrt{im}\rfloor\geqslant j$ is equivalent to $im>j^2-1$, where $j$ is a positive integer not exceeding $2k$, it can be shown that the left-hand side is equal to $$2k^2-\sum_{j=1}^{2k}\left\lfloor\frac{j^2}{m}\right\rfloor+\left\lfloor\frac{2k}{r}\right\rfloor$$ where $r$ is the least positive integer such that $m$ divides $r^2$.
(i) If all the prime factors $p_1,p_2,\dots$ of $m$ satisfy $p_1\equiv p_2\equiv\dots\equiv1\pmod{4}$, then $x^2\equiv-1\pmod{m}$ has a solution. Let $j'$ denotes the unique integer such that $j'\equiv xj\pmod{m}$ and $|j'|\leqslant 2k$ for each $j$. Then $$\left\lfloor\frac{j^2}{m}\right\rfloor + \left\lfloor\frac{j'^2}{m}\right\rfloor = \frac{j^2}{m} + \frac{j'^2}{m} + \epsilon_{j}$$ where $\epsilon_{j} = 0$ if $m$ divides $j^2$, and $\epsilon_{j} = -1$ if $m$ does not divide $j^2$. Since $m$ divides $j^2$ if and only if $r$ divides $j$, then it follows that $$\sum_{j=1}^{2k}\left\lfloor\frac{j^2}{m}\right\rfloor =\frac{1}{2}\sum_{j=1}^{2k}\left(\left\lfloor\frac{j^2}{m}\right\rfloor+\left\lfloor\frac{j'^2}{m}\right\rfloor\right) = \frac{1}{2}\sum_{j=1}^{2k}\left(\frac{j^2}{m}+\frac{j'^2}{m}-1\right) + \frac{1}{2}\sum_{j=1}^{2k}\left(1+\epsilon_{j}\right) =\frac{2k(k-1)}{3}+\frac{1}{2}\left\lfloor\frac{2k}{r}\right\rfloor$$ Thus the identity in the above question holds if and only if $\left\lfloor\frac{2k}{r}\right\rfloor=0$, which is equivalent to $m$ being squarefree.
(ii) If there exists a prime factor $p\equiv3\pmod{4}$ of $m$, I have no idea how to deal with it.
By using Python, I found that the identity holds for $m=245,605,637,1805,2989,6877$. For each of these $m$ and each prime factor $p$ of the form $4n+3$, there exists an integer $l$ such that $p^{2l}$ divides $m$ while $p^{2l+1}$ does not. This leads to the following question:
Question: Let $k$ be a positive integer such that $m=4k+1$ has a prime factor $p\equiv 3\pmod{4}$ with an odd exponent in the prime factorization of $m$. Equivalently, $m$ is not a sum of two square. Is it true that $$\displaystyle\sum_{i=1}^{k}\lfloor\sqrt{im}\rfloor \neq \frac{m^2-1}{12}?$$
Update Dec 31
(a) There are exactly $40$ positive integers $m$ within the range of $1\leqslant k\leqslant 10^6$ (verified via Python), each of which satisfies the identity and has at least one prime factor of the form $4n+3$. All of these $m$ can be expressed as a sum of two integer squares. Below are their prime factorizations.
$$245 = 5\times 7^2$$ $$605 = 5\times 11^2$$ $$637 = 13\times 7^2$$ $$1805 = 5\times 19^2$$ $$2989 = 61\times 7^2$$ $$6877 = 13\times 23^2$$ $$12005 = 5\times 7^4$$ $$13673 = 113\times 11^2$$ $$16577 = 137\times 11^2$$ $$31213 = 13\times 7^4$$ $$53429 = 101\times 23^2$$ $$73205 = 5\times 11^4$$ $$90569 = 41\times 47^2$$ $$106097 = 17\times 79^2$$ $$112993 = 313\times 19^2$$ $$130181 = 29\times 67^2$$ $$146461 = 61\times 7^4$$ $$381517 = 397\times 31^2$$ $$399829 = 181\times 47^2$$ $$588245 = 5\times 7^6$$ $$602213 = 173\times 59^2$$ $$623113 = 337\times 43^2$$ $$651605 = 5\times 19^4$$ $$791437 = 157\times 71^2$$ $$859301 = 389\times 47^2$$ $$1018349 = 461\times 47^2$$ $$1124381 = 509\times 47^2$$ $$1529437 = 13\times 7^6$$ $$1654433 = 113\times 11^4$$ $$1828613 = 293\times 79^2$$ $$2005817 = 137\times 11^4$$ $$2015561 = 449\times 67^2$$ $$2220649 = 1201\times 43^2$$ $$2697889 = 601\times 67^2$$ $$2983217 = 857\times 59^2$$ $$3033397 = 157\times 139^2$$ $$3093161 = 449\times 83^2$$ $$3637933 = 13\times 23^4$$ $$3720925 = 13\times 5^2 \times 107^2$$ $$3887089 = 241\times 127^2$$
In this list, $3720925$ is the smallest integer that has more than 2 distinct prime factors. The other integers are of the form $pq^{2k}$, where $p\equiv1\pmod{4}$, $q\equiv3\pmod{4}$ are primes and $k$ is a positive integer.
It appears that if $pq^{2k}$ is in this list, then so is $pq^{2k+2}$.
(b) As David E Speyer noted, there is a similar question post in MO with several brilliant answers and comments: Conjecture $ \sum _{i=1}^{n-1}\lfloor \frac{i^2}{n}\rfloor \ge \frac{\left(n-1\right)\left(n-2\right)}3$ where $n\in\mathbb N^*$
Based on the answer of David E Speyer in that post and a theorem of Dirichlet, it can be proved that when $m$ is a squarefree number with a prime factor $p\equiv3\pmod{4}$, $$\sum_{i=1}^{k}\lfloor\sqrt{im}\rfloor = \frac{m^2-1}{12} - \frac{S(m)}{2} < \frac{m^2-1}{12}$$
where $S(m)$ is defined as in the answer from Henri Cohen.
