Problem: Does there exist a nonnegative integer $a$ for which the equation
\[\left\lfloor\frac{m}{1}\right\rfloor
  + \left\lfloor\frac{m}{2}\right\rfloor
  + \left\lfloor\frac{m}{3}\right\rfloor
  + \dotsb + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a \]
has more than one million different solutions
$(m, n)$ where $m$ and $n$ are positive integers?

\end{enumerate}

Solution: The answer is yes.
In fact, we prove the following general version:
\begin{claim*}
Consider any function $f \colon \NN \to \NN$
satisfying $f(m) = o(m^2)$.
Then for some $a$, the equation
\[ f(m) = n^2 + a \]
has at least 1000000 solutions $(m,n)$.
\end{claim*}
\begin{proof}
  We consider triples $(m,n,a)$ satisfying the equation
  with the additional property that
  \[ n = \left\lfloor \sqrt{f(m)} \right\rfloor^2
    \implies a = f(m) - n^2 \in [0, 2\sqrt{f(m)}]. \]
  Now, let $M$ be large enough that
  $\max_{m=1}^M f(m) < \frac{M^2}{2 \cdot 10^{12}}$.
  Then there are $M$ triples,
  but $a < \frac{M}{10^6}$ in each triple.
  So some $a$ appears $10^6$ times.
\end{proof}