We do very naive bounding on this problem. We only consider $m>1$, now let $S_m=\sum_{i=1}^{m} \lfloor{\frac{m}{i}}\rfloor\le \sum_{i=1}^{m} {\frac{m}{i}}<100 m\log m$. Thus, if we write this as $n^2+a$, and pick $n=\lfloor{\sqrt{S_m}}\rfloor$, we must have $a\le 2n\implies a\le 20\sqrt{m\log m}$, thus $a$ can take atmost $200 \sqrt{m\log m}$ values for $S_1,S_2,\cdots S_m$. Now, FTSOC assume that no value of $a$ occurs more than $10^6$ times. Then, we have $200\cdot 10^6 \sqrt{m\log m}\ge m$ for all $m$ but this is clearly false for large enough $m$ and we are done.

The left-hand side equals $\tau(1) + \tau(2) + \cdots + \tau(m)$ and now $\tau(k) \leq 2\sqrt{k}$ and $\sqrt{k} \leq \sqrt{m}$ would show that the expression is at most $2m^{3/2}$ and this gives another possibility for bounding as @above. (Everything which is of order strictly below $m^2$ works!)

The answer is yes.

For storage ig: The answer is yes.

)

) But, for any such pair $(m,n)$, we have $$f(m)-n^2 < f(m) \leq f(M) \leq 10^{100} M \ln(M+1)$$ so $f(m)-n^2$ can take at most $10^{100} M \ln(M+1)$ different values. Therefore, there exists an $a \geq 0$ such that $f(m)=n^2+a$ for at least $$\frac{\sqrt{M}}{10^{200} \ln (M+1)}$$ such pairs, and the above quantity goes to $\infty$ as $M$ gets arbitrarily large, and so we are done.

Let $f(m)$ be the expression $\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor$. Then $f(2^{2k+1}) \leq 2^{2k}(4k+4)$ from the known inequalities such as $[n] \leq n$ and $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \cdots + \frac{1}{2^t} \leq t+1$. Then we have $2^{2k+1}$ integers which are the values of the function $f(x)$ and which are not exceeding $2^{2k}(4k+4)$. The numbers $f(2^{2k}+1), f(2^{2k}+2), \cdots, f(2^{2k}+2^{2k})$ are at least $2^{2k}$. Then we can get $2^{2k}*2^{k}$ numbers obtained by subtracting from the set $f(2^{2k}+1), f(2^{2k}+2), \cdots, f(2^{2k}+2^{2k})$ all the squares from $0$ to $2^{k}-1$. These $2^{3k}$ numbers are positive integers with $\leq 2^{2k}(4k+4)$. So, the average number of repeated numbers is equal to $\frac{2^{3k}}{2^{2k}(4k+4)}=\frac{2^{k}}{4k+4}$ which can be arbitrarily large for sufficiently large $k$.

The answer is yes. In fact, we prove the following general version: Claim: Consider any function $f \colon {\mathbb N} \to {\mathbb N}$ satisfying $f(m) = o(m^2)$. Then for some $a$, the equation $$f(m) = n^2 + a$$ has at least 1000000 solutions $(m,n)$. Proof. We consider triples $(m,n,a)$ satisfying the equation with the additional property that $n = \left\lfloor \sqrt{f(m)} \right\rfloor^2$. Note this 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. $\blacksquare$

The answer is yes. Let $f(m)$ denote $\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor$ and $d(m)$ denote the number of divisors of $m$. First, $$\begin{align*} f(m) &> \frac{m}{1} + \frac{m}{2} + \cdots + \frac{m}{m} - (1 + 1 + \cdots + 1)\\ &= \frac{m}{2} + \frac{m}{3} + \cdots + \frac{m}{m}\\ &> m\ln\left(\frac{m+1}{2}\right), \end{align*}$$ where we underestimate $\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{m}$ with $\int_{a=2}^{m+1} \frac{1}{a} = \ln\left(\frac{m+1}{2}\right)$. Now I claim there exists a positive integer $V$ such that if $X$ is the smallest positive integer with $d(X)\ge V$ , then $X > V^2$ and $\ln\left(\frac{X+1}{2}\right) > (10^6+6)^2$. Let $p_1^{k_1}\cdots p_{\ell}^{k_{\ell}}$ be the prime factorization of $X$ with $p_1 < \cdots < p_{\ell}$, so $(k_1 + 1)\cdots(k_{\ell}+1)\ge V$. We have $p^k > (k+1)^2$ for all $k\ge 1$ when $p\ge 5$, $3^k\ge (k+1)^2$ when $k\ge 2$, and $2^{k}\ge (k+1)^2$ when $k\ge 6$. For all sufficiently large $V$, we have $\nu_2(X)\ge 6$, $\nu_3(X)\ge 2$, and $\nu_5(X)\ge 1$, so $$X = p_1^{k_1}\cdots p_{\ell}^{k_{\ell}} > (k_1+1)^2\cdots (k_{\ell}+1)^2\ge V^2.$$ Then $f(X) > X\ln\left(\frac{X+1}{2}\right) > (10^6+6)^2V^2$, so the number of perfect squares $\le f(X)$ is at least $(10^6+6)V$. We also have that $d(X)\le 6V$, because if not, still assuming that $\nu_2(X)\ge 6$, $\frac{X}{2}$ satisfies $d(\tfrac{X}{2})\ge V$, a contradiction. So the number of perfect squares $\le f(X-1)$ is at least $10^6V$. Also $f(m) - f(m-1) = d(m)$, the number of divisors of $m$ because $\left\lfloor\frac{m}{k}\right\rfloor > \left\lfloor\frac{m-1}{k}\right\rfloor$ for $1\le k\le m$ iff $m - 1\equiv -1 \pmod k$, or $k\mid m$. So for each perfect square $n^2\le f(X-1)$, of which there are at least $10^6V$ of, let $S$ be the set of values $f(m) - n^2\le d(m)$ for $1\le m < X$ such that $m$ is the smallest integer for which $f(m)\ge n^2$. All values in $S$ are at most $d(m) < V$, so some value occurs at least $10^6$ times, as desired.

Consider $f(x)$ be $x^{-}$( the greatest square $< x$). One has $f(x)< 2\sqrt {x} .$ Call the $\text {LHS}$ $g(m) ,$ it is at most $m (1+\ln m).$ The numbers $f(g(1)),f(g(2)),\hdots,f(g(N))$ are at most $2\sqrt {N}\sqrt {1+\ln N}< N^{\frac {3}{5}}.$ For $N=10^{17},$ and by $\text {PHP}$ take $N^{\frac {2}{5}}$ such numbers so that they coincide. Such value is required $a ,$ gives at least $10^{\frac {34}{5}}>\text {numerous solutions}.$

Consider a positive integer $m$ and a bipartite graph with vertices on the left side being $1, 2, \ldots, m$, and vertices on the right side being $0,1, \ldots, f(m)-1$. Now connect $i$ from the left side with $j$ from the right side if and only if $f(i)=n^2+j$ for some positive integer $n$. Note that $i$ has $\lfloor \sqrt{f(i)} \rfloor$ neighbours. We can bound $f(m)$ from below with $m$ and from above with $100m \log m$ as someone already showed in some posts . Now the number of edges in our graph is at least $\sqrt{1}-1+\sqrt{2}-1+\ldots+\sqrt{m}-1$, which is eventually strictly larger than $m^{1+\epsilon}$ for some small $\epsilon>0$ (in fact, any $\epsilon<0.5$ would work). Then, by pidgeonhole principle, one of the vertices from the right side has at least $\frac{m^\epsilon}{100 \log m}$, which is eventually larger than $1$ million.

Let, $$\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor = D(m)$$ we know ,if $F(n)=\sum_{d\mid n} f(d)$ then $\sum_{n=1}^N F(n) =\sum_{k=1}^N f(k)\left\lfloor\frac{N}{k}\right\rfloor$ So, $D(m)=\sum_{r=1}^m \tau (r)$. $D(m)$ is known as Dirichlet's fuction of divisors . Using Dirichlet's Hyperbola method it can be shown that , $$D(x)=\sum_{n\le x} \tau (n) =x\log x +(2\gamma -1)x +O(\sqrt{x})$$ for a large enough number $r$ we can say that , $$D(x) < rx\log x$$ if $n^2 +a =D(x)$ then ,$n\le \left\lfloor\sqrt{D(x)}\right\rfloor$ $$\implies D(x) = n^2 +a \ge \left\lfloor\sqrt{D(x)}\right\rfloor ^2 +a$$ $$\ge (\sqrt{D(x)} -1)^2 +a = D(x) -2\sqrt{D(x)} +1 +a$$ $$\Rightarrow a \le 2\sqrt{D(x)} -1 < 2\sqrt{rx\log x}$$ Let, there do not exist a value of $a$ which appear more than $10^6$ times , so $$x\le 10^6 .a \le 10^6 .(\sqrt{rx\log x})$$ but $$\lim_{x\to \infty} \frac{x}{\sqrt{x\log x}} \to \infty$$ contradiction !

anser wrote: Does there exist a nonnegative integer $a$ for which the equation $$\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a$$ has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? The expression $\lfloor x\rfloor$ denotes the integer part (or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor = 1, \lfloor\pi\rfloor =\lfloor 22/7 \rfloor = 3, \lfloor 42\rfloor = 42,$ and $\lfloor 0 \rfloor = 0$. This isn't as bad as I expected, in fact a total bait for a P6 (Apparently doing the dumbest thing possible works really well ) The answer is $\boxed{\textit{yes}}$. For convenience, let $S_m = \sum_{i = 1}^m \left \lfloor \frac{m}{i} \right \rfloor$. For our entire solution, fix $m$ as a very large positive integer. Claim 01. $\sum_{i = 1}^{m} d(i) = \sum_{i = 1}^m \left \lfloor \frac{m}{i} \right \rfloor$ Proof. We will double count the number of triple $(a,b)$, where $a,b \le m$ such that $a \mid b$. Fixing $a$, then there are $\left \lfloor \frac{m}{a} \right \rfloor$ possible values of $b$. Fixing $b$, then there are $d(b)$ possible values of $a$, and we are done. Claim 02. $d(n) \le 2 \sqrt{n}$ for all $n \in \mathbb{N}$. Proof. Let us count the number of tuples $(a,b)$ such that $ab = n$. Notice that there are exactly $d(n)$ of this tuple. Furthermore, since $\min \{ a,b \} \le \sqrt{n}$, and choosing value of one variable uniquely determine the others: then there is a maximum of $2 \sqrt{n}$ such tuple, implying the bound. Now, we'll provide a bound for $S_m$: $$\begin{align*} S_m &= \sum_{i = 1}^m \left \lfloor \frac{m}{i} \right \rfloor = \sum_{i = 1}^m d(i) = \sum_{i = 1}^m 2 \sqrt{i} \\ &< 2 \sqrt{\left( \sum_{i = 1}^m 1\right) \left( \sum_{i = 1}^m i \right)} = m \sqrt{2(m + 1)} \end{align*}$$ Now, choose $n = \lfloor \sqrt{S_m} \rfloor$ for every fixed $m$. Therefore, we get the bound $$a \le 2 \lfloor \sqrt{S_m} \rfloor < 2(S_m)^{\frac{1}{2}} = O(m^{\frac{3}{4}})$$ Therefore, for $S_1, S_2, \dots, S_m$, there are at most $O(m^{\frac{3}{4}})$ possible values of $a$. For the sake of contradiction, there is no such $a$. We will now count the number of $(m,n,a)$ satisfying the equation for a fixed positive integer $m$. Suppose the number of such pair is $X$. Then, by our assumption, start by choosing $a$, then $X \le 10^6 \cdot O(m^{\frac{3}{4}}) = O(m^{\frac{3}{4}})$. However, each $m$ gives a unique solution of $n$ and $a$, and therefore $X \ge O(m)$, which implies a contradiction.

This problem is easier after seeing https://artofproblemsolving.com/community/c6h130811p741367

Asymptotics go brr. The answer is yes. Define $f: \mathbb{Z^+} \to \mathbb{Z^+}$ as: $$f(m)=\left \lfloor \frac{m}{1} \right \rfloor + \left \lfloor \frac{m}{2} \right \rfloor + \cdots + \left \lfloor \frac{m}{m} \right \rfloor.$$ Observe that $f$ is unbounded above and nondecreasing. Further, note that: $$f(m)\leq \frac{m}{1}+\frac{m}{2}+\cdots+\frac{m}{m} \leq m(\ln(m)+1),$$ where the second inequality is well-known from properties of the harmonic series. Also note that $f(m) \geq m$, which is a very stupid bound but ends up working fine. Now consider the equation $f(m)=n^2+a$. Instead of fixing $a$ and counting positive integer solutions $(m,n)$, we will instead fix $m$ and count the number of solutions in $a$. Then, there are clearly exactly $\lfloor \sqrt{f(m)} \rfloor$ possible values that $a$ can take, since there are $\lfloor \sqrt{f(m)} \rfloor$ positive integers $n$ such that $n^2\leq f(m)$, and each one gives a unique nonnegative $a$. This means that there are exactly: $$\lfloor \sqrt{f(k)} \rfloor+\lfloor \sqrt{f(k-1)} \rfloor+\cdots+\lfloor \sqrt{f(1)} \rfloor$$ unique solutions $(m,n,a)$ to the equation such that $m \leq k$. Observe that each of these solutions satisfies $a \leq f(k)$. We now do a bit of bounding. Observe that: $$\begin{align*} \lfloor \sqrt{f(k)} \rfloor+\lfloor \sqrt{f(k-1)} \rfloor+\cdots+\lfloor \sqrt{f(1)} \rfloor &\geq \sqrt{f(k)}+\sqrt{f(k-1)}+\cdots+\sqrt{f(1)}-k\\ &\geq \left(\sum_{i=1}^k \sqrt{i}\right)-k\\ &\geq \left(\int_0^k \sqrt{i}\right)-k\\ &=\frac{2}{3}k^{3/2}-k=O(k^{3/2}) \end{align*}$$ Also, by Pigeonhole, since $a \leq f(k)$, by Pigeonhole there exists some value $0 \leq x \leq f(k)$ such that there are at least $$\left \lceil\frac{\lfloor \sqrt{f(k)} \rfloor+\lfloor \sqrt{f(k-1)} \rfloor+\cdots+\lfloor \sqrt{f(1)} \rfloor}{f(k)+1}\right \rceil$$ solutions $(m,n,a)$ to the equation such that $a=x$. Now applying the bounds previously proven, we have: $$\begin{align*} \left \lceil\frac{\lfloor \sqrt{f(k)} \rfloor+\lfloor \sqrt{f(k-1)} \rfloor+\cdots+\lfloor \sqrt{f(1)} \rfloor}{f(k)+1}\right \rceil &\geq \frac{\lfloor \sqrt{f(k)} \rfloor+\lfloor \sqrt{f(k-1)} \rfloor+\cdots+\lfloor \sqrt{f(1)} \rfloor}{f(k)+1}\\ &=\frac{O(k^{3/2})}{f(k)+1}\\ &\geq\frac{O(k^{3/2})}{O(k\ln(k))}=O\left(\frac{\sqrt{k}}{\ln(k)}\right). \end{align*}$$ But $\lim_{k \to \infty} \frac{\sqrt{k}}{\ln(k)}=\infty$, so for sufficiently large $k$ there exists some value $0 \leq a \leq f(k)$ such that the equation $f(m)=n^2+a$ has over one million distinct solutions $(m,n)$, where $(m,n) \in \mathbb{Z^+}$ as desired. $\blacksquare$

Define $$S_m=\bigg\lfloor\frac{m}{1}\bigg\rfloor+\bigg\lfloor\frac{m}{2}\bigg\rfloor+\cdots +\bigg\lfloor\frac{m}{m}\bigg\rfloor.$$ Let $$f(m)=S_m-\lfloor\sqrt{S_m}\rfloor^2=2\sqrt{S_m}\cdot\{\sqrt{S_m}\}-\{\sqrt{S_m}\}^2\geq 0,$$ where $\{x\}$ denotes the fractional part of $x,$ as usual$.$ Note that $f(m)\leq 2\sqrt{S_m}$ for all $m.$ Next we do some bounding$:$ $$S_m\leq \frac{m}{1}+\frac{m}{2}+\cdots \frac{m}{m}=m\cdot H_m,$$ where $H_m$ is the $\text{m-}$th Harmonic number$.$ Define $\{\gamma_n\}_{n=1}^{\infty}$ as $\gamma_n=H_n-\ln{n}.$ It is a well-known fact that $\{\gamma_n\}_{n=1}^{\infty}$ is a strictly decreasing sequence$,$ which converges to the Euler-Mascheroni constant $\gamma =0,57\dots.$ Therefore $$H_n=\ln{n}+\gamma_n\leq \ln{n}+\gamma_1=\ln{n}+1$$ We conclude $S_m\leq m\cdot H_m\leq m(\ln{m}+1)$ for all $m.$ Now comes the hardest part of the problem for me$:$ the Pigeonhole Principle$.$ Assume that for all $a\geq 0,$ the number of solutions of the equation $S_m=n^2+a$ is less than some positive integer constant $c.$ In particular this implies that the number of solutions to the equation $f(m)=a$ is also less than $c.$ Fix some (very) large integer $M.$ Let $N=\bigg\lfloor \frac{M}{c}\bigg\rfloor$ and consider the numbers $$f(1), f(2), \cdots, f(M)$$ If the number of all the different values they take on is $\leq N,$ then by the Pigeonhole Principle there exists a value that is taken on by at least $\frac{M}{N}\geq c$ of these numbers$,$ which is a contradiction$.$ Therefore these numbers cover more than $N$ non negative integers$.$ This implies that there is some $m\in \{1, 2, \dots, M\},$ for which $f(m)\geq N>\frac{M}{c}-1.$ But this means $$\frac{M}{c}-1<f(m)\leq 2\sqrt{S_m}\leq 2\sqrt{m(\ln{m}+1)}\leq 2\sqrt{2M\ln{M}}$$ We obtain $$2\sqrt{2M\ln{M}}>\frac{M}{c}-1>\frac{M}{2c}$$ $$\Rightarrow 8M\ln{M}> \frac{M^2}{4c^2}$$ $$\Rightarrow\frac{\ln{M}}{M}> \frac{1}{32c^2}$$ for all large $M.$ But recall that $\lim_{x\to\infty} \frac{\ln{x}}{x} = \lim_{x\to\infty} \frac{\frac{1}{x}}{1}=0$ and thus we reach a contradiction$.$

Somehow I find all the write-ups here incredibly unsatisfying, so I will put my write-up over here. FTSOC (Convenience Reasons) let us set $$S_m:=\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor$$ And consider the quantity, $T(m,n):= S_m-n^2$. Note that $$S_m\le m\cdot \left(1+\frac12+\frac13+\ldots+\frac1m\right)<2m\ln (m)$$ We define a map $\mathcal M: (m,n)\mapsto T(m,n)$. Consider the image of the $N$ pairs $\left(j,\left\lfloor \sqrt{S_j}\right\rfloor\right)$ where $j\in [N]$ under $\mathcal M$. Observe that $$T\left(j,\lfloor \sqrt{S_j}\rfloor\right)\le 2\left\lfloor \sqrt{S_j}\right\rfloor\le 2\left\lfloor \sqrt{S_N}\right\rfloor <2\sqrt{2N\ln (N)}$$ So these $N$ pairs correspond to non-negative integers $\le 2\sqrt{2N\ln (N)}$. Hence, there must be some non-negative integer such that at least $$\frac{N}{1+2\sqrt{2N\ln (N)}}$$ pairs are mapped to it by the pigeonhole principle. Now as $$\lim_{N\to\infty} \frac{N}{1+2\sqrt{2N\ln (N)}}=\infty$$ we conclude that the answer is $\boxed {\textrm{YES}}$.

Answer is $\boxed{\text{yes}}$. Let $f(m)$ denote what is on the LHS and $d(m)$ denote the number of positive integer divisors of $m$. Claim: $f(m)=d(1)+d(2)+d(3)+\ldots+d(m-1)+d(m)$. Proof: Note that $\left\lfloor \frac{m}{a} \right\rfloor$ is the number of positive integers $b\le m$ so that $a|b$ and $d(b)$ is the number of positive integers $a\le m$ so that $a|b$. Thus, both of $f(m)$ and $d(1)+d(2)+\ldots +d(m)$ count the number of pairs of positive integers $1\le a,b\le m$ so that $a|b$. Thus, $f(m)=o(m^2)$. So the equation is $f(m)=n^2+a$. We claim that there exists an $a$ where there are at least $10^6$ solutions $(m,n)$. Proof: Consider the triples $(m,n,a)$ satisfying the original equation, with the additional constraint that $n=\left\lfloor \sqrt{f(m)} \right\rfloor$. Then $a=f(m)-\left(\left\lfloor \sqrt{f(m)} \right\rfloor\right)^2\implies 0\le a\le 2\sqrt{f(m)}$ Now since $f(m)=o(m^2)$, we can consider very large $c$ so that $\max_{m=1}^c f(m)<\frac{c^2}{4\cdot 10^{12}}$. Consider the $c$ such triples, with $1\le m\le c$. Then $a< 2\sqrt{\frac{c^2}{4\cdot{10^{12}}}}=\frac{c}{10^6}$. Thus some $a$ must appear at least $10^6$ times.

Let $f(m) = \left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor$. The key observation here is that $\mathcal{O}(f(n))<n^2$. Let $f(n)=S_n^2+a_n$ for $a_n < 2S_n+1$. Let $\lambda(x)$ denotes the number of positive integers $y \leq x$ satisfying $S_y < S_{y+1}$. We claim $\lim_{x \to \infty} \frac{\lambda(x)}{x}=0$. Indeed, this obviously follows from the fact that $\mathcal{O}(f(n))<n^2$. Now, assume for the sake of contradiction that there exists some constant $K$ for which the number of solutions to $a_\ell = c$ for some constant $c$ is at most $K$. Indeed, it must follow that for every constant $c$, there is a positive integer $\ell \leq cK+1$ where $a_\ell \geq c$. Thus, there is a function $t \colon \mathbb{N} \to \mathbb{N}$ where $a_{t(k)} \geq k$ and $\mathcal{O}(t(n)) \leq n$. Recall, $$k \leq a_{t(k)} < 2S_{t(k)}+1$$ So, $S_{t(k)} > \frac{k-1}{2}$ where $\mathcal{O}(t(n)) \leq n$. Thus, $S_i$ grows at least linearly. Contradiction. So, there exists no such constant $K$. Which implies the equation $a_{\ell}=c$ has arbitrarily many solutions for some choice in $c$.

Here is my solution: https://calimath.org/pdf/EGMO2021-6.pdf And I uploaded the solution with motivation to: https://youtu.be/GlX7b9mFQBM

Funny asymptote problem. Note that the LHS is equal to $$c_m = \sum_{i=1}^m d(i) = x \log x + (2\gamma - 1)x + O\left(\sqrt{x}\right)$$ Now, it remains to show by pigeonhole that $$\sum_{i=1}^{m} \left\lfloor \sqrt{c_i} \right\rfloor \ge C \cdot \left\lfloor \sqrt{c_m} \right\rfloor + O(1)$$ eventually holds for $C = 2 \cdot 10^6$. However, by taking $m$ large enough such that $$a^2 < c_{m-C} < c_{m-C+1} < \dots < \sqrt{c_m} < (a + 1)^2$$ (which follows as the growth rate is at most $\ln M$), the result follows.

Let $f(m)=\sum_{i=1}^m \lfloor \frac mi\rfloor$ and $g(i)=i^2-\lfloor \sqrt i \rfloor^2\leq 2\sqrt i+1$. Fix $m$. Consider $g(f(i))$ for $1\leq i \leq m$. Note $f(i)\leq 1434i\log i$, so $$g(f(i))\leq 2\sqrt{1434m\log m}+1\leq 3 \sqrt{1434m\log m}.$$ But for sufficently large $m$, $10^6\cdot 3\sqrt{1434m\log m}<m$ so such $a$ exists.

Let $T=2024^{2024}!!$ denote a ridiculously large number. We show that there exists $a$ such that there are more than $T$ different solutions $(m,n)$. Note that $$f(m)\leq \frac m1+\frac m2+\cdots+\frac mm = m\left(\frac 11+\frac12+\cdots+\frac 1m \right)<7m\log(m)$$ As $m$ gets larger, $\frac{\log(m)}m$ gets frighteningly smaller. The main idea now is to give construction and then use Pigeonhole Principle. Consider triples $(m,n,a)$ such that $n=\left\lfloor \sqrt{f(m)} \right\rfloor$. Then, $0\leq a=f(m)-n^2\leq2\sqrt{f(m)}$. Now, take large $N$ such that $f(N)<\frac{N^2}{2\cdot T^2}$. We can ensure this because $f(n)=o(n\log n)$ and for sufficiently large $n$, we can have $f(n)<cn^2$ for any constant $c>0$. Then, for any $1\leq i\leq N$, we have $f(i)<\frac{N^2}{4\cdot T^2}$, as the function is strictly increasing. Note that for each $1\leq i\leq N$, the (non negative) value of $a$ obtained is at most $2\sqrt{\frac{N^2}{4\cdot T^2}}=\frac NT$. Since there are $N$ values taken, one of the values of $a$ is chosen at least $T$ times by Pigeonhole Principle, as desired. $\blacksquare$