can't you just take 199 integers before a very large square?

ythomashu wrote: can't you just take 199 integers before a very large square? For large $k$, why should $k^2-1$ not be almost square? In fact $k^2-1 = (k-1)(k+1)$, which is almost square for sufficiently large $k$.

We consider set $T$ of positive integers between $N^2$ and $M^2$ including $N^2$ then $|T|=M^2-N^2$. Let $S$ be set of almost-squares numbers in $T$, i.e. $S=\{a_1, a_2, \ldots \}$ where $N^2\le a_1<a_2< \ldots<M^2$. With this, we can partition $T$ into $|S|$ subsets $T_i= \{ n| a_i \le n < \max\{M^2,a_{i+1}\}\}$. Observe that $T_i$ contains $|T_i|-1$ consecutive integers that are not almost-squares. The main idea is that there must exists a subset $T_i \subset T$ such that $|T_i| \ge |T|/|S|$. We wish to find a lower bound for $|T|/|S|$. Since we already know $|T|$, it suffices to find an upper bound for $|S|$. Claim 1. $|S| \le \frac{1}{200}(M-N)(M+N+199)$. Proof. Denote $X_k$ set of almost-squares of the form $ab$ so $a<k, ab \ge k^2$. $Y$ set of almost-squares of the form $ab$ where $N \le a<M$. Then we see that $|S| \le |Y|-|X_M|+|X_N|$. Now it suffices to find some bound for each such set. For any $a$, number of almost-squares of the form $ab$ where $a \le b \le 1.01$ is at most $ \frac{a}{100}+1$. Therefore, we can find \[|Y| \le \sum_{i=N}^{M-1} \left( \frac{i}{100}+1\right) = \frac{1}{200}(M-N)(M+N+199).\]For $X_k$, since $a<k$ and $k^2 \le ab \le 1.01a^2$ so $\frac{k}{\sqrt{1.01}}\le a<k$. For such each $a$, we can choose $\frac{k^2}{a} \le b \le 1.01a$ to obtain $ab \in X_k$. Therefore, $|X_k| \ge \varepsilon k$ for some $\varepsilon>0$. On the other hand, $|X_k|$ is also bounded above since $k^2 \le ab \le 1.01a^2<1.01k^2$ for any $ab \in X_k$. Therefore, we can choose sufficient large $M$ such that $|X_M|>|X_N|$. Thus, we obtain $|S| \le |Y| \le \frac{1}{200}(M-N)(M+N+199)$. $\square$ From the claim and the observation at the beginning, there must exists a subset $T_i \subset T$ such that \[|T_i| \ge |T|/|S| \ge \frac{200(M+N)}{M+N+199}>199.\]Since $|T_i| \ge 200$, there exists $199$ consecutive integers in $T_i$ that are not almost-square. This is true for arbitrarily large $N$.