, 2014

If $N$ is a large enough positive integer, then $N^2 - 1 = (N - 1)(N + 1)$ and $N^2 - 4 = (N - 2)(N + 2)$ are both $\varepsilon$-squarish. Next, if $k$ is a large enough positive integer and $N = (k-1)(k+2) = k^2 + k - 2$, then $N^2 = (k-1)^2(k+2)^2$, $N^2 - 2 = (k^2 - 2)(k^2 + 2k - 1)$ and $N^2 - 5 = (k^2 - k - 1)(k^2 + 3k + 1)$ are all three $\varepsilon$-squarish. Finally, if $n$ is a large enough positive integer and $N = 2n^2 - 2$, then $N^2 - 3 = (2n^2 - 2n - 1)(2n^2 + 2n - 1)$ is $\varepsilon$-squarish.

Consequently, $N^2 - 5$, $N^2 - 4$, $\ldots$, $N^2$ are six consecutive $\varepsilon$-squarish integers, provided that $N = k^2 + k - 2 = 2n^2 - 2$, where $k$ and $n$ are sufficiently large integers. To conclude, write $m = 2k + 1$ to turn the condition into a Pell equation, $m^2 - 8n^2 = 1$, which has arbitrarily large solutions,
$$\begin{pmatrix} m_r \\ n_r \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 1 & 3 \end{pmatrix}^r \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad r \in \mathbb{N}.$$