Consecutive non square free numbers I was thinking to solve this by computer programs but I prefer a solution.
How to obtain a list of 3 consecutive non square free positive integers? In general, how to obtain the same kind of list with $k$ elements? Thanks.

Let $n$ be the first number.  It will work if we can arrange for the following:
$$
n\equiv 0\pmod{4}
$$
$$
n+1\equiv 0\pmod{9}
$$
$$
n+2\equiv 0\pmod{25}
$$
Using the Chinese Remainder Theorem, the first two congruences are equivalent to requiring that $n\equiv 8 \pmod{36}$.  Combining this with the third congruence gives that $n\equiv 548\pmod{900}$.  Thus, three such numbers are $548$, $549$, and $550$.
A similar algorithm works for $k$ consecutive square-free numbers.
