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.
 A: $48 = 3 \times 2^4,49 = 7^2,50 = 2 \times 5^2$ would be the first example of a sequence with $3$ consecutive numbers. I found it simply by looking at them all until I found the first one. Given any set of distinct coprime numbers $a_1, \dots, a_k$, the Chinese Remainder Theorem ensures us of the existence of a solution to $n + i \equiv 0 \pmod {a_i}$, which gives you at least the existence of some large possibility for a sequence of non-square free consecutive integers. 
For the info, the first consecutive non-square free integers I found were $(8,9)$, $(24,25)$ and $(44,45)$. Then I found $(48,49,50)$. Note that you can easily generate infinitely many such triples in the following manner : Since the prime divisors of $(48,49,50)$ that are the square parts of them are $2,5,7$, then the triple ($48 + k(2\times 5 \times 7)^2$, $49 + k(2 \times 5 \times 7)^2$, $50 + k (2 \times 5 \times 7)^2$) is also a sequence of three consecutive non-square free numbers ($k \in \mathbb N$).
Hope that helps,
A: See http://oeis.org/A045882 and references given there.
A: 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.
