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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.

3 Answers 3

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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.

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    It's interesting how your solution is composed by large numbers. See the other solution below, given by Patrick. It starts with multiple of$4$but uses power of$5$and 7, not 3 as yours. I guess that it's impossible to know which primes to use to obtain a small solution.2012-06-20
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$48 = 3 \times 2^4,49 = 7^2,50 = 2 \times 5^2$ 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,

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    @Thiago, oh, lol, my bad. I'll fix this too!2012-06-20
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See http://oeis.org/A045882 and references given there.

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    Thanks so much. The list is very interesting.2012-06-20