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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.

Bill Dubuque
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Sigur
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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.

Jim Belk
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    By subtracting from $900$ you instantly get $350$, $351$ and $352$ as another triplet, and not surprisingly there are six triplets involving multiples of $2^2$, $3^2$ and $4^2$ in every $900$ integers corresponding to the permutations of the three squares. – Henry Jun 19 '12 at 23:31
  • Hi. Why are you supposing the first number $n$ being $n\equiv 0 (\mod 4)$? – Sigur Jun 19 '12 at 23:32
  • @Thiago This is the same as saying that $n$ is a multiple of $4$, which will guarantee that it isn't square-free. – Jim Belk Jun 19 '12 at 23:41
  • Yes, I agree. But if $n$ is not multiple of 4 it could be still non-free square. To find 3 numbers, OK, you got it. But may I think like you in the general case? For example, to find a list with 30 consecutive numbers. May we suppose $n\equiv 0 \mod 4$?.

    Does your solution states that we always find such a list starting with a multiple of 4? Even for $k$ numbers instead of 3? Thanks.

     – Sigur Jun 19 '12 at 23:52
  • @Thiago For $4$ numbers, you would want the same three conditions, plus $n+3\equiv 0\pmod{49}$. For $5$ numbers, you would also add in $n+4 \equiv 0\pmod{121}$. And so forth. – Jim Belk Jun 20 '12 at 01:15
  • 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. – Sigur Jun 20 '12 at 01:25
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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,

Patrick Da Silva
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    Thanks for the first list. I can not believe that I didn't find it... lol – Sigur Jun 19 '12 at 23:26
  • Thiago : Sometimes we must not think that computers do everything, it is possible to just quickly try our own hands at it. Computers are mostly useful when you expect things to get large. There are not many consecutive non-squarefree at first because there are lots of small primes and products of small primes, but as you make the size grow just a little, non-squarefree numbers own the place. – Patrick Da Silva Jun 19 '12 at 23:35
  • Yes. I am trying to find larger lists. Maybe you want to edit your post to correct the index $a_i$ and the parenthesis on the second number of your general list. I tried but it does not work. – Sigur Jun 19 '12 at 23:58
  • @Thiago : I just forgot to put some ${$ $}$, but thanks for noticing. – Patrick Da Silva Jun 20 '12 at 00:10
  • How about $49+k(2\times 5\times 7)$, ... – Sigur Jun 20 '12 at 00:13
  • Thiago : You want to make sure that adding the multiple keeps the non-squarefreeness of the numbers, that is precisely why I add a multiple of the square of the primes. If you would try the triple $48,49,50$ + $2\times 5 \times 7$, note how $48 + 70$ is not divisible by $4$ anymore. – Patrick Da Silva Jun 20 '12 at 00:19
  • Sorry. I was talking about the miss print. You forgot to close the ). Thanks. – Sigur Jun 20 '12 at 00:21
  • @Thiago, oh, lol, my bad. I'll fix this too! – Patrick Da Silva Jun 20 '12 at 00:36
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See http://oeis.org/A045882 and references given there.

Robert Israel
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  • Thanks so much. The list is very interesting. – Sigur Jun 20 '12 at 16:37
  • While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – Тyma Gaidash Sep 06 '23 at 12:09
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Here is a way to obtain infinitely many sets of $3$ consecutive non-square-free positive integers.

Note that there are infinitely many solutions to the negative Pell equation $ n^2−2m^2=−1.$

Given a solution $(n,m)$, $n$ must be odd, so $\color{blue}{n^2-1}$ is divisible by $4=2^2$;

$\color{blue}{n^2}$ is a square; and $\color{blue}{n^2+1=2m^2}$ is twice a square.

J. W. Tanner
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