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As the title says I was wondering whether there are infinitely many squarefree values of $n(n+4)$, $n \in \mathbb N$. Clearly $n \notin 2 \mathbb N$, but that is unfortunately the most I've gotten to so far. I know that questions about squarefree values of polynomials are extremely tough, but perhaps there exists a solution for this particular instance?

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up vote 5 down vote accepted

For an odd number to be squareful, it needs to be divisible by the square of an odd prime. $n$ and $n+4$ cannot share any odd factors, so one of them must be divisible by the square of an odd prime. According to this MathOverflow question $$\sum_{p\text{ prime}}{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots \text{so}\\ \sum_{p\text{odd prime}}{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.2022474200\dots$$ at most $20.225\%$ of odd numbers are squareful. The first term counts the fraction divisible by $3^2$, the second the number divisible by $5^2$, etc. Some odds are divisible by the square of more than one odd prime, so the sum is an overcount. Then at most $40.25\%$ of odd numbers of the form $n(n+4)$ are squareful. Infinitely many odd numbers of the form $n(n+4)$ must be squarefree. Thanks to Hagen von Eitzen for the correction.

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Where does this make use of the specific form $n(n+4)$ of the numbers? (E.g. $n(n+1)(n+2)(n+3)$ is never square-free) – Hagen von Eitzen Feb 20 '14 at 20:27
@HagenvonEitzen: The idea is that $n$ can only be squareful less than $45.225\%$ of the time, same for $n+4$, so the product can only be squareful twice that, and in fact will be less. This fails because if $n \equiv 2 \pmod 4, 4|n(n+4)$. I believe it can be patched up, but will need to work on it. – Ross Millikan Feb 20 '14 at 20:33
Ah, I see - my eyes matched $0.45$ with $(9)0.45$ unconciously. - We can drop $\frac1{2^2}$ from the sum, so we end up with $0.2022472300\ldots$. Thus at most $40.45\%$ of all numbers $n(n+4)$ have an odd square factor. And $50\%$ have an even square factor. That still leaves $9.55\%$ without any square factor. That should fix it. – Hagen von Eitzen Feb 20 '14 at 20:40

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