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I'm reading an article by Dan Carmon on Square-free values of large polynomials over the rational function field and whilst investigating its meaning, I started reading a proof showing that

The 'probability' that a large random integer is square-free is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.

I know I haven't defined probability and density explicitly, but hope you understand its meaning anyway. It made use of the following

Given two random integers from $\{1,2,\ldots, n\}, n\in \mathbb{N}$. The asymptotic probability that they are relative prime as $n\to \infty$ is $\frac{6}{\pi^2}$.

My question is the following:

If we look at polynomials of degree $1$, the above clearly proves that the density of square-free values of the polynomial $f(x) = x$ (over the integers) is $\frac{6}{\pi^2}$. However, why it this the case for every polynomial of degree $1$? (and seeing the comment below, it this even true?)

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    $\begingroup$ I'd expect that being square-free and being in an arithmetic progression are independent events and so the probability of $ax+b$ being square-free should be $\frac{1}{a}\frac{6}{\pi^2}$ $\endgroup$
    – lhf
    May 8 '18 at 12:25
  • $\begingroup$ Any thoughts on how I can prove this? $\endgroup$
    – Whizkid95
    May 8 '18 at 13:21
  • $\begingroup$ I think in particular, the probability that two random integers are relatively prime (coprime) is $\frac{6}{\pi^2}$. But for sufficiently large integers, I believe you are correct :) $\endgroup$
    – Mr Pie
    May 8 '18 at 23:33
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In the paper A Note on Square‐Free Numbers in Arithmetic Progressions, Hooley says:

enter image description here

The paper by Prachar is Über die kleinste quadratfreie Zahl einer arithmetischen Reihe.

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