# Density of square-free values of polynomials of degree $1$ over integers.

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?)

• 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 $\frac{1}{a}\frac{6}{\pi^2}$
– lhf
May 8 '18 at 12:25
• Any thoughts on how I can prove this? May 8 '18 at 13:21
• 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 :) May 8 '18 at 23:33