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This is a reference request.

Assume that $f_1,\ldots,f_r \in \mathbb{Z}[t]$ are non-zero linear polynomial.

Letting $\mu$ be the M\"{o}bius function, is there any work on $$ \sum_{p\leq x} \prod_{i=1}^r \mu(f_i(p))^2 \ \ ?$$

(One needs some obvious local conditions which should be built within the leading constant of the asymptotic I guess.)

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  • $\begingroup$ Am I correct to understand that you are asking for the density of square-free values of the polynomial $f(x) = f_1(x) \cdots f_r(x)$? $\endgroup$ – Stanley Yao Xiao Apr 25 '17 at 17:11
  • $\begingroup$ Almost. For any $p$ I am allowing some of the integers $f_i(p)$ to share a non-unit common divisor. Quite probably the problem you mention and the one I do have an almost identical proof-but I cannot find a reference for either of those. $\endgroup$ – Captain Darling Apr 25 '17 at 20:07
  • $\begingroup$ The two cases are roughly equivalent, since if $p$ divides two of the linear factors it must divide the resultant of those two linear factors, which is a fixed number; hence there are only finitely many primes that can do that. $\endgroup$ – Stanley Yao Xiao Apr 25 '17 at 20:13
  • $\begingroup$ The problem for square-free values at primes for a product of linear polynomials is not the same as your question, but it is close to it. The way one deals with it is reducing to the case of one linear polynomial using a "resultant trick"; then the necessary estimates for one linear polynomial follow from non-trivial results on the distribution of prime numbers in arithmetic progressions. This is well-known to experts. $\endgroup$ – Pasten Apr 26 '17 at 2:13
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Edit: I neglected to see that the OP asked for square-free values at prime arguments, in which some of the arguments below need to be modified. The necessary adjustments are found in the three papers linked below, where in each they dealt with the case of prime arguments.

Let $f(x) = f_1 \cdots f_r(x)$ with $f_j(x) = a_j x + b_j$ pairwise non-proportional linear polynomials with integer coefficients. Define the multiplicative function

$$\displaystyle \rho_f(n) = \# \{m \pmod{n} : f(m) \equiv 0 \pmod{n}\}$$

Define

$$\displaystyle N_f(X) = \#\{n \leq X : f(n) \text{ is square-free}\}$$ and define

$$\displaystyle N_f^{(1)}(X) = \# \{n \leq X : p^2 | f(n) \text{ implies } p > (\log X)/r\}.$$

One can show without much trouble by using simple properties of the Mobius function that, for any $\varepsilon > 0$,

$$\displaystyle N_f^{(1)}(X) = \prod_{p \leq (\log X)/r} \left(1 - \frac{\rho_f(p^2)}{p^2}\right) X + O_{f, \varepsilon}\left(X^{1 - 1/d + \varepsilon}\right).$$

The main term in the above expression tends to the expected main term for $N_f(X)$, which is

$$\displaystyle \prod_p \left(1 - \frac{\rho_f(p^2)}{p^2}\right)X.$$ Note that this product converges absolutely. Thus it suffices to estimate the term

$$\displaystyle N_f^{(2)}(X) = \# \{n \leq X : \exists p \text{ s.t. } p^2 | f(n) \text{ and } p > (\log X)/r\}.$$

To do so, let us examine the possibilities. There can be two reasons for $p^2$ to divide $f(x)$: either $p^2 | f_j(x)$ for some $1 \leq j \leq r$ or $p | f_i(x), p | f_j(x)$ for distinct indices $i$ and $j$. In the second case, $p$ divides the resultant of $f$ and $g$, which is just a rational integer (in fact, a divisor of the discriminant of $f$). Since $\Delta(f)$ is assumed to be non-zero, this can only happen for finitely many primes; and by taking $X$ large enough, we can ignore this possibility.

Thus we assume that $p^2 | f_j(x)$ for some $j$. For each $j$, define

$$\displaystyle N_{f,j}(X) = \# \{n \leq X : \exists p^2 | f_j(n) \text{ with } p > (\log X)/r\}.$$

Observe that $f_j(n) \ll X$ since $n \leq X$, and thus $p \ll \sqrt{X}$. By using the same argument which allows one to show that the number of square-free numbers up to $X$ is asymptotic to $(6/\pi^2)X$, one sees that $N_{f,j}(X) = O_{f_j}(X^{1/2})$ for $j = 1,2, \cdots, r$. Moreover, we see that

$$\displaystyle N_f^{(2)}(X) \leq \sum_{j = 1}^r N_{f,j}(X) = O_f(X^{1/2}),$$

and so $N_f^{(1)}(X)$ approximates $N_f(X)$, obtaining the correct order of magnitude.

For a review of such results, see:

https://arxiv.org/abs/1103.2028

https://people.maths.bris.ac.uk/~matdb/preprints/powerfree.pdf

https://projecteuclid.org/euclid.acta/1485801837

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