A Distribution Problem for Powerfree Values of Irreducible Polynomials

by B. Beasley and M. Filaseta

Theorem 2. Let k ≥ (\sqrt{2} − 1/2)g. Given any γ > 0, there exists a δ = δ(γ) > 0 such that

∑_{\begin{matrix}s_{n+1}≤X \\ s_{n+1}−s_{n}≤X^{δ} \end{matrix}}(s_{n+1} − s_{n})^{γ} ∼ B(γ, f, k)X,

for some constant B(γ, f, k) depending only on γ, f(x), and k.

Definition 1. Given f(x) ∈ ℤ[x], let s_{n} = s_{n}(f) be the nth positive integer m such that f(m) is k-free. Let

L(h) = L(h, X) = L(h, X, f) = | {n ∈ ℤ^{+} : h < s_{n+1} − s_{n} ≤ 2h, X/2 < s_{n+1} ≤ X} |.

(2)  L(2h) ≪ \frac{X}{h^{γ+ε}}

Comment: (2) & Theorem 3 ⟹ Theorems 1 & 2 (sort-of).

Notation: D = R(f, f′), the resultant of f and f′ z = 4g A = \{p : p|D or p ≤ z\} Q = ∏_{p∈A} p^{k} H = \frac{h log h}{8gQ} ρ(q) = ρ(f, q) = |\{a ∈ ℤ_{q} : f(a) ≡ 0 (mod q)\}| T ∈ T = \{2^{j}H : j = 0, 1, …, J\}

Lemma 4. Let

F (n) = F (n, h, f) = ∑_{n<m≤n+h} ∑_{\begin{matrix}p>H \\ p^{k}|f(m) \end{matrix}} 1.

If there are no integers m in (n, n + h] such that f(m) is k-free, then

F (n) ≥ \frac{h}{4Q}.

Lemma 5. For r a positive integer, define

M_{r} = M_{r}(h, X, f) = ∑_{X/2<n≤X} F^{r}(n).

Then

L(2h) ≪_{r} \frac{M_{r}}{h^{r+1}} + 1.

Definition 2. Let S_{r}(T) be the number of 2r-tuples (p_{1}, …, p_{r}, m_{1}, …, m_{r}) with T < p_{1} < p_{2} < ⋯ < p_{r} ≤ 2T such that, for each j ∈ {1, 2, …, r},

|m_{1} − m_{j}| < h, \frac{X}{2} < m_{j} ≤ X + h ≤ 2X, and p_{j}^{k}|f(m_{j}).

Notation: G(n, T) = G(n, T, f) = ∑_{n<m≤n+h} ∑\begin{matrix}p^{k}|f(m) \\ T <p≤2T \end{matrix} 1

Lemma 6. Let r be a positive integer. If r = 1, then

∑_{X/2<n≤X} \binom{G(n, T)}{r} ≤ (2g)^{r}hS_{r}(T).

If r > 1, then

∑_{\begin{matrix}X/2<n≤X \\ G(n,T)≥2gr \end{matrix}} \binom{G(n, T)}{r} ≤ (2g)^{r}hS_{r}(T).

Lemma 7. If T ≥ H and r ∈ ℤ^{+}, then

S_{r}(T) ≪ \frac{h^{r−1}X}{T^{(k−1)r} log^{r} T} + \frac{h^{r−1}T^{r}}{log^{r} T} .

Lemma 8. For T as previously defined,

∑_{\begin{matrix}T ∈T \\ U<T ≤V \end{matrix}} T^{a} ≪ \{ \begin{matrix}U^{a} & if a < 0 \\ V^{a} & if a > 0. \end{matrix}

Lemma 9. Let k ≥ (\sqrt{2} − 1/2)g. Let ε ∈ (0, (k − 1)/k] such that

ε < 1 − \frac{8g(g − 1)}{(2k + g)^{2} − 4}.

If X^{1−ε} < T ≪ X^{g/k}, then there is a ξ = ξ(ε) > 0 such that

S_{1}(T) ≪ X^{1−ξ}.