A NOTE ON POWERFUL NUMBERS IN SHORT INTERVALS

Abstract

We investigate uniform upper bounds for the number of powerful numbers in short intervals $(x, x + y]$. We obtain unconditional upper bounds $O({y}/{\log y})$ and $O(\kern1.3pt y^{11/12})$ for all powerful numbers and $y^{1/2}$-smooth powerful numbers, respectively. Conditional on the $abc$-conjecture, we prove the bound $O({y}/{\log ^{1+\epsilon } y})$ for squarefull numbers and the bound $O(\kern1.3pt y^{(2 + \epsilon )/k})$ for k-full numbers when $k \ge 3$. These bounds are related to Roth’s theorem on arithmetic progressions and the conjecture on the nonexistence of three consecutive squarefull numbers.

1 Introduction and main result

A number n is squarefull if its prime factorisation $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ satisfies $a_i \ge 2$ for all $1 \le i \le r$ . Similarly, a number n is k-full if $a_i \ge k$ for $1 \le i \le r$ . For example, $72 = 2^3 \cdot 3^2$ is squarefull and $243 = 3^5$ is $5$ -full. Let $Q_k(x)$ denote the number of k-full numbers which are less than or equal to x. It is known that

(1.1) $$ \begin{align} Q_k(x) = \prod_{p} \bigg(1 + \sum_{m = k+1}^{2k - 1} \frac{1}{p^{m/h}} \bigg) \kern1pt x^{1/k} + O(x^{1/(k+1)}), \end{align} $$

where the product is over all primes (see, for example, [1, 4]). There are also estimates for the number of k-full numbers in short intervals $(x, x+y]$ with $y = o(x)$ . For moderate size y, there are some asymptotic results. For example, Trifonov [8] and Liu [6] respectively obtained

$$ \begin{align*} Q_2 (x + x^{1/2 + \theta} ) - Q_2(x) \sim \frac{\zeta(3/2)}{2 \zeta(3)} x^\theta \quad\text{for } \frac{19}{154} = 0.12337\ldots < \theta < \frac{1}{2}, \end{align*} $$

and

$$ \begin{align*} Q_3 (x + x^{2/3 + \theta} ) - Q_3(x) \sim \frac{\zeta(4/3)}{3 \zeta(4)} x^\theta \quad\text{for } \frac{5}{42} = 0.11904\ldots < \theta < \frac{1}{3}. \end{align*} $$

What happens when y is very small, say $y \ll x^{1/2}$ or even $y \ll \log x$ ? For such short intervals, one can only expect suitable upper bounds rather than asymptotic formulae. Thus, in this note, we are interested in finding uniform upper bounds for $Q_k(x + y) - Q_k(x)$ with $1 \le y \le x$ that are independent of x. By comparing k-full numbers with perfect kth powers, we suspect the following conjecture to be true.

Conjecture 1.1. Given an integer $k \ge 2$ and a real number $x \ge 1$ , there exists some constant $C_k \ge 1$ such that

$$ \begin{align*} Q_k(x + y) - Q_k(x) \le C_k y^{1/k} \end{align*} $$

uniformly over $1 \le y \le x$ .

We are far from proving this at the moment. The current best upper bound,

(1.2) $$ \begin{align} Q_k(x + y) - Q_k(x) \ll \frac{y \log \log (\kern1.3pt y + 2)}{\log (\kern1.3pt y + 2)}, \end{align} $$

was obtained by De Koninck et al. [3]. We improve (1.2) slightly.

Theorem 1.2. Given an integer $k \ge 2$ and a real number $x \ge 1$ , we have

(1.3) $$ \begin{align} Q_k(x + y) - Q_k(x) \ll \frac{y}{\log (\kern1.3pt y + 1)} \end{align} $$

uniformly over $1 \le y \le x$ .

In fact, we shall prove the following more general result concerning squarefull numbers in arithmetic progression over short intervals which gives Theorem 1.2 immediately, as k-full numbers are included in squarefull numbers.

Theorem 1.3. Given real numbers $x \ge 1$ and $0 < \alpha < 1$ and integers $q> 0$ and r with $\gcd (r, q) = 1$ , we have

$$ \begin{align*} \sum_{\substack{x < n \le x + y\\ n\ {\mathrm{squarefull}}\\ n \equiv r \!\pmod{q}}} 1 \ll_\alpha \frac{y}{\phi(q) \log (\kern1.3pt y + 1)} \end{align*} $$

uniformly over $1 \le y \le x$ and $1 \le q \le y^{1 - \alpha }$ .

Using a similar technique, we can obtain some power savings over (1.3) for smooth k-full numbers in short intervals.

Theorem 1.4. Given an integer $k \ge 2$ and a real number $x \ge 1$ , we have

(1.4) $$ \begin{align} \sum_{\substack{x < n \le x + y\\ n\ k\mbox{-}\mathrm{full}\\ p^{+}(n) \le y^{1/2}}} 1 \le \sum_{\substack{x < n \le x + y\\ n\ {\mathrm{squarefull}}\\ p^{+}(n) \le y^{1/2}}} 1 \ll y^{11/12} \end{align} $$

uniformly over $1 \le y \le x$ . Here $p^{+}(n)$ stands for the largest prime factor of n.

One may increase the exponent $1/2$ up to $1$ and obtain a similar power saving upper bound.

The bound (1.4) lends evidence towards Conjecture 1.1 and shows that the difficulty lies with nonsmooth k-full numbers. Another piece of evidence comes from the famous $abc$ -conjecture. It was proved in [3] that, given any $\delta> 0$ , the interval

(1.5) $$ \begin{align} (x, x + x^{1 - (2 + \delta) / k}] \end{align} $$

contains at most one k-full number for sufficiently large x under the $abc$ -conjecture. From this, one has the following result.

Theorem 1.5. Assume the $abc$ -conjecture. Given an integer $k \ge 2$ and real numbers $\delta> 0$ and $x \ge 1$ , we have

(1.6) $$ \begin{align} Q_k(x + y) - Q_k(x) \ll_{\epsilon, k} y^{(2 + \delta) / k} \end{align} $$

uniformly over $1 \le y \le x$ .

We shall modify the proof in [3] concerning (1.5) slightly to correct an inaccuracy (since the a, b, c in the application of the $abc$ -conjecture might not be relatively prime). Then we apply it to obtain Theorem 1.5. Observe that (1.5) or (1.6) give us nothing nontrivial when $k = 2$ . To remedy this, we shall prove the following conditional result which improves (1.3) slightly by a small power of a logarithm.

Theorem 1.6. The $abc$ -conjecture implies that for some absolute constant $c> 0$ ,

$$ \begin{align*} Q_2(x+y) - Q_2(x) \ll \frac{y}{\log^{1+c} (\kern1.3pt y+1)} \end{align*} $$

uniformly over $1 \le y \le x$ .

The proof relies on the following recent breakthrough result of Bloom and Sisask on the density of integer sequences without three-term arithmetic progressions.

Theorem 1.7 (Bloom–Sisask, [2])

Let $N \ge 2$ and $A \subset \{1, 2, \ldots , N \}$ be a set with no nontrivial three-term arithmetic progressions, that is, solutions to $x + y = 2z$ with $x \,{\neq}\, y$ . Then

$$ \begin{align*} |A| \ll \frac{N}{(\log N)^{1+c}}, \end{align*} $$

where $c> 0$ is an absolute constant.

This paper is organised as follows. First, we will prove Theorems 1.3 and 1.4 using the Brun–Titchmarsh inequality and ideas from Shiu’s generalisation [7]. Then we will prove Theorem 1.5 using the $abc$ -conjecture. Finally, we will prove Theorem 1.6 by establishing the nonexistence of three-term arithmetic progressions for squarefull numbers in short intervals.

Notation. We use $| A |$ to denote the number of elements in a finite set A and $\lfloor x \rfloor $ to denote the greatest integer less than or equal to x. We let $p_{-}(n)$ and $p^{+}(n)$ be the smallest and the largest prime factor of n, respectively. The symbols $f(x) = O(g(x))$ and $f(x) \ll g(x)$ are equivalent to $|f(x)| \leq C g(x)$ for some constant $C> 0$ . Also, $f(x) = O_{\lambda _1, \ldots , \lambda _r} (g(x))$ and $f(x) \ll _{\lambda _1, \ldots , \lambda _r} g(x)$ mean that the implicit constant may depend on $\lambda _1, \ldots , \lambda _r$ . Furthermore, $f(x) = o(g(x))$ means $\lim _{x \rightarrow \infty } f(x)/g(x) = 0$ and $f(x) \sim g(x)$ means $\lim _{x \rightarrow \infty } f(x)/g(x) = 1$ . Finally, the summation symbol $\mathop {\sum \nolimits '}$ signifies that a sum is over squarefull numbers only.

2 Some preparations

Lemma 2.1. For any $X \ge 1$ ,

$$ \begin{align*} \mathop{\sum\nolimits'}_{X < n \le X^2} \frac{1}{n} \ll X^{-1/2}. \end{align*} $$

Proof. From (1.1), $Q_2(X) \ll X^{1/2}$ . By partial summation, the above sum is

$$ \begin{align*} \hspace{-22pt}\int_{X}^{X^2} \!\frac{1}{u} \,d Q(u) = \frac{Q(X^2)}{X^2} - \frac{Q(X)}{X} + \!\int_{X}^{X^2} \frac{Q(u)}{u^2} \,du \ll \frac{1}{X^{1/2}} + \!\int_{X}^{X^2} \frac{1}{u^{3/2}} \,du \ll \frac{1}{X^{1/2}}.\\[-42pt] \end{align*} $$

Lemma 2.2 (Brun–Titchmarsh inequality)

Let $q \ge 1$ and r be integers satisfying $\gcd (r, q) = 1$ . Suppose $q < y \le x$ and $z \ge 2$ . Then,

$$ \begin{align*} \sum_{\substack{x < n \le x + y\\ n \equiv r \!\pmod{q}\\ p_{-}(n)> z}} 1 \ll \frac{y}{\phi(q) \log z} + z^2. \end{align*} $$

The above bound is still true when $y \le q$ or $y < 1$ since there is at most one term in the sum. The estimate follows from the Selberg upper bound sieve method (see, for example, [5, page 104]).

Finally, let us recall the $abc$ -conjecture. For any nonzero integer m, the kernel of m is

$$ \begin{align*} \kappa(m) := \prod_{p | m} p. \end{align*} $$

Conjecture 2.3 ( $abc$ -conjecture)

For any $\epsilon> 0$ , there exists a constant $C_\epsilon> 0$ such that, for any integers $a, b, c$ with $a + b = c$ and $\gcd (a,b) = 1$ , we have

$$ \begin{align*} \max\{ |a|, |b|, |c| \} \le C_\epsilon \kappa(a b c)^{1 + \epsilon}. \end{align*} $$

3 Proof of Theorem 1.3

Our proof is inspired by Shiu [7] on the Brun–Titchmarsh theorem for multiplicative functions. We may assume that $y \ge 2^{2/\alpha }$ for the theorem is clearly true when $1 \le y < 2^{2/\alpha }$ by choosing a large enough implicit constant. Recall that $1 \le q \le y^{1 - \alpha }$ for some $\alpha> 0$ . Let $z = y^{\alpha / 2} \ge 2$ . Any squarefull number n in $[x, x + y]$ can be factored as

$$ \begin{align*} n = \underbrace{p_1^{a_1} \cdots p_j^{a_j}}_{b_n} \underbrace{p_{j+1}^{a_{j+1}} \cdots p_s^{a_s}}_{d_n}\quad \text{with } p_1 < p_2 < \cdots < p_s, \end{align*} $$

where j is the greatest index such that $ p_1^{a_1} \cdots p_j^{a_j} \le z$ . Hence, $b_n \le z < b_n p_{j+1}^{a_{j+1}}$ . Note that j may be $0$ (the product is an empty product) if $p_1^{a_1}> z$ . In this case, $b_n = 1$ and $d_n = n$ . Also, since $n \equiv r \pmod {q}$ with $\text {gcd}(r,q) = 1$ , we must have $\text {gcd}(b_n, q) = 1 = \text {gcd}(d_n, q)$ .

Case 1: $b_n> z^{1/2}$ . As $q \le y^{1 - \alpha }$ and $z = y^{\alpha /2}$ , the number of such squarefull numbers is bounded by

(3.1) $$ \begin{align} \mathop{\sum\nolimits'}_{\substack{z^{1/2} < b \le z\\ \gcd(b, q) = 1}} \sum_{\substack{x < n \le x + y\\ b \mid n\\ n \equiv r \!\pmod{q}}} 1 \le \mathop{\sum\nolimits'}_{z^{1/2} < b \le z} \bigg( \frac{y/b}{q} + 1 \bigg) \ll \frac{y}{q z^{1/4}} + z^{1/2} \ll_\alpha \frac{y}{\phi(q) \log y} \end{align} $$

by (1.1) and Lemma 2.1.

Case 2: $b_n \le z^{1/2}$ and $p_{-}(d_n) \le z^{1/2}$ . Then $p_{j+1} \le z^{1/2}$ and $p_{j+1}^{a_{j+1}}> z^{1/2}$ which implies $p_{j+1}^{-a_{j+1}} \le \min (z^{-1/2}, p_{j+1}^{-2})$ as $a_{j+1} \ge 2$ . Hence, the sum

$$ \begin{align*} \sum_{p_{j+1} \le z^{1/2}} \frac{1}{p_{j+1}^{a_{j+1}}} \le \sum_{p_{j+1} \le z^{1/4}} z^{-1/2} + \sum_{z^{1/4} < p_{j+1} \le z^{1/2}} \frac{1}{p_{j+1}^2} \ll \frac{1}{z^{1/4}}. \end{align*} $$

Therefore, by replacing $p_{j+1}^{a_{j+1}}$ with a generic $p^a$ , the number of squarefull numbers in this case is bounded by

(3.2) $$ \begin{align} \sum_{\substack{p \le z^{1/2}\\ \gcd(p,q) = 1}} \sum_{\substack{x < n \le x + y\\ p^a \mid n\\ n \equiv r \!\pmod{q}}} 1 \le \sum_{p \le z^{1/2}} \bigg(\frac{y / p^a}{q} + 1 \bigg) \ll \frac{y}{q z^{1/4}} + z^{1/2} \ll_\alpha \frac{y}{\phi(q) \log y}, \end{align} $$

since $q \le y^{1 - \alpha }$ and $z = y^{\alpha /2}$ .

Case 3: $b_n \le z^{1/2}$ and $p_{-}(d_n)> z^{1/2}$ . As $q \le y^{1 - \alpha }$ and $z = y^{\alpha /2}$ , the number of such squarefull numbers is bounded by

(3.3) $$ \begin{align} \mathop{\sum\nolimits'}_{\substack{b \le z^{1/2}\\ \gcd(b,q) = 1}} \sum_{\substack{x/b < n/b \le (x + y)/b\\ p_{-}(n / b)> z^{1/2}\\ (n/b) \equiv r \overline{b} \!\pmod q}} 1 \ll \mathop{\sum\nolimits'}_{b \le z} \bigg( \frac{y / b}{\phi(q) \log z} + z \bigg) \ll \frac{y}{\phi(q) \log z} + z^{3/2} \ll_\alpha \frac{y}{\phi(q) \log y} \end{align} $$

by (1.1), Lemma 2.2 and the convergence of the sum of reciprocals of squarefull numbers (which follows from Lemma 2.1 for instance). Here $\overline {b}$ denotes the multiplicative inverse of $b \;(\bmod {q})$ , that is, $b \overline {b} \equiv 1 \pmod {q}$ .

Combining (3.1), (3.2) and (3.3), we have Theorem 1.3.

4 Proof of Theorem 1.4

This is very similar to the proof of Theorem 1.3, so we just highlight the necessary adjustments. We set $q = 1$ and $z = y^{1/3}$ . The arguments for Case 1 and Case 2 are exactly the same as (3.1) and (3.2), and we get the bound

$$ \begin{align*} \frac{y}{z^{1/4}} + z^{1/2} \ll y^{11/12}. \end{align*} $$

It remains to deal with Case 3, where $b_n \le z^{1/2}$ and $z^{1/2} < p_{-}(d_n) \le y^{1/2}$ as the squarefull numbers are assumed to be $y^{1/2}$ -smooth. Thus, with $p := p_{-}(d_n)$ and $d_n := p^2 d$ , the number of squarefull numbers in this case is bounded by

$$ \begin{align*} \mathop{\sum\nolimits'}_{b \le z^{1/2}} \sum_{z^{1/2} < p \le y^{1/2}} & \sum_{\substack{x/b < p^2 d \le (x + y)/b\\ p_{-}(d)> z^{1/2}}} 1 = \mathop{\sum\nolimits'}_{b \le z^{1/2}} \sum_{z^{1/2} < p \le y^{1/2}} \sum_{\substack{{x}/{b p^2} < d \le {(x + y)}/{b p^2}\\ p_{-}(d)> z^{1/2}}} 1 \\ \ll& \mathop{\sum\nolimits'}_{b \le z^{1/2}} \sum_{z^{1/2} < p \le y^{1/2}} \bigg( \frac{y/ (b p^2)}{\log z} + z \bigg) \ll \frac{y}{z^{1/2} \log z} + \frac{z^{5/4} y^{1/2}}{\log y} \ll y^{11/12} \end{align*} $$

by (1.1), Lemma 2.2 and the convergence of the sum of reciprocals of squarefull numbers. The above bounds together yield Theorem 1.4.

5 Proof of (1.5) and Theorem 1.5

Given an integer $k \ge 2$ and a small real number $\delta> 0$ , we claim that the interval from (1.5), namely

$$ \begin{align*} (x, x + x^{1 - (2 + \delta)/k}], \end{align*} $$

contains at most one k-full number for all sufficiently large $x> C$ (in terms of $\delta $ and k) under the $abc$ -conjecture.

Following De Koninck et al. [3], we suppose that the interval $(x, x + x^{1 - (2 + \delta )/k}]$ contains two k-full numbers, $b < c$ . Then $c = a + b$ for some integer a with $0 < a \le x^{1 - (2 + \delta )/k}$ . With $d = \text {gcd}(a, b)$ , the integers $a/d$ , $b/d$ and $c/d$ are pairwise relatively prime. Note that $\kappa (n) \le n^{1/k}$ for any k-full number. Applying the $abc$ -conjecture with $\epsilon = \delta / k$ to the equation $a/d + b/d = c/d$ , we get

$$ \begin{align*} \frac{x}{d} < \frac{c}{d} &\le C_{\delta/k} \bigg( \kappa \bigg(\frac{a}{d} \bigg) \kappa \bigg( \frac{b}{d} \bigg) \kappa \bigg( \frac{c}{d} \bigg) \bigg)^{1 + \delta/k} \le C_{\delta / k} \bigg( \frac{a}{d} \cdot \kappa(b) \kappa(c) \bigg)^{1 + \delta/k} \\ &\le C_{\delta/k} \bigg( \frac{x^{1 - (2 + \delta)/k}}{d} (2x)^{2/k} \bigg)^{1 + \delta/k} = 2^{({2}/{k}) (1 + {\delta}/{k})} C_{\delta/k} \frac{x^{1- \delta^2 / k^2}}{d^{1 + \delta /k}} \\ &\le 2^{({2}/{k} )(1 + {\delta}/{k})} C_{\delta/k} \frac{x^{1- \delta^2 / k^2}}{d}. \end{align*} $$

This implies

$$ \begin{align*} x^{\delta^2 / k^2} \le 2^{({2}/{k}) (1 + {\delta}/{k})} C_{\delta/k} \quad\text{or} \quad x \le ( 2^{({2}/{k}) (1 + {\delta}/{k})} C_{\delta/k} )^{k^2 / \delta^2} =: C \end{align*} $$

and the claim follows.

Clearly, Theorem 1.5 is true for $1 \le y \le C$ by picking the implicit constant to be C. Now, for $C < y \le x$ , the above claim implies that the interval

$$ \begin{align*} (x, x + y^{1 - (2 + \delta) / k}] \end{align*} $$

contains at most one k-full number. By dividing the interval $(x, x+y]$ into subintervals of length $y^{1 - (2 + \delta )/k}$ , we obtain

$$ \begin{align*} Q_k(x + y) - Q_k(x) \ll \frac{y}{y^{1 - (2 + \delta) / k}} \cdot 1 = y^{{(2 + \delta)}/{k}}, \end{align*} $$

which gives Theorem 1.5.

6 Proof of Theorem 1.6

First, we suppose $y \le x^{0.2}$ . We claim that there is no nontrivial three-term arithmetic progression among the squarefull numbers in the interval $(x, x + y]$ under the $abc$ -conjecture. Suppose the contrary. Then we have three squarefull numbers $x < a_1^2 b_1^3 < a_2^2 b_2^3 < a_3^2 b_3^3 \le x + y$ such that

$$ \begin{align*} a_1^2 b_1^3 = a_2^2 b_2^3 - d \quad \text{and} \quad a_3^2 b_3^3 = a_2^2 b_2^3 + d \end{align*} $$

for some positive integer d with $2 d \le y$ . Multiplying the above two equations, we get

$$ \begin{align*} a_1^2 a_3^2 b_1^3 b_3^3 = a_2^4 b_2^6 - d^2 \quad \text{or} \quad a_1^2 a_3^2 b_1^3 b_3^3 + d^2 = a_2^4 b_2^6. \end{align*} $$

Say $D^2 = \text {gcd}(a_2^4 b_2^6, d^2)$ as the numbers are perfect squares. Then, the three integers

$$ \begin{align*} \frac{a_1^2 a_3^2 b_1^3 b_3^3}{D^2}, \; \; \frac{d^2}{D^2}, \; \; \frac{a_2^4 b_2^6}{D^2} \end{align*} $$

are pairwise relatively prime and we have the equation

$$ \begin{align*} \frac{a_1^2 a_3^2 b_1^3 b_3^3}{D^2} + \frac{d^2}{D^2} = \frac{a_2^4 b_2^6}{D^2}. \end{align*} $$

Now, by the $abc$ -conjecture,

$$ \begin{align*} \frac{x^2}{D^2} \le \frac{a_2^4 b_2^6}{D^2} &\ll_\epsilon \kappa \bigg(\frac{a_1^2 a_3^2 b_1^3 b_3^3}{D^2} \frac{d^2}{D^2} \frac{a_2^4 b_2^6}{D^2} \bigg)^{1+\epsilon} \\ &\ll_\epsilon \kappa ( a_1^2 a_3^2 b_1^3 b_3^3 )^{1 + \epsilon} \kappa \bigg(\frac{d^2}{D^2} \bigg)^{1 + \epsilon} \kappa (a_2^4 b_2^6 )^{1 + \epsilon} \\ &\ll_\epsilon (a_1 b_1 a_2 b_2 a_3 b_3)^{1 + \epsilon} \bigg(\frac{d}{D}\bigg)^{1 + \epsilon} \ll_\epsilon x^{3/2 + 3\epsilon/2} \frac{y^{1 + \epsilon}}{D^{1 + \epsilon}}. \end{align*} $$

Since $1 \le D \le d \le y$ , this implies $x^{1/2 - 3\epsilon /2} \ll _\epsilon D^{1 - \epsilon } y^{1+\epsilon } \ll y^2 \le x^{0.4}$ , which is a contradiction for small enough $\epsilon $ , say $\epsilon = 0.01$ , and sufficiently large $x> C$ (in terms of the implicit constant).

Clearly, the theorem is true for $1 \le y \le C$ by picking an appropriate implicit constant. So, we may assume $y> C$ . Since arithmetic progressions are invariant under translation, we may shift the interval $(x, x+y]$ to $(0, y]$ . Therefore, by Theorem 1.7, we have

$$ \begin{align*} Q_2(x + y) - Q_2(x) \ll \frac{y}{\log^{1+c} y}, \end{align*} $$

which gives the theorem.

Now, if $y> x^{0.2}$ , one can simply divide the interval $(x, x + y]$ into subintervals of length $x^{0.2}$ :

$$ \begin{align*} (x, x + x^{0.2}] \cup (x + x^{0.2}, x + 2 x^{0.2}] \cup \cdots \cup \bigg(x + \bigg\lfloor \frac{y}{x^{0.2}}\bigg\rfloor x^{0.2}, x + \bigg( \bigg\lfloor \frac{y}{x^{0.2}} \bigg\rfloor + 1 \bigg) x^{0.2} \bigg]. \end{align*} $$

Then, over each interval $(x + i x^{0.2}, x + (i+1) x^{0.2}]$ , we have the bound

$$ \begin{align*} Q_2(x + (i+1) x^{0.2}) - Q_2(x + i x^{0.2}) \ll \frac{x^{0.2}}{\log^{1+c} x}. \end{align*} $$

Summing over $\lfloor \,{y}/{x^{0.2}} \rfloor + 1$ of these intervals, we have

$$ \begin{align*} Q_2(x + y) - Q_2(x) \ll \frac{y}{x^{0.2}} \cdot \frac{x^{0.2}}{\log^{1+c} x} \ll \frac{y}{\log^{1+c} y}, \end{align*} $$

which gives the theorem as well.