2 Reduction to Diophantine equations

We will work with a variant of $S_{\alpha ,\beta ,\gamma }(X)$

Definition 2.1

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Let $S^*_{\alpha ,\beta ,\gamma }(X)$ to be the number of $(a,b,c)\in \mathbb {N}^3$ with $\gcd (a,b,c)=1$ and

\begin{align*} c\in [X/2,X],\quad a+b=c,\quad \operatorname{rad}(a)\sim X^{\alpha },\quad \operatorname{rad}(b)\sim X^{\beta },\quad \operatorname{rad}(c)\sim X^{\gamma }. \end{align*}

We begin by noting that by the pigeonhole principle,

Lemma 2.2

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We have

\begin{equation} \label{eq:step1} N_\lambda (X)\ll (\log X)^4 \max _{\substack { \alpha ,\beta ,\gamma {\gt}0\\ \alpha +\beta +\gamma \leq \lambda }} S^*_{\alpha ,\beta ,\gamma }(X). \end{equation}

2.0.1

Proof ▶

Theorem 2.3

There exists $\varepsilon {\gt}0$ such that for all $\mathbf{c}\in \mathbb {Z}^3$ and $\mathbf{X},\mathbf{Y},\mathbf{Z}\in \mathbb {R}_{{\gt}0}^{d}$. we have

\begin{align} B_d({\bf c, X,Y,Z}) \ll X^{0.66-\varepsilon }. \end{align}

Proof ▶

Proof of Theorem 1.4 ▶

The following result allows us to bound $S^*_{\alpha ,\beta ,\gamma }(X)$ in terms of the number of solutions to certain monomial Diophantine equations. In order to state it, we need to introduce the quantity $B_d$

Definition 2.4

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For $\mathbf{c}\in \mathbb {Z}^3$ and $\mathbf{X},\mathbf{Y},\mathbf{Z}\in \mathbb {R}_{{\gt}0}^{d}$. we have

\begin{equation} \label{eq:Bk} B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z}) := \# \left\{ (\mathbf{x},\mathbf{y},\mathbf{z})\in \mathbb {N}^{3d}\; :\; \begin{array}{l} x_i\sim X_i,\, y_i\sim Y_i,\, z_i\sim Z_i \\ c_1\prod _{j\leq d}x_j^j+c_2\prod _{j\leq d}y_j^j=c_3\prod _{j\leq d}z_j^j\\ \gcd (c_1\prod _{j\leq d}x_j,c_2\prod _{j\leq d}y_j , c_3\prod _{j\leq d}z_j)=1 \end{array} \right\} . \end{equation}

2.0.3

Lemma 2.5

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Let $\varepsilon \in (0,1/2)$, and let $2\leq n\leq X$ be an integer. Then there exists a factorisation

\begin{align*} n=c\prod _{j\leq \frac{5}{2}\varepsilon ^{-2}}x_j^{j}, \end{align*}

for positive integers $x_j,c$ such that $c\leq X^{\varepsilon /2}$, the $x_j$ are pairwise coprime, and

\begin{align*} X^{-\varepsilon }\prod _{j\leq \frac{5}{2}\varepsilon ^{-2}}x_j \leq \operatorname{rad}(n) \leq X^{\varepsilon }\prod _{j\leq \frac{5}{2}\varepsilon ^{-2}}x_j. \end{align*}

Proof ▶

Fix $2\le n\le X$ and let $K=2\lceil \varepsilon ^{-1}\rceil $, $M=\lfloor \frac{5}{2}\varepsilon ^{-2}\rfloor $. Define

\begin{align*} y_j:=\prod _{p^j\| n} p. \end{align*}

For $j\leq M$, we set

\begin{alignat*}{2} x_j:=\begin{cases} y_j & \text{for $j\neq K$},\\ y_j\prod _{m{\gt}M}y_m^{\lfloor m/K\rfloor } & \text{for $j=K$}, \end{cases}\qquad \text{and}\qquad c\, :=\prod _{m{\gt}M}y_m^{m-K\lfloor m/K\rfloor }. \end{alignat*}

All the $x_j$ are pairwise coprime, since the $y_j$ are pairwise coprime.

Note that by definition $c\prod _{j\leq M}x_j^j=\prod _{m\geq 1}y_m^m=n\leq X$. In particular,

\begin{align*} \prod _{m\geq M}y_m\leq \Big(\prod _{m\geq M}y_m^{m}\Big)^{1/M}\leq X^{1/M}. \end{align*}

Then, since $m-K\lfloor m/K\rfloor \le K$, it follows from the definition of $c$ that

\begin{align*} c\leq \prod _{m\geq M}y_m^K\leq X^{K/M}\leq X^{\varepsilon /2}. \end{align*}

Thus

\[ \operatorname{rad}(n)\leq \operatorname{rad}(c)\prod _{j\leq M}\operatorname{rad}(x_j)\leq X^{\varepsilon /2} \prod _{j\leq M} x_j. \]

On the other hand, we have

\begin{align*} x_K = y_K\prod _{m{\gt} M}y_m^{\lfloor m/K\rfloor }\leq \Big(y_K^K\cdot \prod _{m{\gt} M}y_m^m\Big)^{1/K}\leq n^{1/K}\leq X^{\varepsilon /2}. \end{align*}

Recalling that the $y_j$ are squarefree and pairwise coprime for $j\neq K$, gives the lower bound

\[ \operatorname{rad}(n)=\prod _{m\geq 1}y_m \geq \prod _{\substack {j\leq M\\ j\neq K}}x_j\geq X^{-\varepsilon /2}\prod _{j\leq M}x_j, \]

as claimed.

Proposition 2.6

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Let $\alpha ,\beta ,\gamma \in (0,1]$ be fixed and let $X\geq 2$. For any $\varepsilon {\gt}0$ there exists an integer $d=d(\varepsilon )\geq 1$ such that the following holds. There exist $X_1,\ldots , X_d,Y_1,\ldots , Y_d, Z_1,\ldots , Z_d\geq 1$ satisfying

\begin{align} \label{eq:xiyizi_1} X^{\alpha -\varepsilon }\ll _{\varepsilon } \prod _{j=1}^{d}X_j\leq 2 X^{\alpha +\varepsilon },\quad X^{\beta -\varepsilon }\ll _{\varepsilon }\prod _{j=1}^{d}Y_j\leq 2 X^{\beta +\varepsilon },\quad X^{\gamma -\varepsilon }\ll _{\varepsilon }\prod _{j=1}^{d}Z_j\leq 2 X^{\gamma +\varepsilon } \end{align}

and

\begin{align} \label{eq:xiyizi_2} \prod _{j=1}^d X_j^j \leq X, \quad \prod _{j=1}^d Y_j^j\leq X,\quad X^{1-\varepsilon ^2}\ll _{\varepsilon } \prod _{j=1}^d Z_j^j\leq X \end{align}

and pairwise coprime integers $1\leq c_1,c_2,c_3\leq X^{\varepsilon }$, such that

\begin{align*} S^*_{\alpha ,\beta ,\gamma }(X) \ll _{\varepsilon } X^{\varepsilon }B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z}). \end{align*}

Proof of Proposition 2.6 ▶

We may assume that $X$ is large enough in terms of $\varepsilon $, since otherwise the claim is trivial. Let $(a,b,c)$ be a triple counted by $S^*_{\alpha ,\beta ,\gamma }(X)$. Apply Lemma 2.5 (with $\varepsilon ^2/2$ in place of $\varepsilon $) to each of $a,b,c$ to obtain factorisations of the form

\begin{align*} a=c_1\prod _{j\leq d}x_j^j,\quad b=c_2\prod _{j\leq d}y_j^j,\quad c=c_3\prod _{j\leq d}z_j^j, \end{align*}

where $d=\lfloor 10\varepsilon ^{-4}\rfloor $ and $1\leq c_1,c_2,c_3\leq X^{\varepsilon ^2/4}$. Since $(a,b,c)$ is counted by $S^*_{\alpha ,\beta ,\gamma }(X)$, we have $\gcd (a,b,c)=1$ and $a+b=c$, so $a,b,c$ are pairwise coprime. Hence, all the $3d+3$ numbers $x_i,y_i,z_i,c_1,c_2,c_3$ are pairwise coprime. Note also that by the properties of the factorisation given by Lemma 2.5, we have

\begin{align*} & X^{-\varepsilon /2}\prod _{j\leq d}x_j\leq \operatorname{rad}(a)\leq X^{\varepsilon /2}\prod _{j\leq d}x_j,\quad X^{-\varepsilon /2}\prod _{j\leq d}y_j\leq \operatorname{rad}(b)\leq X^{\varepsilon /2}\prod _{j\leq d}y_j,\\ & X^{-\varepsilon /2}\prod _{j\leq d}z_j\leq \operatorname{rad}(c)\leq X^{\varepsilon /2}\prod _{j\leq d}z_j. \end{align*}

Since $\operatorname{rad}(a)\sim X^{\alpha },\operatorname{rad}(b)\sim X^{\beta },\operatorname{rad}(c)\sim X^{\gamma }$ for all triples under consideration, this implies

\begin{align*} X^{\alpha -\varepsilon }\leq \prod _{j\leq d}x_j\leq X^{\alpha +\varepsilon },\quad X^{\beta -\varepsilon }\leq \prod _{j\leq d}y_j\leq X^{\beta +\varepsilon },\quad X^{\gamma -\varepsilon }\leq \prod _{j\leq d}z_j\leq X^{\gamma +\varepsilon }. \end{align*}

By dyadic decomposition, we can now find some $X_i, Y_i,Z_i$ such that ?? and ?? hold, and such that

\[ S^*_{\alpha ,\beta ,\gamma }(X)\ll _{\varepsilon } (\log X)^{3d} \sum _{\substack {\mathbf{c}\in \mathbb {N}^3\\ c_1,c_2,c_3 \leq X^{\varepsilon /4}}}B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z}). \]

Now the claim follows from the pigeonhole principle.