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\begin{center}
\vskip 1cm{\LARGE\bf 
Arithmetic Progressions Among \\
\vskip .1in
Powerful Numbers 
}
\vskip 1cm
\large
Tsz Ho Chan\\
Mathematics Department \\
Kennesaw State University \\
Marietta, GA 30060 \\
USA \\
\href{mailto: tchan4@kennesaw.edu}{\tt tchan4@kennesaw.edu}
\end{center}

\vskip .2 in


\begin{abstract}
In this paper, we study $k$-term arithmetic progressions $N, N+d, \ldots, N+(k-1)d$ of powerful numbers. Unconditionally, we exhibit infinitely many $3$-term arithmetic progressions of powerful numbers with $d \le 5 N^{1/2}$. Assuming the $abc$-conjecture, we obtain a nearly tight lower bound on the common difference. We also prove some partial results when $k \ge 4$ and pose some open questions.
\end{abstract}

\section{Introduction and main results}

For any integer $k \ge 1$, a non-trivial $k$-term arithmetic progression (abbreviated as $k$-AP) is a sequence of the form
\[
N, N+d, N+2d, \ldots ,  N+(k-1)d
\]
with initial term $N$ and common difference $d > 0$. Any single number or two different numbers can be considered as a $1$-AP or $2$-AP respectively. So, from now on, we will assume $k \ge 3$. It is well-known that there are infinitely many $3$-APs among perfect squares (e.g., $1, 25, 49$) but there is no $4$-AP of perfect squares (first discovered by Fermat). One may ask if there exist $k$-APs among other interesting arithmetic or polynomial sequences. For instance, Green and Tao \cite{GT} recently proved that there are arbitrarily long arithmetic progressions among the prime numbers. In this paper, we are interested in studying arithmetic progressions of powerful numbers \seqnum{A001694} which are square-like.
\begin{definition}
A number $n$ is {\it powerful} if $p^2 \mid n$ whenever $p \mid n$ (i.e., its prime factorization $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ satisfies $a_i \ge 2$ for all $1 \le i \le r$.)
\end{definition}
For example, $72 = 2^3 \cdot 3^2$ is powerful but $24 = 2^3 \cdot 3$ is not. Another common name for powerful number is {\it squarefull} number. A closely related concept is {\it squarefree} number. 
\begin{definition}
A number $n$ is {\it squarefree} if $p^2 \nmid n$ for all prime $p \mid n$ (i.e., its prime factorization $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ satisfies $a_i = 1$ for all $1 \le i \le r$.).
\end{definition}
For example, $30 = 2 \cdot 3 \cdot 5$ is squarefree but $24 = 2^3 \cdot 3$ is not. By unique prime factorization, one can show that any positive integer and any powerful number can be factored uniquely as $n = a^2 b$ and $n = a^2 b^3$ respectively for some integer $a \ge 1$ and squarefree number $b \ge 1$. Unlike perfect squares, there are arbitrarily long arithmetic progressions among powerful numbers.
\begin{theorem} \label{thm-longAP}
For any integer $k \ge 3$, there is a $k$-term arithmetic progression of powerful numbers.
\end{theorem}

The above theorem is not new (see remark 2 in \cite{H} for example) but we include a proof for completeness. For $k = 3$, there is a folklore conjecture concerning $3$-AP of powerful numbers which seems
to have been first posed by Erd\H{o}s \cite{E}.
\begin{conjecture} \label{conj-3}
There are no three consecutive powerful numbers.
\end{conjecture}
Later, Mollin and Walsh \cite{MW} and Granville \cite{G} reiterated the same conjecture and provided some evidence and interesting consequences. Currently, we are far from being able to prove Conjecture \ref{conj-3}. However, Conjecture \ref{conj-3} follows from the famous $abc$-conjecture (see the first half of the proof of Theorem 1.6 with $d = 1$ and $y =3$ in \cite{C} for example). Let
\[
\kappa(m) := \prod_{p | m} p
\]
denote the {\it squarefree kernel} or {\it radical} of an integer $m$.
\begin{conjecture}[$abc$-conjecture] \label{abc}
For any $\epsilon > 0$, there exists a constant $C_\epsilon > 0$ such that, for any integers $a, b, c$ with $a + b = c$ and $\text{gcd}(a,b) = 1$, the bound
\[
\max\{ |a|, |b|, |c| \} \le C_\epsilon \cdot \kappa(a b c)^{1 + \epsilon}
\]
holds.
\end{conjecture}
In other words, there is no $3$-AP of powerful numbers with common difference $d = 1$ under the $abc$-conjecture. Recently, the author \cite{C} studied powerful numbers in short intervals, and it can be deduced that, for any $\epsilon > 0$, there is no $3$-AP of powerful numbers with $d \le N^{1/4 - \epsilon}$ for sufficiently large $N$ under the $abc$-conjecture. On the other hand, for any integer $m \ge 1$, the following three expressions
\begin{equation} \label{3APsquare}
(2m^2 - 1)^2,  (2m^2 + 2m + 1)^2,  (2m^2 + 4m + 1)^2
\end{equation}
form a $3$-AP of perfect squares with common difference $d = 8m^3 + 12m^2 + 4m$. Hence, there are infinitely many $3$-APs of powerful numbers with $d \le 6 N^{3/4}$. Thus, we are led to the following natural question.
\begin{question}
We say that $0 < \theta < 1$ is an {\it admissible exponent} if there exists $C_\theta > 0$ such that there are infinitely many $3$-APs of powerful numbers $N, N+d, N+2d$ with common difference $d \le C_\theta N^\theta$. Find the infimum, $\theta_3$, among all such admissible exponents.
\end{question}

The above discussion yields $\frac{1}{4} \le \theta_3 \le \frac{3}{4}$. We shall prove the following optimal result.
\begin{theorem} \label{thm-3APbounds}
Assuming the $abc$-conjecture, we have $\theta_3 = \frac{1}{2}$.
\end{theorem}

Analogously, one can define $\theta_k$ for $k$-AP of powerful numbers when $k \ge 4$. We have the following partial results.
\begin{theorem} \label{thm-4APbounds}
Assuming the $abc$-conjecture, we have
\[
\frac{1}{2} \le \theta_4 \le \frac{4}{5},  \frac{1}{2} \le \theta_5 \le \frac{9}{10}, \text{ and }  \frac{1}{2} \le \theta_k \le 1 - \frac{1}{10 \cdot 3^{k-5}}
\]
for $k \ge 5$.
\end{theorem}
Note that the upper bounds in Theorems \ref{thm-3APbounds} and \ref{thm-4APbounds} hold unconditionally and it is their lower bounds that require the $abc$-conjecture.

\bigskip

It would be interesting to see if one can prove $\theta_4 > 1/2$ under the $abc$-conjecture. Another future direction would be narrowing the above ranges for $\theta_k$ when $k \ge 4$. One can also ask if there exist infinitely many $3$-APs of powerful numbers with common difference $d = o(\sqrt{N})$.

\subsection{Notation} 

Throughout the paper, the letters $N$, $k$, $m$, $n$, $a$, $b$, $c$ and $d$ stand for positive integers while the letters $p$, $p_{i j}$ and $q_{i j'}$ stand for prime numbers. The symbol $a \mid b$ means that $a$ divides $b$, the symbol $a \nmid b$ means that $a$ does not divide $b$, and the symbol $p^n || a$ means that $p^n \mid a$ but $p^{n+1} \nmid a$. The function $\nu_p(a)$ stands for the $p$-adic valuation of $a$ (i.e., $\nu_p(a) = n$ when $p^n || a$). The symbol $f(x) \ll g(x)$ is equivalent to $|f(x)| \leq C g(x)$ for some constant $C > 0$. The symbol $f(x) \ll_{\lambda} g(x)$ means that the implicit constant may depend on $\lambda$. Finally, the symbol $f(x) = o(g(x))$ means that $\lim_{x \rightarrow \infty} f(x)/g(x) = 0$.



\section{Proof of Theorem \ref{thm-longAP}: long AP of powerful numbers}

\begin{proof}
We apply induction on $k$. The base case $k = 3$ follows from \eqref{3APsquare} on $3$-AP among perfect squares. Suppose, for some $k \ge 3$, there is a $k$-AP among powerful numbers, say
\[
a_1^2 b_1^3 < a_2^2 b_2^3 < \cdots < a_k^2 b_k^3 \quad \text{ with common difference }  d \ge 1.
\]
Consider the number $a_k^2 b_k^3 + d = a^2 b$ for some integer $a$ and squarefree number $b$. Then
\[
a_1^2 b_1^3 b^2 < a_2^2 b_2^3 b^2 < \cdots < a_k^2 b_k^3 b^2 < a^2 b^3
\]
is a $(k+1)$-AP of powerful numbers with common difference $d b^2$. This finishes the induction proof.
\end{proof}


\section{Proof of Theorem \ref{thm-3APbounds}: $3$-AP upper bound}

\begin{proof}
For the upper bound $\theta_3 \le 1/2$, we first consider the following $3$-AP:
\[
x^2 - 2x - 1,  x^2,  x^2 + 2 x + 1.
\]
The last two terms are perfect squares. We want the first term to contain a large square factor. The Pell equation
\[
X^2 - 2 Y^2 = -1
\]
has infinitely many integer solutions given by
\[
X_m + \sqrt{2} Y_m = (1 + \sqrt{2})^{2m + 1}  \text{ with integer } m \ge 1.
\]
By setting $n = X$ and $\frac{x-1}{2} = Y$, the generalized Pell equation
\begin{equation} \label{gen-Pell}
(x-1)^2 - 2n^2  = 2
\end{equation}
has infinitely many integer solutions given by
\[
x - 1 = 2 Y_m  \text{ and } n = X_m.
\]
Then, equation \eqref{gen-Pell} gives us infinitely many integers $x$ such that $x^2 - 2x - 1 = 2 n^2$ for some integer $n$. Therefore, we have infinitely many $3$-APs of powerful numbers, namely
\[
N = 2^2 (x^2 - 2x - 1) = 2^3 n^2,  N + d = 2^2 x^2 = (2 x)^2,  N + 2d = 2^2 (x^2 + 2 x + 1) = (2 (x+1))^2
\]
with common difference
\[
d = 2^2 (2x + 1) = 8 x + 4 \le 5 N^{1/2}.
\]
Hence, we have $\theta_3 \le 1/2$.
\end{proof}


\section{Proof of Theorem \ref{thm-3APbounds}: $3$-AP lower bound}

First, we need a simple observation.
\begin{lemma} \label{lem-valuation}
Suppose  $a$ and $b$ are positive integers and $p^\delta \mid a^2 b^3$ for some prime $p$ and integer $\delta \ge 1$. Then $\nu_p (a b) \ge \delta / 3$.
\end{lemma}

\begin{proof}
From the definitions of divisibility and $p$-adic valuation, we have $\delta \le 2 \nu_p(a) + 3 \nu_p(b)$. Dividing everything by $3$, we have $\delta / 3 \le 2 \nu_p(a) / 3 + \nu_p(b) \le \nu_p(a) + \nu_p(b) = \nu_p(a b)$.
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{thm-3APbounds}]
Consider any $3$-AP of powerful numbers $N, N+d, N+2d$ with
\[
N = a_1^2 b_1^3,  N + d = a_2^2 b_2^3,  \text{ and } N + 2d = a_3^2 b_3^3
\]
for some integers $a_1, a_2, a_3$ and squarefree numbers $b_1, b_2, b_3$. If some prime $p$ divides $b_1$, $b_2$ and $b_3$, then we can consider the reduced $3$-AP of powerful numbers
\[
\frac{N}{p^3},   \frac{N}{p^3} + \frac{d}{p^3},   \frac{N}{p^3} + \frac{2d}{p^3}.
\]
If one could prove a lower bound $d/p^3 \ge C_\theta (N/p^3)^\theta$ with some $0 < \theta < 1$ and $C_\theta > 0$ for the reduced $3$-AP, one would also have $d \ge C_\theta N^\theta$ for the original $3$-AP. Hence, we may assume $\text{gcd}(b_1, b_2, b_3) = 1$.

\bigskip

Since $(N+d)^2 = N(N+2d) + d^2$, we have $a_2^4 b_2^6 = a_1^2 b_1^3 a_3^2 b_3^3 + d^2$. Let $D^2 = \text{gcd}(a_2^4 b_2^6, d^2)$ which also equals to $\text{gcd}(a_2^4 b_2^6, a_1^2 b_1^3 a_3^2 b_3^3)$ and $\text{gcd}(a_1^2 b_1^3 a_3^2 b_3^3, d^2)$. Note that as $D \mid a_2^2 b_2^3$ and $D \mid d$, we also have $D \mid a_1^2 b_1^3$ and $D \mid a_3^2 b_3^3$. Dividing everything by $D^2$, we have the equation
\[
\Bigl( \frac{a_2^2 b_2^3}{D} \Bigr)^2 = \Bigl( \frac{a_1^2 b_1^3}{D} \frac{a_3^2 b_3^3}{D} \Bigr) + \Bigl( \frac{d}{D} \Bigr)^2
\]
where the three terms are pairwise relatively prime. By the $abc$-conjecture, we have
\begin{equation} \label{abc-eq1}
\frac{N^2}{D^2} \le \Bigl( \frac{a_2^2 b_2^3}{D} \Bigr)^2 \le C_\epsilon \Bigl( \kappa \Bigl(  \frac{a_1^2 b_1^3}{D} \frac{a_2^2 b_2^3}{D} \frac{a_3^2 b_3^3}{D} \Big) \kappa \Bigl( \frac{d}{D} \Bigr) \Bigr)^{1 + \epsilon}
\end{equation}
as $\kappa(m n) \le \kappa(m) \kappa(n)$. If one simply bounds the right-hand side of \eqref{abc-eq1} by
\[
\le C_\epsilon \Bigl(a_1 b_1 a_2 b_2 a_3 b_3 \frac{d}{D} \Bigr)^{1 + \epsilon} \ll C_\epsilon \Bigl( \frac{N^{3/2} d}{D} \Bigr)^{1 + \epsilon},
\] 
solves for $d$, and applies $D \le d$ as in \cite[Theorem1.6]{C}, one would obtain $N^{1/4 - \epsilon} \ll_\epsilon d$ and the lower bound $\theta_3 \ge 1/4$ only. So, in order to prove Theorem \ref{thm-3APbounds}, we need a finer analysis. We claim that
\begin{equation} \label{abc-eq2}
\kappa \Bigl(  \frac{a_1^2 b_1^3}{D} \frac{a_2^2 b_2^3}{D} \frac{a_3^2 b_3^3}{D} \Big) \le \frac{a_1 b_1 a_2 b_2 a_3 b_3}{D}
\end{equation}
which would follow from
\begin{equation} \label{abc-eq3}
\nu_p \Bigl( \kappa\Bigl( \frac{a_1^2 b_1^3}{D} \frac{a_2^2 b_2^3}{D} \frac{a_3^2 b_3^3}{D} \Big) \Bigr) \le \nu_p (a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p (D)
\end{equation}
for any prime $p$. Firstly, if a prime $p$ does not divide $a_1 b_1 a_2 b_2 a_3 b_3$, then \eqref{abc-eq3} is true as both sides are $0$. Secondly, if a prime $p \mid a_1 b_1 a_2 b_2 a_3 b_3$ but $p \nmid D$, then left hand side of \eqref{abc-eq3} is exactly $1$ while the right-hand side of \eqref{abc-eq3} is $\ge 1 - 0$. So, inequality \eqref{abc-eq3} is true for such primes. Thus, it remains to consider those primes $p$ which divide both $a_1 b_1 a_2 b_2 a_3 b_3$ and $D$. Notice that the left-hand side of \eqref{abc-eq3} is at most $1$ for such primes. Suppose we have the following prime factorizations
\begin{align*}
b_1 &= p_{1 1} \cdots p_{1 r_1},  a_1 = p_{1 1}^{\alpha_{1 1}} \cdots p_{1 r_1}^{\alpha_{1 r_1}} \cdot q_{11}^{\beta_{11}} \cdots q_{1 s_1}^{\beta_{1 s_1}} \\
b_2 &= p_{2 1} \cdots p_{2 r_2},  a_2 = p_{2 1}^{\alpha_{2 1}} \cdots p_{2 r_2}^{\alpha_{2 r_2}} \cdot q_{2 1}^{\beta_{2 1}} \cdots q_{2 s_2}^{\beta_{2 s_2}} \\
b_3 &= p_{3 1} \cdots p_{3 r_3},  a_3 = p_{3 1}^{\alpha_{3 1}} \cdots p_{3 r_3}^{\alpha_{3 r_3}} \cdot q_{3 1}^{\beta_{3 1}} \cdots q_{3 s_3}^{\beta_{3 s_3}}
\end{align*}
for some integers $r_1, r_2, r_3, s_1, s_2, s_3 \ge 0$, $\alpha_{i j} \ge 0$, $\beta_{i j'} \ge 1$ and primes $p_{i j}, q_{i j'}$ with $q_{i j'} \neq p_{i j}$. Now, consider a fixed prime $p \mid D$ with $\delta := \nu_p(D)$. Note that $p$ does not divide all of the $b_1, b_2, b_3$ as $\text{gcd}(b_1, b_2, b_3) = 1$.

\bigskip

Case 1: Suppose $p$ does not divide any of the $b_1, b_2, b_3$. Since $D \mid a_i^2 b_i^3$ for $i = 1, 2, 3$, we must have $p \mid a_1, a_2, a_3$. Say $p = q_{1 j_1} = q_{2 j_2} = q_{3 j_3}$ for some $1 \le j_m \le s_m$ for $m = 1, 2, 3$. As $p^{2 \delta} || D^2 = \text{gcd}(a_2^4 b_2^6, a_1^2 b_1^3 a_3^2 b_3^3)$ and $\text{gcd}(p, b_1 b_2 b_3) = 1$, we have $2 \delta = \min(4 \beta_{2, j_2}, 2 (\beta_{1, j_1} + \beta_{3  j_3})) \ge 4$. Hence, we have $2 \le \delta = \min(2 \beta_{2 j_2}, \beta_{1 j_1} + \beta_{3 j_3})$ and
\[
\nu_p(a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p(D) \ge \beta_{1 j_1} + \beta_{2 j_2} + \beta_{3 j_3} - (\beta_{1 j_1} + \beta_{3 j_3}) = \beta_{2 j_2} \ge 1.
\] 

Case 2: Suppose $p$ divides exactly one of the $b_1, b_2, b_3$. 

\bigskip

Subcase 1: Suppose $\delta$ is even. Without loss of generality, say $p \mid b_1$, $p \nmid b_2$, $p \nmid b_3$ as the other cases are similar. Then, we have $p \mid a_2, a_3$ and $p = q_{2 j_2} = q_{3 j_3}$ for some $1 \le j_2 \le s_3$ and $1 \le j_3 \le s_3$. As $p^\delta \mid a_2^2 b_2^3, a_3^2 b_3^3$, we have $\delta/2 \le \beta_{2 j_2}$ and $\delta/2 \le \beta_{3 j_3}$. By Lemma \ref{lem-valuation}, we have
\begin{align*}
\nu_p(a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p(D) &\ge \max (\delta / 3, 1) + \beta_{2 j_2} + \beta_{3 j_3} - \delta \\
&\ge \left\{ \begin{array}{ll} 1 + \delta/2 + \delta/2 - \delta \ge 1, & \text{ when } \delta = 2; \\
\delta/3 + \delta/2 + \delta/2 - \delta > 1, & \text{ when } \delta \ge 4.
\end{array} \right.
\end{align*}

Subcase 2: Suppose $\delta$ is odd. If $p \mid b_1$, $p \nmid b_2$, $p \nmid b_3$, then $p \mid b_1, a_2, a_3$. We have $\nu_p(a_2^4 b_2^6) \ge 4$ and $\nu_p(a_1^2 b_1^3 a_3^2 b_3^3) \ge 5$. Hence, we get $\delta \ge 3$ as $p^{2 \delta} || D^2 = \text{gcd}(a_2^4 b_2^6, a_1^2 b_1^3 a_3^2 b_3^3)$ and $\delta$ is odd. If $p \mid b_2$, $p \nmid p_1$, $p \nmid p_3$, then $p \mid b_2, a_1, a_3$. We have $\nu_p(a_2^4 b_2^6) \ge 6$ and $\nu_p(a_1^2 b_1^3 a_3^2 b_3^3) \ge 4$. Hence, we get $\delta \ge 3$ by similar reasoning. If $p \mid b_3$, $p \nmid b_1$, $p \nmid b_2$, we also get $\delta \ge 3$ as it is similar to $p \mid b_1$, $p \nmid b_2$, $p \nmid b_3$. Therefore, we obtain $\delta \ge 3$ in all circumstances.

\bigskip

Suppose $p \nmid b_i, b_{i'}$ for some $1 \le i < i' \le 3$. Then $p \mid a_i, a_{i'}$. Say $p = q_{i j_i} = q_{i' j_{i'}}$ for some $1 \le j_i \le s_i$ and $1 \le j_{i'} \le s_{i'}$. Thus, we have $3 \le \delta \le 2 \beta_{i j_i} - 1$ and $3 \le \delta \le 2 \beta_{i' j_{i'}} - 1$ as $\delta$ is odd. Hence, we have $\beta_{i j_i}, \beta_{i' j_{i'}} \ge \delta/2 + 1/2$ and
\[
\nu_p(a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p(D) \ge 1 + \beta_{i j_i} + \beta_{i' j_{i'}} - \delta > 1.
\]


Case 3: Suppose $p$ divides exactly two of the $b_1, b_2, b_3$.

\bigskip

Subcase 1: Suppose $\delta$ is even. Without loss of generality, say $p \mid b_1$, $p \mid b_2$, $p \nmid b_3$ as the other cases are similar. Then $p \mid a_3$ and $p = p_{1 j_1} = p_{2 j_2} = q_{3 j_3}$ for some $1 \le j_1 \le r_1$, $1 \le j_2 \le r_2$ and $1 \le j_3 \le s_3$. As $p^\delta \mid a_3^2 b_3^3$, we have $\delta/2 \le \beta_{3 j_3}$. Also, as $p^\delta \mid a_1^2 b_1^3, a_2^2 b_2^3$, we have $\delta \le 2 \alpha_{1 j_1} + 2$ and $\delta \le 2 \alpha_{2 j_2} + 2$ since $\delta$ is even. Hence, we have
\[
\nu_p(a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p(D) \ge (\alpha_{1 j_1} + 1) + (\alpha_{2 j_2} + 1) + \beta_{3 j_3} - \delta \ge \beta_{3 j_3} \ge 1.
\]

Subcase 2: Suppose $\delta$ is odd. If $p \mid b_1$, $p \mid b_2$, $p \nmid b_3$, then $p \mid b_1, b_2, a_3$. We have $\nu_p(a_2^4 b_2^6) \ge 6$ and $\nu_p(a_1^2 b_1^3 a_3^2 b_3^3) \ge 5$. Hence, we get $\delta \ge 3$ as $p^{2 \delta} || D^2 = \text{gcd}(a_2^4 b_2^6, a_1^2 b_1^3 a_3^2 b_3^3)$ and $\delta$ is odd. If $p \mid b_1$, $p \nmid b_2$, $p \mid b_3$, then $p \mid b_1, a_2, b_3$. We have $\nu_p(a_2^4 b_2^6) \ge 4$ and $\nu_p(a_1^2 b_1^3 a_3^2 b_3^3) \ge 6$. Hence, we get $\delta \ge 3$ by similar reasoning. If $p \nmid b_1$, $p \mid b_2$, $p \mid b_3$, we also get $\delta \ge 3$ as it is similar to the case $p \mid b_1$, $p \mid b_2$, $p \nmid b_3$. Therefore, we obtain $\delta \ge 3$ in all circumstances.

\bigskip

Suppose $p \nmid b_i$ for some $1 \le i \le 3$. Then $p \mid a_i$ and $p = q_{i j_i}$ for some $1 \le j_i \le s_i$. Thus, we have $3 \le \delta \le 2 \beta_{i j_i} - 1$ as $\delta$ is odd, and $\beta_{i j_i} \ge \delta/2 + 1/2$. By Lemma \ref{lem-valuation}, we have
\[
\nu_p(a_1 b_1 a_2 b_2 a_3 b_3) - \nu_p(D) \ge \frac{\delta}{3} + \frac{\delta}{3} + \beta_{i j_i} - \delta \ge \frac{1}{2} + \frac{\delta}{6} \ge \frac{1}{2} + \frac{3}{6} = 1.
\]

Consequently, the right-hand side of \eqref{abc-eq3} is at least $1$  in all of the above cases. As a result, we have inequalities \eqref{abc-eq3} and \eqref{abc-eq2}. Putting \eqref{abc-eq2} into \eqref{abc-eq1}, we obtain
\[
\frac{N^2}{D^2} \le C_\epsilon \Bigl( \frac{a_1 b_1 a_2 b_2 a_3 b_3 d}{D^2} \Bigr)^{1 + \epsilon} \ll C_\epsilon \Bigl( \frac{N^{3/2} d}{D^2} \Bigr)^{1 + \epsilon}
\]
as $a_1^2 b_1^3, a_2^2 b_2^3, a_3^2 b_3^2 \ll N$. Together with $D \ge 1$, we have $N^{1/2 - 2 \epsilon} \ll_\epsilon d$ and $\theta_3 \ge 1/2$ as $\epsilon$ can be arbitrarily small.
\end{proof}


\section{Proof of Theorem \ref{thm-4APbounds}}

\begin{proof}
For $k \ge 4$, the lower bound $\theta_k \ge 1/2$ follows from the observation that $\theta_k \ge \theta_3$ and $\theta_3 \ge 1/2$ from Theorem \ref{thm-3APbounds} under the $abc$-conjecture.

\bigskip

For the upper bound $\theta_4 \le 4/5$, we construct $4$-APs of powerful numbers as follows. With positive integer $a$, the following four expressions
\begin{equation} \label{poly4}
(x - a)^3 (x + a)^2,  (x - a)^2 x (x + a)^2,  (x - a)^2 (x + a)^3,  (x - a)^2 (x + a)^2 (x + 2a)
\end{equation}
form a $4$-AP with common difference $d = a (x - a)^2 (x + a)^2$. Note that the first and third terms give powerful numbers for any integer $x$. If $x$ and $x + 2a$ are powerful, then all four polynomials would result in powerful numbers. We can pick $a = 2$. Note that the Pell equation
\[
X^2 - 2Y^2 = 1  \text{ or }  2 X^2 - 4 Y^2 = 2  \text{ or }  4 X^2 = 8 Y^2 + 4 
\]
has solutions
\[
X_m + \sqrt{2} Y_m = (3 + 2 \sqrt{2})^m  \text{ for positive integer } m.
\]
By choosing $x = 8 Y_m^2$ and $x + 2a = x + 4 = 4 X_m^2$, we turn \eqref{poly4} into our desired $4$-AP of powerful numbers. Observe that the common difference
\[
d = 2 (x - 2)^2 (x + 2)^2 \le 3 ((x - 2)^3 (x + 2)^2)^{4/5} = 3 N^{4/5}
\]
for large enough $m$ (and hence $N$). We have $\theta_4 \le 4/5$.

\bigskip

For the upper bound $\theta_5 \le 9/10$, one can build upon our $3$-AP and $4$-AP constructions. With positive integer $a$, the following five expressions
\begin{equation} \label{poly5}
(y - 2a) (y - a)^2 (y + a)^2,  (y - a)^3 (y + a)^2,  (y - a)^2 y (y + a)^2,  (y - a)^2 (y + a)^3,  (y - a)^2 (y + a)^2 (y + 2a)
\end{equation}
form a $5$-AP with common difference $d = a (y - a)^2 (y + a)^2$. Note that the second and fourth terms give powerful numbers for any integer $y$. If $y - 2a$, $y$ and $y + 2a$ are powerful, then all five terms would be powerful. From our $3$-AP construction, we can find infinitely many $3$-APs of powerful numbers
\[
y - 2a = 2^2 (x^2 - 2x - 1) = 2^3 n^2,  y = 2^2 x^2 = (2 x)^2,  y + 2a = 2^2 (x^2 + 2 x + 1) = (2 (x+1))^2
\]
with
\[
2a = 2^2 (2x + 1) = 8 x + 4.
\]
With the above choices, we obtain the desired $5$-AP of powerful numbers with common difference
\begin{align*}
d &= (4x + 2) (4 x^2 - 4 x - 2)^2 (4x^2 + 4x + 2)^2 \\
&\le 3 \Bigl( (4 x^2 - 8 x - 4) (4 x^2 - 4x - 2)^2 (4 x^2 + 4x + 2)^2 \Bigr)^{9/10} = 3 N^{9/10}
\end{align*}
for large enough $x$ (and hence $N$). Thus, we have $\theta_5 \le 9/10$.

\bigskip

For the general upper bound $\theta_k \le 1 - \frac{1}{10 \cdot 3^{k-5}}$, we use induction on $k \ge 5$ similar to Theorem \ref{thm-longAP}. The base case $\theta_5 \le 1 - \frac{1}{10 \cdot 3^{5-5}}$ is true from $\theta_5 \le 9/10$.  Suppose, for some $k \ge 5$, there are infinitely many $k$-APs among powerful numbers with $d \le C_k N^{1 - \frac{1}{10 \cdot 3^{k-5}}}$. Say one such AP is
\[
N = a_1^2 b_1^3 < a_2^2 b_2^3 < \cdots < a_k^2 b_k^3 \quad \text{ with common difference }  1 \le d \le C_k N^{1 - \frac{1}{10 \cdot 3^{k-5}}}.
\]
Consider the number $a_k^2 b_k^3 + d = a^2 b$ for some integer $a$ and squarefree number $b$. Multiply everything by $b^2$, the following $k+1$ numbers
\[
N_1 := N b^2 = a_1^2 b_1^3 b^2 < a_2^2 b_2^3 b^2 < \cdots < a_k^2 b_k^3 b^2 < a^2 b^3
\]
form a $(k+1)$-AP of powerful numbers with common difference $d b^2$. Note that
\[
b \le a^2 b = N + k d \le (1 + k C_k) N.
\]
Hence, we have
\begin{align*}
d b^{\frac{2}{10 \cdot 3^{k-4}}} &\le d (1 + k C_k)^{\frac{2}{10 \cdot 3^{k-4}}} N^{\frac{2} {10 \cdot 3^{k-4}}} \\
&\le C_k  (1 + k C_k)^{\frac{2}{10 \cdot 3^{k-4}}} N^{1 - \frac{1}{10 \cdot 3^{k-5}} + \frac{2}{10 \cdot 3^{k-4}}} = C_k (1 + k C_k)^{\frac{2}{10 \cdot 3^{k-4}}} N^{1 - \frac{1}{10 \cdot 3^{k-4}}}
\end{align*}
and
\[
d b^2 \le C_k (1 + k C_k)^{\frac{2}{10 \cdot 3^{k-4}}} (N b^2)^{1 - \frac{1}{10 \cdot 3^{k-4}}} = C_{k+1} N_1^{1 - \frac{1}{10 \cdot 3^{(k+1) - 5}}}
\]
with $C_{k+1} := C_k (1 + k C_k)^{\frac{2}{10 \cdot 3^{k-4}}}$. This completes the induction proof.
\end{proof}

\section{Acknowledgment}

The author would like to thank Prof.\ Lajos Hajdu for pointing out that Theorem \ref{thm-longAP} has long been known and is even true for perfect powers only. The author also wants to thank the anonymous referee for some helpful corrections.

\begin{thebibliography}{9}

\bibitem{C} T. H. Chan, A note on powerful numbers in short intervals, {\it Bull. Australian Math. Soc.}, 2022.  Available at \url{https://doi.org/10.1017/S0004972722000995}.

\bibitem{E} P. Erd\H{o}s, Consecutive integers, {\it Eureka} {\bf 38} (1975-76), 3--8.

\bibitem{H} L. Hajdu, Perfect powers in arithmetic progression. A note on the inhomogeneous case, {\it Acta Arith.} {\bf 113} (2004), 343--349.

\bibitem{G} A. Granville, Powerful numbers and Fermat's last theorem, {\it C. R. Math. Rep. Acad. Sci. Canada} {\bf 8} (1986), 215--218.

\bibitem{GT} B. J. Green and T. C. Tao, The primes contain arbitrarily long arithmetic progressions, {\it Ann. of Math. (2)} {\bf 167} (2008), 481--547.

\bibitem{MW} R. A. Mollin and P. G. Walsh, A note of powerful numbers, quadratic fields and the Pellian, {\it C. R. Math. Rep. Acad. Sci. Canada} {\bf 8} (1986), 109--114.


\end{thebibliography}


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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11N25.

\noindent \emph{Keywords:} powerful number, $k$-full number, arithmetic progression, $abc$-conjecture.

\bigskip
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\noindent (Concerned with sequence
\seqnum{A001694}.)

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\vspace*{+.1in}
\noindent
Received September 30 2022;
revised versions received October 1 2022; October 3 2022;
December 13 2022.
Published in {\it Journal of Integer Sequences}, December 14 2022.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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