Arithmetic Progressions Among Powerful Numbers

Tsz Ho Chan Mathematics Department Kennesaw State University Marietta, GA 30060 USA

                      tchan4@kennesaw.edu

Abstract

In this paper, we study k-term arithmetic progressions N, N + d, …, N + (k − 1)d of powerful numbers. Unconditionally, we exhibit infinitely many 3-term arithmetic progressions of powerful numbers with d ≤ 5N^{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 ≥ 4 and pose some open questions.

1 Introduction and main results

For any integer k ≥ 1, a non-trivial k-term arithmetic progression (abbreviated as k-AP) is a sequence of the form

N, N + d, N + 2d, …, 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 ≥ 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 [5] 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 \underline{A001694} which are square-like.

Definition 1. A number n is powerful if p^{2} | n whenever p | n (i.e., its prime factorization n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{r}^{a_{r}} satisfies a_{i} ≥ 2 for all 1 ≤ i ≤ r.)

For example, 72 = 2^{3} · 3^{2} is powerful but 24 = 2^{3} · 3 is not. Another common name for powerful number is squarefull number. A closely related concept is squarefree number.

Definition 2. A number n is squarefree if p^{2} ∤ n for all prime p | n (i.e., its prime factorization n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{r}^{a_{r}} satisfies a_{i} = 1 for all 1 ≤ i ≤ r.).

For example, 30 = 2·3·5 is squarefree but 24 = 2^{3}·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 ≥ 1 and squarefree number b ≥ 1. Unlike perfect squares, there are arbitrarily long arithmetic progressions among powerful numbers.

Theorem 3. For any integer k ≥ 3, there is a k-term arithmetic progression of powerful numbers.

The above theorem is not new (see remark 2 in [3] 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 Erdős [2].

Conjecture 4. There are no three consecutive powerful numbers.

Later, Mollin and Walsh [6] and Granville [4] reiterated the same conjecture and provided some evidence and interesting consequences. Currently, we are far from being able to prove Conjecture 4. However, Conjecture 4 follows from the famous abc-conjecture (see the first half of the proof of Theorem 1.6 with d = 1 and y = 3 in [1] for example). Let

κ(m) := ∏_{p|m} p

denote the squarefree kernel or radical of an integer m.

Conjecture 5 (abc-conjecture). For any ϵ > 0, there exists a constant C_{ϵ} > 0 such that, for any integers a, b, c with a + b = c and gcd(a, b) = 1, the bound

max{|a|, |b|, |c|} ≤ C_{ϵ} · κ(abc)^{1+ϵ}

holds.

In other words, there is no 3-AP of powerful numbers with common difference d = 1 under the abc-conjecture. Recently, the author [1] studied powerful numbers in short intervals, and it can be deduced that, for any ϵ > 0, there is no 3-AP of powerful numbers with d ≤ N^{1/4−ϵ} for sufficiently large N under the abc-conjecture. On the other hand, for any integer m ≥ 1, the following three expressions

(2m^{2} − 1)^{2}, (2m^{2} + 2m + 1)^{2}, (2m^{2} + 4m + 1)^{2}  (1)

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 ≤ 6N^{3/4}. Thus, we are led to the following natural question.

Question 6. We say that 0 < θ < 1 is an admissible exponent if there exists C_{θ} > 0 such that there are infinitely many 3-APs of powerful numbers N, N + d, N + 2d with common difference d ≤ C_{θ}N^{θ}. Find the infimum, θ_{3}, among all such admissible exponents.

The above discussion yields \frac{1}{4} ≤ θ_{3} ≤ \frac{3}{4}. We shall prove the following optimal result.

Theorem 7. Assuming the abc-conjecture, we have θ_{3} = \frac{1}{2}.

Analogously, one can define θ_{k} for k-AP of powerful numbers when k ≥ 4. We have the following partial results.

Theorem 8. Assuming the abc-conjecture, we have

\frac{1}{2} ≤ θ_{4} ≤ \frac{4}{5}, \frac{1}{2} ≤ θ_{5} ≤ \frac{9}{10}, and \frac{1}{2} ≤ θ_{k} ≤ 1 − \frac{1}{10 · 3^{k−5}}

for k ≥ 5.

Note that the upper bounds in Theorems 7 and 8 hold unconditionally and it is their lower bounds that require the abc-conjecture.

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

1.1 Notation

Throughout the paper, the letters N, k, m, n, a, b, c and d stand for positive integers while the letters p, p_{ij} and q_{ij′} stand for prime numbers. The symbol a | b means that a divides b, the symbol a ∤ b means that a does not divide b, and the symbol p^{n}||a means that p^{n} | a but p^{n+1} ∤ a. The function ν_{p}(a) stands for the p-adic valuation of a (i.e., ν_{p}(a) = n when p^{n}||a). The symbol f(x) ≪ g(x) is equivalent to |f(x)| ≤ Cg(x) for some constant C > 0. The symbol f(x) ≪_{λ} g(x) means that the implicit constant may depend on λ. Finally, the symbol f(x) = o(g(x)) means that lim_{x→∞} f(x)/g(x) = 0.

2 Proof of Theorem 3: long AP of powerful numbers

Proof. We apply induction on k. The base case k = 3 follows from (1) on 3-AP among perfect squares. Suppose, for some k ≥ 3, there is a k-AP among powerful numbers, say

a_{1}^{2}b_{1}^{3} < a_{2}^{2}b_{2}^{3} < ⋯ < a_{k}^{2}b_{k}^{3} with common difference d ≥ 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} < ⋯ < a_{k}^{2}b_{k}^{3}b^{2} < a^{2}b^{3}

is a (k + 1)-AP of powerful numbers with common difference db^{2}. This finishes the induction proof.

3 Proof of Theorem 7: 3-AP upper bound

Proof. For the upper bound θ_{3} ≤ 1/2, we first consider the following 3-AP:

x^{2} − 2x − 1, x^{2}, x^{2} + 2x + 1.

The last two terms are perfect squares. We want the first term to contain a large square factor. The Pell equation

X^{2} − 2Y^{2} = −1

has infinitely many integer solutions given by

X_{m} + \sqrt{2}Y_{m} = (1 + \sqrt{2})^{2m+1} with integer m ≥ 1.

By setting n = X and \frac{x−1}{2} = Y, the generalized Pell equation

(x − 1)^{2} − 2n^{2} = 2  (2)

has infinitely many integer solutions given by

x − 1 = 2Y_{m} and n = X_{m}.

Then, equation (2) gives us infinitely many integers x such that x^{2} − 2x − 1 = 2n^{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} = (2x)^{2}, N + 2d = 2^{2}(x^{2} + 2x + 1) = (2(x + 1))^{2}

with common difference

d = 2^{2}(2x + 1) = 8x + 4 ≤ 5N^{1/2}.

Hence, we have θ_{3} ≤ 1/2.

4 Proof of Theorem 7: 3-AP lower bound

First, we need a simple observation.

Lemma 9. Suppose a and b are positive integers and p^{δ} | a^{2}b^{3} for some prime p and integer δ ≥ 1. Then ν_{p}(ab) ≥ δ/3.

Proof. From the definitions of divisibility and p-adic valuation, we have δ ≤ 2ν_{p}(a) + 3ν_{p}(b). Dividing everything by 3, we have δ/3 ≤ 2ν_{p}(a)/3 + ν_{p}(b) ≤ ν_{p}(a) + ν_{p}(b) = ν_{p}(ab).

Proof of Theorem 7. 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}, 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} ≥ C_{θ}(N/p^{3})^{θ} with some 0 < θ < 1 and C_{θ} > 0 for the reduced 3-AP, one would also have d ≥ C_{θ}N^{θ} for the original 3-AP. Hence, we may assume gcd(b_{1}, b_{2}, b_{3}) = 1.

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} = gcd(a_{2}^{4}b_{2}^{6}, d^{2}) which also equals to gcd(a_{2}^{4}b_{2}^{6}, a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) and gcd(a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}, d^{2}). Note that as D | a_{2}^{2}b_{2}^{3} and D | d, we also have D | a_{1}^{2}b_{1}^{3} and D | a_{3}^{2}b_{3}^{3}. Dividing everything by D^{2}, we have the equation

(\frac{a_{2}^{2}b_{2}^{3}}{D} )^{2} = (\frac{a_{1}^{2}b_{1}^{3}}{D} \frac{a_{3}^{2}b_{3}^{3}}{D} ) + (\frac{d}{D})^{2}

where the three terms are pairwise relatively prime. By the abc-conjecture, we have

\frac{N^{2}}{D^{2}} ≤ (\frac{a_{2}^{2}b_{2}^{3}}{D} )^{2} ≤ C_{ϵ}(κ(\frac{a_{1}^{2}b_{1}^{3}}{D} \frac{a_{2}^{2}b_{2}^{3}}{D} \frac{a_{3}^{2}b_{3}^{3}}{D} )κ(\frac{d}{D}))^{1+ϵ}  (3)

as κ(mn) ≤ κ(m)κ(n). If one simply bounds the right-hand side of (3) by

≤ C_{ϵ}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3} \frac{d}{D})^{1+ϵ} ≪ C_{ϵ}(\frac{N^{3/2}d}{D})^{1+ϵ},

solves for d, and applies D ≤ d as in [1, Theorem1.6], one would obtain N^{1/4−ϵ} ≪_{ϵ} d and the lower bound θ_{3} ≥ 1/4 only. So, in order to prove Theorem 7, we need a finer analysis. We claim that

κ(\frac{a_{1}^{2}b_{1}^{3}}{D} \frac{a_{2}^{2}b_{2}^{3}}{D} \frac{a_{3}^{2}b_{3}^{3}}{D} ) ≤ \frac{a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}}{D}  (4)

which would follow from

ν_{p}(κ(\frac{a_{1}^{2}b_{1}^{3}}{D} \frac{a_{2}^{2}b_{2}^{3}}{D} \frac{a_{3}^{2}b_{3}^{3}}{D} )) ≤ ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D)  (5)

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 (5) is true as both sides are 0. Secondly, if a prime p | a_{1}b_{1}a_{2}b_{2}a_{3}b_{3} but p ∤ D, then left hand side of (5) is exactly 1 while the right-hand side of (5) is ≥ 1−0. So, inequality (5) 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 (5) is at most 1 for such primes. Suppose we have the following prime factorizations

b_{1} = p_{11} ⋯ p_{1r_{1}}, a_{1} = p_{11}^{α_{11}} ⋯ p_{1r_{1}}^{α_{1r_{1}}} · q_{11}^{β_{11}} ⋯ q_{1s_{1}}^{β_{1s_{1}}}
b_{2} = p_{21} ⋯ p_{2r_{2}}, a_{2} = p_{21}^{α_{21}} ⋯ p_{2r_{2}}^{α_{2r_{2}}} · q_{21}^{β_{21}} ⋯ q_{2s_{2}}^{β_{2s_{2}}}
b_{3} = p_{31} ⋯ p_{3r_{3}}, a_{3} = p_{31}^{α_{31}} ⋯ p_{3r_{3}}^{α_{3r_{3}}} · q_{31}^{β_{31}} ⋯ q_{3s_{3}}^{β_{3s_{3}}}

for some integers r_{1}, r_{2}, r_{3}, s_{1}, s_{2}, s_{3} ≥ 0, α_{ij} ≥ 0, β_{ij′} ≥ 1 and primes p_{ij}, q_{ij′} with q_{ij′} ≠ p_{ij}. Now, consider a fixed prime p | D with δ := ν_{p}(D). Note that p does not divide all of the b_{1}, b_{2}, b_{3} as gcd(b_{1}, b_{2}, b_{3}) = 1.

Case 1: Suppose p does not divide any of the b_{1}, b_{2}, b_{3}. Since D | a_{i}^{2}b_{i}^{3} for i = 1, 2, 3, we must have p | a_{1}, a_{2}, a_{3}. Say p = q_{1j_{1}} = q_{2j_{2}} = q_{3j_{3}} for some 1 ≤ j_{m} ≤ s_{m} for m = 1, 2, 3. As p^{2δ}||D^{2} = gcd(a_{2}^{4}b_{2}^{6}, a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) and gcd(p, b_{1}b_{2}b_{3}) = 1, we have 2δ = min(4β_{2,j_{2}}, 2(β_{1,j_{1}} + β_{3j_{3}})) ≥ 4. Hence, we have 2 ≤ δ = min(2β_{2j_{2}}, β_{1j_{1}} + β_{3j_{3}}) and

ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D) ≥ β_{1j_{1}} + β_{2j_{2}} + β_{3j_{3}} − (β_{1j_{1}} + β_{3j_{3}}) = β_{2j_{2}} ≥ 1.

Case 2: Suppose p divides exactly one of the b_{1}, b_{2}, b_{3}.

Subcase 1: Suppose δ is even. Without loss of generality, say p | b_{1}, p ∤ b_{2}, p ∤ b_{3} as the other cases are similar. Then, we have p | a_{2}, a_{3} and p = q_{2j_{2}} = q_{3j_{3}} for some 1 ≤ j_{2} ≤ s_{3} and 1 ≤ j_{3} ≤ s_{3}. As p^{δ} | a_{2}^{2}b_{2}^{3}, a_{3}^{2}b_{3}^{3}, we have δ/2 ≤ β_{2j_{2}} and δ/2 ≤ β_{3j_{3}}. By Lemma 9, we have

ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D) ≥ max(δ/3, 1) + β_{2j_{2}} + β_{3j_{3}} − δ

≥ \{ \begin{matrix}1 + δ/2 + δ/2 − δ ≥ 1, & when δ = 2; \\ δ/3 + δ/2 + δ/2 − δ > 1, & when δ ≥ 4. \end{matrix}

Subcase 2: Suppose δ is odd. If p | b_{1}, p ∤ b_{2}, p ∤ b_{3}, then p | b_{1}, a_{2}, a_{3}. We have ν_{p}(a_{2}^{4}b_{2}^{6}) ≥ 4 and ν_{p}(a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) ≥ 5. Hence, we get δ ≥ 3 as p^{2δ}||D^{2} = gcd(a_{2}^{4}b_{2}^{6}, a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) and δ is odd. If p | b_{2}, p ∤ p_{1}, p ∤ p_{3}, then p | b_{2}, a_{1}, a_{3}. We have ν_{p}(a_{2}^{4}b_{2}^{6}) ≥ 6 and ν_{p}(a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) ≥ 4. Hence, we get δ ≥ 3 by similar reasoning. If p | b_{3}, p ∤ b_{1}, p ∤ b_{2}, we also get δ ≥ 3 as it is similar to p | b_{1}, p ∤ b_{2}, p ∤ b_{3}. Therefore, we obtain δ ≥ 3 in all circumstances.

Suppose p ∤ b_{i}, b_{i′} for some 1 ≤ i < i′ ≤ 3. Then p | a_{i}, a_{i′}. Say p = q_{ij_{i}} = q_{i′j_{i′}} for some 1 ≤ j_{i} ≤ s_{i} and 1 ≤ j_{i′} ≤ s_{i′}. Thus, we have 3 ≤ δ ≤ 2β_{ij_{i}} − 1 and 3 ≤ δ ≤ 2β_{i′j_{i′}} − 1 as δ is odd. Hence, we have β_{ij_{i}}, β_{i′j_{i′}} ≥ δ/2 + 1/2 and

ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D) ≥ 1 + β_{ij_{i}} + β_{i′j_{i′}} − δ > 1.

Case 3: Suppose p divides exactly two of the b_{1}, b_{2}, b_{3}.

Subcase 1: Suppose δ is even. Without loss of generality, say p | b_{1}, p | b_{2}, p ∤ b_{3} as the other cases are similar. Then p | a_{3} and p = p_{1j_{1}} = p_{2j_{2}} = q_{3j_{3}} for some 1 ≤ j_{1} ≤ r_{1}, 1 ≤ j_{2} ≤ r_{2} and 1 ≤ j_{3} ≤ s_{3}. As p^{δ} | a_{3}^{2}b_{3}^{3}, we have δ/2 ≤ β_{3j_{3}}. Also, as p^{δ} | a_{1}^{2}b_{1}^{3}, a_{2}^{2}b_{2}^{3}, we have δ ≤ 2α_{1j_{1}} + 2 and δ ≤ 2α_{2j_{2}} + 2 since δ is even. Hence, we have

ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D) ≥ (α_{1j_{1}} + 1) + (α_{2j_{2}} + 1) + β_{3j_{3}} − δ ≥ β_{3j_{3}} ≥ 1.

Subcase 2: Suppose δ is odd. If p | b_{1}, p | b_{2}, p ∤ b_{3}, then p | b_{1}, b_{2}, a_{3}. We have ν_{p}(a_{2}^{4}b_{2}^{6}) ≥ 6 and ν_{p}(a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) ≥ 5. Hence, we get δ ≥ 3 as p^{2δ}||D^{2} = gcd(a_{2}^{4}b_{2}^{6}, a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) and δ is odd. If p | b_{1}, p ∤ b_{2}, p | b_{3}, then p | b_{1}, a_{2}, b_{3}. We have ν_{p}(a_{2}^{4}b_{2}^{6}) ≥ 4 and ν_{p}(a_{1}^{2}b_{1}^{3}a_{3}^{2}b_{3}^{3}) ≥ 6. Hence, we get δ ≥ 3 by similar reasoning. If p ∤ b_{1}, p | b_{2}, p | b_{3}, we also get δ ≥ 3 as it is similar to the case p | b_{1}, p | b_{2}, p ∤ b_{3}. Therefore, we obtain δ ≥ 3 in all circumstances.

Suppose p ∤ b_{i} for some 1 ≤ i ≤ 3. Then p | a_{i} and p = q_{ij_{i}} for some 1 ≤ j_{i} ≤ s_{i}. Thus, we have 3 ≤ δ ≤ 2β_{ij_{i}} − 1 as δ is odd, and β_{ij_{i}} ≥ δ/2 + 1/2. By Lemma 9, we have

ν_{p}(a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}) − ν_{p}(D) ≥ \frac{δ}{3} + \frac{δ}{3} + β_{ij_{i}} − δ ≥ \frac{1}{2} + \frac{δ}{6} ≥ \frac{1}{2} + \frac{3}{6} = 1.

Consequently, the right-hand side of (5) is at least 1 in all of the above cases. As a result, we have inequalities (5) and (4). Putting (4) into (3), we obtain

\frac{N^{2}}{D^{2}} ≤ C_{ϵ}(\frac{a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}d}{D^{2}})^{1+ϵ} ≪ C_{ϵ}(\frac{N^{3/2}d}{D^{2}})^{1+ϵ}

as a_{1}^{2}b_{1}^{3}, a_{2}^{2}b_{2}^{3}, a_{3}^{2}b_{3}^{2} ≪ N. Together with D ≥ 1, we have N^{1/2−2ϵ} ≪_{ϵ} d and θ_{3} ≥ 1/2 as ϵ can be arbitrarily small.

5 Proof of Theorem 8

Proof. For k ≥ 4, the lower bound θ_{k} ≥ 1/2 follows from the observation that θ_{k} ≥ θ_{3} and θ_{3} ≥ 1/2 from Theorem 7 under the abc-conjecture.

For the upper bound θ_{4} ≤ 4/5, we construct 4-APs of powerful numbers as follows. With positive integer a, the following four expressions

(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) (6)

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 or 2X^{2} − 4Y^{2} = 2 or 4X^{2} = 8Y^{2} + 4

has solutions

X_{m} + \sqrt{2}Y_{m} = (3 + 2\sqrt{2})^{m} for positive integer m.

By choosing x = 8Y_{m}^{2} and x + 2a = x + 4 = 4X_{m}^{2}, we turn (6) into our desired 4-AP of powerful numbers. Observe that the common difference

d = 2(x − 2)^{2}(x + 2)^{2} ≤ 3((x − 2)^{3}(x + 2)^{2})^{4/5} = 3N^{4/5}

for large enough m (and hence N). We have θ_{4} ≤ 4/5.

For the upper bound θ_{5} ≤ 9/10, one can build upon our 3-AP and 4-AP constructions. With positive integer a, the following five expressions

(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)

(7)

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} = (2x)^{2}, y + 2a = 2^{2}(x^{2} + 2x + 1) = (2(x + 1))^{2}

with

2a = 2^{2}(2x + 1) = 8x + 4.

With the above choices, we obtain the desired 5-AP of powerful numbers with common difference

d = (4x + 2)(4x^{2} − 4x − 2)^{2}(4x^{2} + 4x + 2)^{2}
≤ 3((4x^{2} − 8x − 4)(4x^{2} − 4x − 2)^{2}(4x^{2} + 4x + 2)^{2})^{9/10} = 3N^{9/10}

for large enough x (and hence N). Thus, we have θ_{5} ≤ 9/10.

For the general upper bound θ_{k} ≤ 1 − \frac{1}{10·3^{k−5}} , we use induction on k ≥ 5 similar to Theorem 3. The base case θ_{5} ≤ 1 − \frac{1}{10·3^{5−5}} is true from θ_{5} ≤ 9/10. Suppose, for some k ≥ 5, there are infinitely many k-APs among powerful numbers with d ≤ C_{k}N^{1− \frac{1}{10·3^{k−5}}} . Say one such AP is

N = a_{1}^{2}b_{1}^{3} < a_{2}^{2}b_{2}^{3} < ⋯ < a_{k}^{2}b_{k}^{3} with common difference 1 ≤ d ≤ C_{k}N^{1− \frac{1}{10·3^{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} := Nb^{2} = a_{1}^{2}b_{1}^{3}b^{2} < a_{2}^{2}b_{2}^{3}b^{2} < ⋯ < a_{k}^{2}b_{k}^{3}b^{2} < a^{2}b^{3}

form a (k + 1)-AP of powerful numbers with common difference db^{2}. Note that

b ≤ a^{2}b = N + kd ≤ (1 + kC_{k})N.

Hence, we have

db^{\frac{2}{10·3^{k−4}}} ≤ d(1 + kC_{k})^{\frac{2}{10·3^{k−4}}} N^{\frac{2}{10·3^{k−4}}}
≤ C_{k}(1 + kC_{k})^{\frac{2}{10·3^{k−4}}} N^{1− \frac{1}{10·3^{k−5}} + \frac{2}{10·3^{k−4}}} = C_{k}(1 + kC_{k})^{\frac{2}{10·3^{k−4}}} N^{1− \frac{1}{10·3^{k−4}}}

and

db^{2} ≤ C_{k}(1 + kC_{k})^{\frac{2}{10·3^{k−4}}} (Nb^{2})^{1− \frac{1}{10·3^{k−4}}} = C_{k+1}N_{1}^{1− \frac{1}{10·3^{(k+1)−5}}}

with C_{k+1} := C_{k}(1 + kC_{k})^{\frac{2}{10·3^{k−4}}} . This completes the induction proof.

6 Acknowledgment

The author would like to thank Prof. Lajos Hajdu for pointing out that Theorem 3 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.

References

[1] T. H. Chan, A note on powerful numbers in short intervals, Bull. Australian Math. Soc.,

   2022. Available at https://doi.org/10.1017/S0004972722000995.

[2] P. Erdős, Consecutive integers, Eureka 38 (1975-76), 3–8.

[3] L. Hajdu, Perfect powers in arithmetic progression. A note on the inhomogeneous case, Acta Arith. 113 (2004), 343–349.

[4] A. Granville, Powerful numbers and Fermat’s last theorem, C. R. Math. Rep. Acad. Sci. Canada 8 (1986), 215–218.

[5] B. J. Green and T. C. Tao, The primes contain arbitrarily long arithmetic progressions, Ann. of Math. (2) 167 (2008), 481–547.

[6] R. A. Mollin and P. G. Walsh, A note of powerful numbers, quadratic fields and the Pellian, C. R. Math. Rep. Acad. Sci. Canada 8 (1986), 109–114.

2020 Mathematics Subject Classification: Primary 11N25. Keywords: powerful number, k-full number, arithmetic progression, abc-conjecture.

(Concerned with sequence \underline{A001694}.) Received September 30 2022; revised versions received October 1 2022; October 3 2022; December 13 2022. Published in Journal of Integer Sequences, December 14 2022. Return to Journal of Integer Sequences home page.