Gaps between divisible terms in $a^2 (a^2 + 1)$

TL;DR

This paper investigates the divisibility properties of the sequence defined by a^2(a^2+1), establishing a lower bound on the gap between terms and exploring implications under the abc conjecture.

Contribution

It improves previous results by establishing a new gap principle for divisibility in the sequence without extra assumptions and under the abc conjecture.

Findings

Proves a new lower bound on the gap between divisible terms.

Establishes a conditional bound under the abc conjecture.

Advances understanding of divisibility gaps in quadratic sequences.

Abstract

Suppose $a^2 (a^2 + 1)$ divides $b^2 (b^2 + 1)$ with $b > a$. In this paper, we improve a previous result and prove a gap principle, without any additional assumptions, namely $b \gg a (\log a)^{1/8} / (\log \log a)^{12}$. We also obtain $b \gg_\epsilon a^{15/14 - \epsilon}$ under the abc conjecture.

Equations

$$F_1=1,F_2=1,F_{n+1}=F_n+F_{n-1}\text{ for }n\ge 2.$$

$$b\gg \frac{a(\log a{)}^{1/8}}{(\log \max (e,\log a){)}^{12}}.$$

$$b{\gg }_{ϵ}{a}^{15/14-ϵ}.$$

$$t{a}^2(a^2+1)=b^2(b^2+1)$$

$$t{x}^2\frac{D^2{x}^2+1}{T}=y^2\frac{D^2{y}^2+1}{T}.$$

$$t(D^2{x}^2+1)=mT{y}^2$$

$$D^2{y}^2+1=mT{x}^2.$$

$$[(mT{)}^2-t{D}^4]x^2=t{D}^2+mT.$$

$$[(mT{)}^2-t{D}^4]y^2=t{D}^2+tmT.$$

$$s(y^2-x^2)=(t-1)mT$$

$$s=(mT{)}^2-t{D}^4.$$

$$(s,mT)=((mT{)}^2-t{D}^4,mT)=(t{D}^4,mT)=(t,mT).$$

$$Y^2=t{X}^4+s$$

$$\max (X,Y)<e^{C_1\lambda (\max (1,\log \lambda ){)}^{96}}$$

$$D=X<e^{C_1\frac{1}{C_1}\frac{\log D}{(\log \max (e,\log D){)}^{96}}(\log \max (e,\log D){)}^{96}}=D$$

$$t\gg \frac{\sqrt{\log D}}{(\log \max (e,\log D){)}^{48}}.$$

$$t\gg \frac{\sqrt{\log a}}{(\log \max (e,\log a){)}^{48}}$$

$$c<C_{ϵ}R(abc{)}^{1+ϵ}.$$

$$a=Dx,b=Dy,\text{ and }s+t{D}^4=(mT{)}^2$$

$$(t,(mT{)}^2)=d\text{ and }d=d_1{d}_2^2\text{ with }{d}_1\text{ squarefree}.$$

$$\frac{s}{d}+\frac{t}{d}{D}^4=d_1{S}^2.$$

$$\frac{t}{d}D^{4}<C_{\epsilon}\Bigl{(}\frac{(t-1)t}{d}Dd_{1}S\Bigr{)}^{1+\epsilon}$$

$$(d_{1}S{)}^2\ll t{D}^4\text{ or }{d}_{1}S\ll t^{1/2}{D}^2.$$

$$\frac{t}{d}D^{4}\ll_{\epsilon}\Bigl{(}\frac{t^{5/2}}{d}D^{3}\Bigr{)}^{1+\epsilon}.$$

$$D{\ll }_{ϵ}{t}^{3/2+8ϵ}.$$

$$x^2\ll \frac{t{D}^2}{s}\text{ or }x\ll t^{1/2}D.$$

$$a=Dx\ll t^{1/2}{D}^2{\ll }_{ϵ}{t}^{7/2+16ϵ}\text{ or }t{\gg }_{ϵ}{a}^{2/7-2ϵ}$$

$$10D^4<t\le 2(Dx{)}^{4/5}.$$

$$x>\frac{10}{2^{5/4}}{D}^4.$$

$$mT\le 2D^2{y}^2/x^2\le 2\sqrt{2}{t}^{1/2}{D}^2.$$

Full text

1 Introduction and Main Results

We are interested in increasing sequence of positive integers $(a_{n})_{n\geq 0}$ with each term dividing the next one (i.e. $a_{n}|a_{n+1}$). A simple example is $(2^{n})_{n\geq 0}$. Another example is $(n!)_{n\geq 1}$. These are simple recursively defined sequences. It is more interesting and challenging to require each term of the sequence to have a special form. For example, $(3^{n})_{n\geq 0}$ has all terms odd. For another special example, let us consider the Fibonacci numbers

[TABLE]

By the well-known fact that $F_{m}|F_{n}$ if and only if $m|n$, the sequence $(F_{2^{n}})_{n\geq 0}$ has all terms of Fibonacci-type and each term dividing the next one. One may restrict the sequence to numbers of the form $n^{2}+1$ or other polynomials, and we are interested in the growth of such sequences. In this paper, we shall focus on numbers of the form $n^{2}(n^{2}+1)$ and study the following

Question 1

Suppose $a^{2}(a^{2}+1)$ divides $b^{2}(b^{2}+1)$. Must it be true that there is some gap between $a$ and $b$? More preciously, is it true that $b>a^{1+\epsilon}$ for some small $\epsilon>0$?

In [References], the author studied the above question with some additional restrictions on $a$ and $b$. In this paper, we remove all these restrictions and prove

Theorem 2

Let $a$ and $b$ be positive integers with $a<b$. Suppose $a^{2}(a^{2}+1)$ divides $b^{2}(b^{2}+1)$. Then

[TABLE]

Recently, Stephen Choi, Peter Lam and the author [References] defined gap principle of order $n$ for polynomials with integer coefficients as follows:

Definition 3

Let $n$ be a positive integer and $f(x)$ be a polynomial with integral coefficients. Consider the set of all positive integers $a_{0}<a_{1}<a_{2}<\ldots<a_{n}$ such that $f(a_{i})$ divides $f(a_{i+1})$ for $0\leq i\leq n-1$. We say that $f(x)$ satisfies the gap principle of order $n$ if $\lim a_{n}/a_{0}=\infty$ as $a_{0}\rightarrow\infty$.

Hence, Theorem 2 implies that the polynomial $f(x)=x^{2}(x^{2}+1)$ satisfies the gap principle of order $1$.

Assuming the abc conjecture, we can obtain a better gap result than that in [References], without any extra assumptions.

Theorem 4

Let $a$ and $b$ be positive integers with $a<b$. Suppose $a^{2}(a^{2}+1)$ divides $b^{2}(b^{2}+1)$. Then, under the abc conjecture with any small $\epsilon>0$,

[TABLE]

Thus, this answers question 1 in the affirmative under the abc conjecture. We hope this article would inspire readers to study questions of similar nature.

Some Notations The symbol $a|b$ means that $a$ divides $b$. The notations $f(x)\ll g(x)$ and $g(x)\gg f(x)$ are all equivalent to $|f(x)|\leq Cg(x)$ for some constant $C>0$. Finally $f(x)=O_{\lambda}(g(x))$, $f(x)\ll_{\lambda}g(x)$ or $g(x)\gg_{\lambda}f(x)$ mean that the implicit constant $C$ may depend on $\lambda$.

2 Proof of Theorem 2

Since $a^{2}(a^{2}+1)$ divides $b^{2}(b^{2}+1)$, say

[TABLE]

for some integer $t>1$. We may assume $t\leq\log a$, for otherwise the theorem is true automatically. Let $D$ be the greatest common divisor of $a$ and $b$. Suppose $a=Dx$ and $b=Dy$ with $(x,y)=1$, and let $T=(D^{2}x^{2}+1,D^{2}y^{2}+1)$. Then

[TABLE]

Since $(x,y)=1$, $x^{2}$ must divide $(D^{2}y^{2}+1)/T$. Say $(D^{2}y^{2}+1)/T=mx^{2}$ for some integer $m$. Then $t(D^{2}x^{2}+1)/T=my^{2}$, and we have

[TABLE]

and

[TABLE]

Multiplying (2) by $D^{2}$, (3) by $mT$ and combining, we have

[TABLE]

Similarly, multiplying (2) by $mT$, (3) by $tD^{2}$ and combining, we have

[TABLE]

Subtracting (4) from (5), we get

[TABLE]

with

[TABLE]

From (3), we have $(mT,D)=1$. Hence,

[TABLE]

Therefore, by combining this with (6), we have $s|(t-1)t$ and (7) gives us an hyperelliptic curve

[TABLE]

by setting $Y=mT$ and $X=D$. Let $\lambda:=(t-1)t$. By Theorem 1 of Voutier [References] on the study of integral solutions to hyperelliptic curves using transcendental number theory, we have

[TABLE]

for some constant $C_{1}\geq 1$. Suppose $\lambda\leq\frac{1}{C_{1}}\frac{\log D}{(\log\max(e,\log D))^{96}}$. Then

[TABLE]

which is a contradiction. Therefore, $(t-1)t>\frac{1}{C_{1}}\frac{\log D}{(\log\max(e,\log D))^{96}}$ and

[TABLE]

From (2), we have $2tD^{2}x^{2}\geq mTy^{2}$ and $tD^{2}\gg mT$. This together with (4) gives $tD^{2}\gg x^{2}$. Hence, as $t\leq\log a$ and $a=Dx$, we have $\log D\gg\log a$. Therefore, (8) gives

[TABLE]

and we have Theorem 2 as $b^{4}/a^{4}\gg t$.

3 Proof of Theorem 4

Firstly, for any integer $n$, let $R(n):=\prod_{p|n}p$ be the radical or kernel of an integer $n$. The abc-conjecture is

Conjecture 5

For every $\epsilon>0$, there exists a constant $C_{\epsilon}$ such that for all triples $(a,b,c)$ of coprime positive integers, with $a+b=c$, then

[TABLE]

Secondly, let us state a lemma which follows easily from the unique prime factorization of numbers.

Lemma 6

Suppose $a|A^{2}$ and $a=a_{1}a_{2}^{2}$ with $a_{1}$ squarefree. Then $a_{1}a_{2}|A$.

Basically, we follow the proof of Theorem 2. Using the same notation as in section 2,

[TABLE]

with $(mT,D)=1$. Let

[TABLE]

By Lemma 6, we have $mT=d_{1}d_{2}S$. Then

[TABLE]

Here, the three terms are integers and pairwise relatively prime. We can apply the abc conjecture and obtain

[TABLE]

since $s|(t-1)t$. Suppose $t\leq 10D^{4}$. As $d_{1}\leq d$, (9) gives

[TABLE]

Putting this into (10), we have

[TABLE]

Hence,

[TABLE]

As $s+tD^{4}=(mT)^{2}$, $s|(t-1)t$ and $t\leq D^{4}$, we have $mT\ll t^{1/2}D^{2}$. This together with (4) gives

[TABLE]

Therefore, by (11),

[TABLE]

which gives Theorem 4 as $b^{4}/a^{4}\gg t$.

We are left with the case $t>10D^{4}$. Suppose $y/x=b/a\leq a^{1/5}=(Dx)^{1/5}$. From (1), we have $t^{1/4}\leq\sqrt[4]{2}y/x$. Hence,

[TABLE]

This gives

[TABLE]

On the other hand, (3) and (1) give

[TABLE]

Putting this into (4), we have

[TABLE]

Combining (12) and (14), we obtain

[TABLE]

which contradicts (13). Therefore, we must have $b/a>a^{1/5}$ which also gives Theorem 4.