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On weaker forms of the abc conjecture from the theory of Hölder and logarithmic means

MathOverflow Asked by user142929 on December 18, 2020

In this post (the content of this post is now cross-posted from Mathematics Stack Exchange see below) we denote the radical of an integer $n>1$ as the product of disctinct primes dividing it $$operatorname{rad}(n)=prod_{substack{pmid n\ptext{ prime}}}p,$$ with the definition $operatorname{rad}(1)=1$. The abc conjecture is an important problem in mathematics as one can to see from the Wikipedia abc conjecture. In this post I mean the formulation ABC conjecture II stated in previous link.

I was inspired in the theory of generalized mean or Hölder mean (see [1]) to state the following claim (Mathematics Stack Exchange 3648776 with title A weak form of the abc conjecture involving the definition of Hölder mean asked Apr 28 ’20).

Claim. On assumption of the abc conjecture $forall varepsilon>0$ there exists a constant $mu(epsilon)>0$ such that for triples of positive integers $a,b,cgeq 1$ satisfying $gcd(a,b)=gcd(a,c)=gcd(b,c)=1$ and $a+b=c$ ones has for real numbers $q>0$ that the following inequality holds
$$c<mu(varepsilon)left(frac{operatorname{rad}(a)^q+operatorname{rad}(b)^q+operatorname{rad}(c)^q}{3}right)^{3(1+varepsilon)/q}.tag{1}$$

Remark 1. Thus as $qto 0$ from the theory of Hölder mean we recover the abc conjecture.

In a similar way I was inspired in the definition of the logarithmic mean and its relationship to the arithmetic mean to pose the following conjecture (Mathematics Stack Exchange 3580506 with title Weaker than abc conjecture invoking the inequality between the arithmetic and logarithmic means asked Mar 14 ’20).

Conjecture. For every real number $varepsilon>0$, there exists a positive constant $mu(varepsilon)$ such that for all pairs $(a,b)$ of coprime positive integers $1leq a<b$ the following inequality holds $$2,frac{b-a}{logleft(frac{b}{a}right)}leq mu(varepsilon)operatorname{rad}(ab(a+b))^{1+varepsilon}.tag{2}$$

Remark 2. Thus I think that previous conjecture is weaker than the abc conjecture by virtue of the relation between the artihmetic and logarithmic means.

Question. I wondered what work can be done to prove/discuss unconditionally (I mean on assumption of the cited requirements/conditions, but without invoking any formulations of the abc conjecture) the veracity of previous Claim for the smallest $q>0$ close to* $0$ that you are able to prove. Similarly**, is it possible to prove Conjecture? Many thanks.

*I’m curious to know what is the smallest $q>0$ close to $0$ such that the inequality in Claim is true, I think that the right discussion is for $q>0$ but if you want to discuss $|q|$ very close to $0$ because you think that it makes sense, feel free to study our inequality for real numbers $|q|$ very close to $0$.

$^{**}$On the other hand I think that should be possible to prove the Conjecture, since I think that this statement is much weaker than the abc conjecture.

I was inspired in the Wikipedia articles for Generalized mean and Logarithmic mean. I add references to bilbiography. I know the statement of formulation ABC conjecture II for example from [3].

References:

[1] P. S. Bullen, Handbook of Means and Their Inequalities, Dordrecht, Netherlands: Kluwer (2003).

[2] B. C. Carlson, Some inequalities for hypergeometric functions, Proc. Amer. Math. Soc., 17: in page 36 (1966).

[3] Andrew Granville and Thomas J. Tucker, It’s As Easy As abc, Notices of the AMS, Volume 49, Number 10 (November 2002).

One Answer

abc implies your conjecture with $b-a$.

Case 1 Let $a,b,c=a+b$ be bad abc triple,i.e. $c < rad(ab(a+b))$.

We have $rad(ab(a+b)) > c > b - a$.

Case 2 Let $a,b,c=a+b$ be good abc triple,i.e. $c>rad(ab(a+b))$.

Then $T : (b-a)^2,4ab,(a+b)^2$ is good abc triple too.

The radical is divisor of $ab(a+b)(b-a)$ and we have $(a+b)^2 > (a+b)(b-a)$.

If $log(b-a) < (1-C) log(b+a)$ this will give infinitely many good abc triples with quality $2/(2-C)$, which contradicts abc.

In summary, abc implies there are only finitely many good abc triples satisfying $log(b-a) < (1-epsilon) log(b+a)$

Answered by joro on December 18, 2020

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