Shinichi Mochizuki has given a long proof of the \(abc\) conjecture, it was recently published in a peer-reviewed journal, but only a tiny number of people in the world have a justifiable reason to be certain about the validity (or invalidity) of the proof.LHCb anomaly strengthens: the decay \(B^{0}\rightarrow K^{*0}\mu^{+}\mu^{-}\) has seen the increase from 3 sigma to 3.4 sigma. It is not the only flavor anomaly at the LHCb. See the preprint and Particle Bites

Swampland: a fun Harvard-Cornell paper unifying the Weak Gravity and Distance Conjectures using BPS black holes

Maybe there is a more elementary proof (as well)?

The \(abc\) conjecture says the following.

For any (arbitrarily small) \(\epsilon\gt 0\), there exists a (large enough but fixed) constant \(C_\epsilon\) such that each triplet of relatively prime (i.e. having no common divisors) integers \(a,b,c\) that satisfies \[

a+b=c

\] the following inequality still holds:\[

\Large \max (\abs a, \abs b, \abs c) \leq C_\epsilon \prod_{p|(abc)} p^{1+\epsilon}.

\] That's it. I used larger fonts because it's still a key inequality of this blog entry. Morally, the conjecture says that the ratio of \(c\), the maximum of \(|a|,|b|,|c|\) where we assume \(a+b=c\) but also positiveness of all these numbers, and \(abc\), with all copies of prime factors removed and exponentiated to the power of \(1+\epsilon\) arbitrarily close to 1, is bounded.

It's almost equivalent to vaguely saying that there is just a finite number of exceptions of triplets \((a,b,c)\in \ZZ\) obeying \(a+b=c\) for which the "overgrown" primes in \(abc\) are so numerous that the removal of these extra primes lowers \(abc\) to a number that may compete with the size of \(a,b,c\) themselves (after a modest adjustment by the power).

Fine. It's not completely elementary. But what if there exists a nice function of a complex variable that turns this condition into some simple convergence of a series for this function? OK, let us assume that \(a,b\in\{1,2,3,\dots\}\) and \(c=a+b\) is therefore the largest integer among the three. The inequality in the \(abc\) conjecture said \[

\Large \frac{a+b}{\prod_{p|(ab(a+b))} p^{1+\epsilon}} \leq C_\epsilon

\] where the constant \(C_\epsilon\) must be fixed and finite for a given small \(\epsilon\). But maybe what is finite is actually the sum over all pairs of relatively prime \(a,b\): I am strengthening the hypothesis a little bit. It could mean that for every \(\epsilon\), the sum\[

\zeta_{abc}(\epsilon) = \sum_{a,b=1}^{\infty,RP} \frac{a+b}{\prod_{p|(ab(a+b))} p^{1+\epsilon}}

\] is simply convergent. The superscript \(RP\) indicates that the sum only goes over the positive, relatively prime pairs \((a,b)\). Now, I believe that this sum could be rather easily related to a similar sum over all positive pairs \((a,b)\), ignoring the relatively prime conditions, because the summand for \((A,B)=q(a,b)\) is rather easily related to that for \((a,b)\).

Note that the function that I defined has \(\epsilon\) in the exponent and in this sense, it is a cousin of the Riemann zeta function \(\zeta(\epsilon)=\sum_{n=1}^\infty (1/n^\epsilon)\).

So the most problematic thing is the "product of primes", i.e. the removal of the copies of primes in \(ab(a+b)\). But there could exist some nonlinear (and non-polynomial) function \(g\) of \(\zeta_{abc}(\epsilon)\), expressed via a Taylor series, something like the exponential, and if you expressed \(g[\zeta_{abc}(\epsilon)]\) as a sum of products, you could effectively undo the restriction that the bases of power are squarefree integers – by some logic that is analogous to that of the Euler product.

The whole expression could be reduced to something simple, like summing over all possible partitions of integers times the sum over all possible factorizations of the integers, and \(g[\zeta_{abc}(\epsilon)]\) could therefore be "calculable" in terms of simpler functions, perhaps the familiar zeta function(s). If that were true, \(\zeta_{abc}(\epsilon)\) itself could be written as the inverse function \(g^{-1}[m(\epsilon)]\) with some simpler function \(m\) substituted as the argument.

The validity of the \(abc\) conjecture could be reduced to the convergence of these functions for a tiny positive \(\epsilon\).

Unfortunately, I don't have enough room in this blog post to write the complete proof for you. ;-)

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