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Famous so-called abc-conjecture imply that every (except only finite amount of exceptions,) $(a,b,c)$ that $a+b=c$, satisfies $c<rad(abc)^{1+\epsilon}$ for fixed positive $\epsilon$. Here, $rad(x)$ is multiple of all prime divisors.

Above $\epsilon$ is crucial, because there are infinitely many $a,b,c$ that satisfies $a+b=c$ and $rad(abc)<c$. For example, Take $a=1, b=2^{p(p-1)n}-1, c=2^{p(p-1)n}$. Since $b$ is divisible by $p^2$, we have, $rad(abc)=rad(bc)=2rad(b)/p<2c/p$.

Now since $np(p-1)=\log_{2}c$, we might as well have that

"There are infinitely many counterexamples to $c <rad(abc) \sqrt{logc}$"

which explains why we need that $\epsilon$ term even more because having "constant term bigger" won't help in above case. This was all found on Wikipedia,(abc conjecture page) which naturally lead me to ask this question.

First, Term $\sqrt{\log c}$ can be improved replace $p(p-1)$ with $p^l(p-1)$, with simillar approach, one can get

" For every positive $\epsilon$, There exist infinitely many counterexamples to $c<rad(abc) (\log c)^{1-\epsilon}$ "

If the term $\epsilon$ have the greatest role possible, one can expect, even more, that,

Question: " For every positive $\epsilon$, There exist infinitely many counterexamples to $c<rad(abc) (\log c)^{\epsilon}$ "

Would this be true? Does anyone have any ideas? I'm pretty sure that I'm not the only one to have thought about this. Thank you for taking your time.

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Ok, it turns out that this is indeed the case.

There exists infinitely many $a,b,c$ with $$ c > rad(abc) \exp{(k \sqrt{\log c} / \log \log c )}$$

which is stronger than the conjecture I asked

For more info: https://core.ac.uk/download/pdf/82016238.pdf

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