Question. Is there an integer $n_0 \geq 2$ such that $$\left\{\frac{c}{rad(abc)^{n_0}}: a, b >0,\; c=a+b,\; \gcd(a, b)=1\right \}$$ is bounded?
The abc conjecture can directly deduce this conjecture.
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1$\begingroup$ Both these "conjectures" are questions, not conjectures. $\endgroup$– Carl-Fredrik Nyberg BroddaCommented Aug 23, 2021 at 13:12
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2$\begingroup$ It's not a great idea to edit your question in such a way as to make it difficult to figure out how the answers relate to the question. You original question had 2 parts, so my answer refers to those two parts. So at least you should change the current version of your question to something like: Question 1 Deleted because it is a well-known open conjecture. Question 2 Is there an integer ... That way the answer that I posted makes sense! $\endgroup$– Joe SilvermanCommented Aug 23, 2021 at 14:31
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1$\begingroup$ Translating LMP comments: Hello! First of all thank you for your answer. I only registered an account on this website in the last week, so I am not familiar with the various functions. My English is not good. It is not easy for me to write some sentences in English. Moreover, my major is not number theory, it can be regarded as a hobby. I didn't know that the problem was well-known before. The second question, after listening to your explanation, it is just a question of elementary number theory, just an exercise. Anyway, thank you. --- $\endgroup$– Joe SilvermanCommented Aug 23, 2021 at 15:46
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2$\begingroup$ Translating LMP comments: For some of the more difficult natural problems in number theory, different people often think of the same problem. It's nothing. After you read my reply, I will delete everything. It is recommended that you use a simple software to translate, Chinese into your native language. $\endgroup$– Joe SilvermanCommented Aug 23, 2021 at 15:46
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$\begingroup$ These sorts of questions will get a better reception on MathStackExchange. Good luck with your investigations. $\endgroup$– Joe SilvermanCommented Aug 23, 2021 at 15:47
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1 Answer
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You Question 1 is a weak form of the $abc$ conjecture that is known to imply asymptotic Fermat, among many other applications. So no, it is not known to be true, but it is a well-known conjecture, and thus unfortunately not an appropriate question for this site. For your second question, if there is a solution, note that $c=rad(abc)$ implies that $c$ is square-free. But then $rad(abc)\ge c$, with equality unless every prime dividing $ab$ also divides $c$. But if $p$ divides $a$, then $p$ divides $c$, which (from $c=a+b$) implies that $p$ divides $b$, contradicting your $\gcd(a,b)=1$ assumption. So it seems that your set in Question 2 is the empty set.
answered Aug 23, 2021 at 13:41
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7$\begingroup$ The last set is not quite empty - it contains $2$ coming from an equality $1+1=2$, but your argument implies none of $a,b$ can be greater than $1$. $\endgroup$– WojowuCommented Aug 23, 2021 at 13:48
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4$\begingroup$ @Wojowu Ah, yes, that silly number $1$ that has no prime divisors. :) Thanks for the correction. $\endgroup$ Commented Aug 23, 2021 at 14:28