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We define $A=\{ \frac{c}{rad(abc)}: a, b > 0, c=a+b, gcd(a, b)=1 \}$. Is the set $A$ dense in $[0, +\infty)$? Does $\overline{A}$ have interior? Here $\overline{A}$ is the closure of $A$.

A well-known fact is that $\inf A=0$ and $\sup A=+\infty$.

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    $\begingroup$ The only accumulation point I can come up with (besides $0$ and $\infty$) is $1/2$ by using taking $a=1$ and $b$ a Mersenne prime, but conditional on there being infinitely many such primes. I plotted the ratios for all $abc$-triples below $10^18 (math.leidenuniv.nl/~desmit/abc/abctriples_below_1018.gz) and found no noticeable gaps (until the tail end). Last, I plotted the ratios for all triples $a+b=c$ for $c\le 10^5$ and ratio $<1$ - and found a few small gaps approaching $1$, the most noticeable a gap of size $>0.013$ centered near $0.87$ - but is that significant? $\endgroup$ Aug 30, 2021 at 9:10

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This is not a full answer, but a pair of soft arguments suggesting that $A$ is dense in $[0, +\infty)$.


First Argument

Given any triple $(a,b,c)$, let $\displaystyle r(a,b,c)=\frac{c}{\text{rad}(abc)}$. One can generate two new triples

$$t_1=(a(c+b),b^2,c^2)$$ $$t_2=(a^2,b(c+a),c^2)$$ with ratios $$r(t_1)=r(a,b,c)\cdot\frac{c}{c+b}\cdot\frac{c+b}{\text{rad}(c+b)}$$ $$r(t_2)=r(a,b,c)\cdot\frac{c}{c+a}\cdot\frac{c+a}{\text{rad}(c+a)}$$

Clearly $\displaystyle\frac{1}{2}<\frac{c}{c+a}, \frac{c}{c+b}<1$. On the other hand $\displaystyle \frac{n}{\text{rad}(n)}$ is on multiplicative average equal to $\prod_{p} p^{1/(p^2-p)}\approx2.128$. So the tree of triples generated by repeated application of the transformations above, will form paths of ratios that drift (mostly slowly) to increasingly higher values. $0$ being an obvious accumulation point of $A$, one has infinitely many starting points for the process above, making it plausible that any presumed gap in $A$ will be cut by infinitely many paths.


Second Argument

This is via an example specifically targeting the gap around $0.87$ that was mentioned in a comment to the question. Start by approximating $0.87$ with a rational with squarefree denominator (and preferably multiple small prime factors). $61/70$ is a good candidate. Take the triple $(1, 2^{100}\cdot 5^7\cdot 7^4, 2^{100}\cdot 5^7\cdot 7^4+1)$ where the exponents $100,7,4$ are picked so that $61^2$ divides $c$. There is then a good chance that $c/\text{rad}(c)=61$, which is indeed the case here, and the ratio of the triple is therefore $61/70$.

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  • $\begingroup$ Thanks for your discussions. 本问题只是抛砖引玉,还请各位志士仁人洒潘江、倾陆海。 $\endgroup$
    – LMP
    Aug 31, 2021 at 1:06
  • $\begingroup$ I conjecture that even the subset of $A$ derived from triples with $a=1$ is dense in $[0,+\infty)$. $\endgroup$ Aug 31, 2021 at 8:28
  • $\begingroup$ By looking at all Pythagorean triples with $c$ up to $637460^2$ I conjecture that that subset of $A$ too has ratios dense in the positive reals. $\endgroup$ Aug 31, 2021 at 8:39

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