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The numbers $\,a,b,c\,$ is a $abc$-triplet if they are coprime and $a+b=c$. One version of the abc-conjecture is then:

For all $\varepsilon>0$ the set $E_\varepsilon$ of all $abc$-triplets with $c>\text{rad}(a\cdot b\cdot c)^{1+\varepsilon}$ is finite. where $\,\text{rad}(p_1^{n1}\cdots p_k^{n_k})=p_1\cdots p_k$ and $\,p_1,\dots ,p_k$ are arbitrary primes.

But it is known that $E_0$ is infinite. Conjecture:

For $abc$-triplets,
$(a,b,c),(a',b',c')\in E_0\,$ and $\,a^2+b^2=a'^2+b'^2\implies \{a,b\}=\{a',b'\}$

So far tested for all $a,b$ less than $10^5$. Can this be proved?

enter image description here

Diagram for the 9 exceptional abc-triplets with least distance to origo: $(1,8),(5,27),(1,48),(32,49),(1,63),(1,80),(4,121),(3,125),(81,175)$

http://forthmath.blogspot.se/2016/08/the-abc-conjecture-1.html

Lehs
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  • As detailled in wikipedia :for a positive integer $n$, the radical of $n$, denoted $rad(n)$, is the product of the distinct prime factors of $n$. For example $rad(16) = rad(2^4) = 2$ , $rad(17) = 17$, $rad(18) = rad(2 ⋅ 3^2) = 2 · 3 = 6$, etc –  Aug 26 '16 at 18:58
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    I checked the conjecture up to $a,b\le300$, and it holds. Now if I only understood how to make my naive python script more efficient... – Daniel Robert-Nicoud Aug 26 '16 at 21:12
  • @DanielRobert-Nicoud: I use Pollard Rho to factorize and calculate the radical and it's rather fast. But looking over my routine http://forthmath.blogspot.se/2016/08/the-abc-conjecture-1.html I see there is a lot to be done to optimize. – Lehs Aug 27 '16 at 05:45
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    @Lehs I'll take a look at that. My simplistic radical computation took up something like 99.99% of the total running time... – Daniel Robert-Nicoud Aug 27 '16 at 10:55

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A list of triples $(A,B,C)$ (for $C<10^{18}$) such that $$A+B=C,\\rad(ABC)<C,\tag{1}$$ is accessible by the link [Bart de Smit/ABC triples/by size][1], file [abctriples_below_1018.gz][2] .

A quick check yields that your statement is true for ABC-triples with $C< 10^{18}$.

Rolandb
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