3

My question A conjecture with connection to the $abc$-conjecture is about a conjectured injection from exceptional $abc$-triplets $(a,b,c)\mapsto a^2+b^2$, but this question is about a conjectured injection from exceptional $abc$-triplets $(a,b,c)\mapsto a+2b$. It's similar but not a duplicate.

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, where $a<b$,
$(a,b,c),(a',b',c')\in E_0\,$ and $\,a+2b=a'+2b'\implies \{a,b\}=\{a',b'\}$

So far tested for all $a,b$ less than $100,000$. Can this be proved?

enter image description here

There is no uniqueness for $a+b$, $a-b$ or $2a+b$.

Community
  • 1
Lehs
  • 14,252
  • 4
  • 28
  • 82

1 Answers1

2

The conjecture is not true. For a counterexample consider $$ (a,b,c) = (1,386560,386561) \quad (a',b',c') = (4913,384104,389017) $$ which are $abc$-triplets with $a<b$, $a'<b'$, $a+2b=a'+2b'$ and $\{a,b\}\neq\{a',b'\}$.

Zander
  • 11,118
  • 1
  • 27
  • 55