Q:

Conjecture: injection from exceptional abc-triplets to natural numbers

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?

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

A:

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 $\{a,b\}\neq\{a',b'\}$.

