Let $d(a,b) = 1-\frac{2 \gcd(a,b)}{a+b}$, $d_{ABC}(a,b) = 1-\frac{2\gcd(a,b)^3}{ab(a+b)}$ be two metrics on natural numbers.
The abc-conjecture can be formulated using these two metrics as:
For every $\epsilon >0 $ there exists a $K_{\epsilon}$ such that:
$$d(a,b) < D_{\epsilon}(a,b):=1-\frac{2}{K_{\epsilon}\text{rad}(\frac{2}{1-d_{ABC}(a,b)})^{1+\epsilon}}$$
Trying make $D$ to a metric on natural numbers, leads to $K_{\epsilon} = \frac{1}{2^\epsilon}$. One can prove that with this choice of $K_{\epsilon}$, $D$ is a metric on natural numbers.
Now this is a very explicit version of the abc-conjecture, and for example for $\epsilon = 1/2$ this version is wrong but my conjecture is that for $\epsilon \ge 1$ this is true.
My question is:
1) There are various results which build upon the abc-conjecture. Is there $0 < \epsilon < 1$ needed, or is it enough to have $\epsilon \ge 1$?
I have tested this with the Schoenberg-Criterion for some $\epsilon$ and some numbers $n$:
2) Is the metric $D_{\epsilon}$ for every $\epsilon>0$ embeddable in Euclidean Space? (Maybe this is too much to ask.)
(This seems very counterintuitive, because of the radical function one would not expect the $D$ to be so nice to behave as to embedd in Euclidean Space.)
Thanks for your help!
Edit: Making the following Ansatz:
Suppose that $\phi(a)$ is the vector which embedds the natural number $a$. Then $ D_1(a,b)^2 = |\phi(a)|^2+|\phi(b)|^2 - 2 \left< \phi(a), \phi(b) \right>$
If we assume that:
$$\left< \phi(a), \phi(b) \right> = \frac{-2}{R(a,b)^2}(\frac{4}{R(a,b)^2}-2)$$
where $R(a,b) = \text{ rad }(\frac{ab(a+b)}{\gcd(a,b)^3})$
then we get, substituting $a=b$ that $|\phi(a)| = \frac{1}{\sqrt{2}}$
So this is of course no proof, but if $\phi$ existed, then the matrix
$$G_n = (\frac{-2}{R(a,b)^2}(\frac{4}{R(a,b)^2}-2))_{1 \le a,b \le n}$$
should be positive definite, which I have checked for $n$ up to $20$ in Sagemath, and it is so.
So to be more precise about the question 2):
3) Is the matrix $G_n$ as defined above, positive definite?
Related question: