Let $d(a,b) = 1 - \frac{2\gcd(a,b)^3}{ab(a+b)}$ be a metric on natural numbers. 

The metric space $X = \{x_0,x_1,\cdots,x_n\},n>2$ is isometric embeddable in $\mathbb{R}^n$ if and only if the matrix:
$$M(x_0,x_1,\cdots,x_n) = (1/2 (d(x_0,x_i)^2+d(x_0,x_j)^2-d(x_i,x_j)^2))_{1 \le i,j \le n}$$
is positive semidefinite.


So my question is:

> Is the matrix above for $d$ as above positive semidefinite for all
> choices of $x_i \in \mathbb{N}$? (Maybe it is possible to prove this using quadratic
> forms and then transform it to $\sum_{i} a_{ii} y_i^2$ showing then
> that $a_{ii}\ge 0$?


If it is so, then this would one allow to do euclidean geometry of natural numbers.
For instance for three points / natural numbers we would have:

1) a triangle
2) Sine law
3) Cosine law
4) All other theorems concerning triangles

Then in the limit three consecutive numbers / primes would build an equilateral triangle of side length $1$. Hence one could imagine primes ("in the limit") as an infinite dimensional simplex, which would be a funny thing to think about.

Thanks for your help.

Related question:
https://math.stackexchange.com/questions/3385102/is-this-metric-matrix-positive-semidefinite

See Theorem 2.4 in https://books.google.de/books?id=7_DuCAAAQBAJ&printsec=frontcover&hl=de&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false for isometrically embedding of $(\mathbb{N},d)$ in a Hilbert space.