Let $d(a,b) = 1 - \frac{2\gcd(a,b)^3}{ab(a+b)}$ be a metric on natural numbers without $0$.

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 (pairwise distinct) 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.