The abc-conjecture is:


For every $\epsilon > 0$ there exists $K\_{\epsilon}$ such that for all natural numbers $a \neq b$ we have:


$$ \frac{a+b}{\gcd(a,b)}\,\ <\,\ K\_{\epsilon}\cdot \text{rad}\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^{1+\epsilon} $$


I have two questions after doing some experiments with SAGEMATH:


1) Is the matrix 
$$L\_n = \left( \frac{\gcd(a,b)}{a+b}\right)\_{1\le a,b \le n}$$
positive definite?


2) Is the matrix:
$$ R\_n = \left(
 \frac{1}{\text{rad}\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)}
 \right)\_{1\le a,b \le n} $$
positive definite?


If both of the questions can be answered with yes, then we would have "mappings" 


$$\psi ,\phi: \mathbb{N} \rightarrow \mathbb{R}^n$$


and the abc-conjecture might be stated as an inequality in the inner-product of these mappings:


$$\left< \psi(a),\psi(b) \right>^{1+\epsilon} < K\_{\epsilon} \left < \phi(a), \phi(b) \right >$$


which I think would be very interesting.


**Edit**:
I realized that it is better to ask the following question:


Is 


$$R^{(\epsilon)}\_n := (\frac{2^{\epsilon}}{\text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^{1+\epsilon}})\_{1\le a,b\le n}$$


positive definite for all $\epsilon \ge 0$?


If "yes", then we would have:


For all $\epsilon \ge 1$ and all $a \neq b$ the following are equivalent:


$$1) d\_R^{(\epsilon)}(a,b) = \sqrt{1-\frac{2^{1+\epsilon}}{\text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^{1+\epsilon}}}>d\_L(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$$


$$2) \left < \psi^{(\epsilon)}\_R(a),\psi^{(\epsilon)}\_R(b) \right > < \left < \psi\_L(a),\psi\_L(b) \right >$$


3) The abc conjecture for $\epsilon \ge 1$ with $K\_{\epsilon} = \frac{1}{2^{\epsilon}}$


Related question [Two questions around the $abc$-conjecture](https://mathoverflow.net/questions/351847/two-questions-around-the-abc-conjecture)


Also the metrics $d\_R^{(\epsilon)},d\_L$ would be embedded in Euclidean space.


**Yet another edit**:


It seems that 
$$\frac{\phi(n)}{n} = \sum\_{d|n} \frac{\mu(d)}{\text{rad}(d)}$$


wher $\mu, \phi$ are the Moebius function and the Euler totient function. 


From this it would follow using Moebius inversion, that :


$$\frac{1}{\text{ rad}(n)} = \sum\_{d|n} \frac{\mu(d)\phi(d)}{d}$$


which *could* (I am not sure about that) be helpful for question 2).


**Edit with proof that $k(a,b)$ is a kernel**:
Let 
$$k(a,b) := \frac{1}{\frac{ab(a+b)}{\gcd(a,b)^3}} = \frac{\gcd(a,b)^3}{ab(a+b)} = \frac{\gcd(a,b)^2}{ab} \cdot \frac{\gcd(a,b)}{a+b} = k\_1(a,b) \cdot k\_2(a,b)$$


It is known that:


$$\int\_0^1 \psi(at)\psi(bt) dt = \frac{1}{12} \frac{(a,b)^2}{ab} = \frac{1}{12} k\_1(a,b).$$
Where $\psi(t) = t - \lfloor t \rfloor - \frac{1}{2}$ is the sawtooth function.
Hence $k\_1(a,b)$ is a kernel.


On the other hand, it is known for example by the answer of @DenisSerre, that $k\_2(a,b)$ is also a kernel.


Hence the product $k(a,b) = k\_1(a,b) \cdot k\_2(a,b)$ is also a kernel.


**Update**:
I found this paper online which is interesting (Set there: $X\_a = \{ a/k | 1 \le k \le a \}$ then: $|X\_a \cap X\_b| = |X\_{\gcd(a,b)}| = \gcd(a,b)$ ) and may be of use for the questions above:


<https://www.researchgate.net/publication/326212690_On_the_positive_semi-definite_property_of_similarity_matrices>


Setting in the paper above $A\_i = \{ i/k | 1 \le k \le i \}$ we see that $|A\_i \cap A\_j| = |A\_{\gcd(i,j)}| = \gcd(i,j)$ and $|A\_i|=i$.
Since in the paper it is proved that:


1) The Sorgenfrei similarity $\frac{|A\_i \cap A\_j|^2}{|A\_i||A\_j|}$ is a (positive definite $\ge0$, symmetric) kernel, we have another proof, that $\frac{\gcd(a,b)^2}{ab}$ is a kernel.


2) The Gleason similarity $\frac{2|A\_i \cap A\_j|}{|A\_i|+|A\_j|}$ is a (positive definite $\ge0$, symmetric) kernel, we have another proof, that $\frac{\gcd(a,b)}{a+b}$ is a kernel.


Using the product of these kernels, we get the new kernel $\frac{\gcd(a,b)^3}{ab(a+b)}$.



The abc-conjecture as an inequality for inner-products?
The matrix $L\_n$ is positive definite.


**Proof**. The matrix $G\_n$ with entries ${\rm gcd}(a,b)$ is positive definite because of $G=D^T\Phi D$ where $\Phi={\rm diag}(\phi(1),\ldots,\phi(n))$ ($\phi$ the Euler's totient function) and $d\_{ij}=1$ if $i|j$ and $0$ otherwise. Then the matrix $H\_n$ with entries $\frac1{a+b}$ is positive definite because
$$h\_{ij}=\int\_0^1 x^{i+j-1}dx$$
and the matrix with entries $x^{i+j-1}$ is positive semi-definite for $x>0$. Finally $L\_n=G\_n\circ H\_n$ (Hadamard product) is positive definite.


