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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

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 $\frac{\gcd(a,b)^3}{ab(a+b)}$.

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    $\begingroup$ What is the maximum value of $n$ that you tried with your experiments? $\endgroup$ – Francesco Polizzi Feb 6 at 10:32
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    $\begingroup$ @FrancescoPolizzi I have tried up to $n=50$. Have you found a counterexample? $\endgroup$ – orgesleka Feb 6 at 10:38
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    $\begingroup$ no, I just asked out of curiosity $\endgroup$ – Francesco Polizzi Feb 6 at 13:02
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    $\begingroup$ $L_n$ is clearly positive definite as it is the Schur product of two PSD matrices (or are you looking for "strict" positivity? that should also be doable). And moreover, it seems $L_n$ is also infinitely divisible because both the gcd matrix and the Cauchy matrix ($1/(a+b)$) are infinitely divisible, thus leading to a Hilbert space embedding. As for $R_n$, I'm not sure what 'rad' does so cannot have a simple answer... $\endgroup$ – Suvrit Feb 6 at 18:48
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    $\begingroup$ Please restrict to one question per post. BTW the relation $\phi(n)/n=\sum_{d\mid n}\mu(d)/\mathrm{rad}(d)$ is straightforward to verify: both sides are multiplicative functions of $n$, and they agree at prime powers. In the same way, it is straightfoward to verify directly that $1/\mathrm{rad}(n)=\sum_{d\mid n}\mu(d)\phi(d)/d$. $\endgroup$ – GH from MO Feb 8 at 9:09
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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.

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  • $\begingroup$ thanks @DenisSerre for your answer $\endgroup$ – orgesleka Feb 6 at 19:13
  • $\begingroup$ My comment already provides a proof of $L_n$ being not only positive definite, but infinitely divisible :-) $\endgroup$ – Suvrit Feb 7 at 2:52
  • $\begingroup$ @DenisSerre Is it possible to give an explicit matrix $A^TA=L_n$? $\endgroup$ – orgesleka Feb 7 at 7:05
  • $\begingroup$ @orgesleka. In principle it is possible to calculate an upper triangular $A$ such that $A^TA=L_n$. This is called the Choleski factorization. It costs $O(n^3)$ operations. $\endgroup$ – Denis Serre Feb 7 at 13:08
  • $\begingroup$ @DenisSerre: I know of the Cholesky factorization. I was hoping for something more in the lines of $G = D^T \Phi D$ which you gave in your answer. $\endgroup$ – orgesleka Feb 7 at 13:18

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