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It is a little bit fun and crazy to think about what conjectures this function $\eta$ might also satisfy:

$$\forall a,b : \frac{a+b}{\gcd(a,b)} < \eta\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^2$$

which is obviously inspired by the abc-conjecture, and as a consequence, would maybe have an application for FLT:

$$z^n < \eta((xyz)^n)^2 = n^2 \eta(xyz)^2 = n^2( \eta(x) + \eta(y) + \eta(z))^2 $$ $$\le^{\text{ because } \eta(a)\le a} n^2 ( x+y+z)^2 \le n^2(3z)^2 = 9 n^2 z^2$$

and since $n>2$ we would get:

$$z^{n-2} \le 9n^2$$

and again since $n-2>0$ we would derive:

$$z < \exp\left(\frac{\log(5n^2)}{n-2}\right)$$

But the right hand side of this inequality is $<2$ for $n\ge 13$. So this would mean that

$$x^n+y^n = z^n, 1 \le x \le y \le z, \gcd(x,y)=1, n>2$$

would have no solution for $n\ge 13$. This is of course wild speculation, because the abc-inequality for $\eta$ has yet to be tested if it is true or not, but let me get a little bit crazier and get back to my favourite topic, p.s.d kernels defined over the natural numbers, and point to the symmetric function, which also seems to be a p.s.d kernel:

$$K(a,b):=\frac{a+b}{\gcd(a,b) \eta( \frac{ab(a+b)}{\gcd(a,b)})^2}$$

The abc-conjecture for $\eta$ might than be restated as:

$$\forall a,b \in \mathbb{N}: K(a,b) < 1$$

Q3) It would be nice to check this inequality for many numbers and report with or without code a counterexample or that it has succeded for these examples?

Edit: The p.s.d conjecture for $K$ is wrong for $N=110$ and the matrix $K_N := (K(a,b))_{ 1 \le a,b \le N}$ is not positive semidefinite.

Edit: Deleted the comment about abc-conjecture of $\eta$ because it is wrong for $a=4095,b=1$.

Edit: The p.s.d conjecture for $K$ is wrong for $N=110$ and the matrix $K_N := (K(a,b))_{ 1 \le a,b \le N}$ is not positive semidefinite.

Consider the completely additive function $\eta(n) := \sum_{p|n} v_p(n)p$ defined on natural numbers, with values in natural numbers. For literature, on this function, see the corresponding OEIS sequence. Consider the Gram matrix:

is a positive semi definite kernel on the natural numbers, because with $\phi(n) := \sum_{p|n} \sqrt{v_p(n)p} e_p$ we can embedd $n$ in the Hilbert space of series over $\mathbb{R}$ or $\mathbb{C}$ and we have $k(a,b) = \langle \phi(a),\phi(b) \rangle$

$$d_s(a,b) = \eta( \frac{ab}{\gcd(a,b)^2} ) = \eta(a)+\eta(b)-2\eta(\gcd(a,b)) = |\phi(a) - \phi(b)|_{H}^2 = d_H(a,b)^2$$

where $\sigma_p := \{ \lambda \in \sigma(G(n)) | \lambda \in \mathbb{N} , n/2 < \lambda \le n \}$ and

$\sigma_c := \{ \lambda \in \sigma(G(n)) | \lambda \notin \mathbb{N} , n/2 < \lambda \le n \}$ and $\sigma_0 := \{ \lambda \in \sigma(G(n)) | 0 \le \lambda \le n/2 \}$

$$A = \int_{\mathbb{R}} \lambda d E_A(\lambda)$$

1. $\gcd(a,b) = 1 \iff k(a,b) = 0$

2. $k(p^a,p^a) = ap$ for all primes $p$ and $a\ge 0$

3. $k(\frac{ac}{\gcd(a,c)^2},b) = k(\frac{a}{\gcd(a,c)},b) + k(\frac{c}{\gcd(a,c)},b)$

$$\forall a,b : \frac{a+b}{\gcd(a,b)} < \eta(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$

and again since $n-2>0$ we woud derive:

$$z < \exp(\frac{\log(5n^2)}{n-2})$$

Consider the completely additive function $\eta(n) := \sum_{p\mid n} v_p(n)p$ defined on natural numbers, with values in natural numbers. For literature, on this function, see the corresponding OEIS sequence. Consider the Gram matrix:

is a positive semi definite kernel on the natural numbers, because with $\phi(n) := \sum_{p\mid n} \sqrt{v_p(n)p} e_p$ we can embedd $n$ in the Hilbert space of series over $\mathbb{R}$ or $\mathbb{C}$ and we have $k(a,b) = \langle \phi(a),\phi(b) \rangle$

$$d_s(a,b) = \eta( \frac{ab}{\gcd(a,b)^2} ) = \eta(a)+\eta(b)-2\eta(\gcd(a,b)) = |\phi(a) - \phi(b)|_H^2 = d_H(a,b)^2$$

where $\sigma_p := \{ \lambda \in \sigma(G(n)) \mid \lambda \in \mathbb{N} , n/2 < \lambda \le n \}$ and

$\sigma_c := \{ \lambda \in \sigma(G(n)) \mid \lambda \notin \mathbb{N} , n/2 < \lambda \le n \}$ and $\sigma_0 := \{ \lambda \in \sigma(G(n)) \mid 0 \le \lambda \le n/2 \}$

$$A = \int_{\mathbb{R}} \lambda \, d E_A(\lambda)$$

1. $\gcd(a,b) = 1 \iff k(a,b) = 0$

2. $k(p^a,p^a) = ap$ for all primes $p$ and $a\ge 0$

3. $k\left(\frac{ac}{\gcd(a,c)^2},b\right) = k\left(\frac{a}{\gcd(a,c)},b\right) + k\left(\frac{c}{\gcd(a,c)},b\right)$

$$\forall a,b : \frac{a+b}{\gcd(a,b)} < \eta\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^2$$

and again since $n-2>0$ we would derive:

$$z < \exp\left(\frac{\log(5n^2)}{n-2}\right)$$