One formulation of the abc-conjecture is:
$$\forall a,b \in \mathbb{N}: \frac{a+b}{\gcd(a,b)}< \operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2 $$
Let us define:
$$K(a,b) := \frac{2(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} $$
I have checked in Sagemath up to $n=100$ that the matrix
$$(K(i,j))_{1 \le i,j \le n} \text{ is positive definite }$$
and up to $1\le a,b \le 100$ that :
$$K(a,b)\le 1$$
In the Encyclopedia of Distances the word "similarity" is described as:
A similarity $s:X\times X \rightarrow \mathbb{R}$ is defined in the Encyclopedia of Distances as:
$s(x,y) \ge 0 \forall x,y \in X$
$s(x,y) = s(y,x) \forall x,y \in X$
$s(x,y) \le s(x,x) \forall x,x \in X$
$s(x,y) = s(x,x) \iff x=y$
So one possible formulation of the empirical observation above might be:
$$K \text{ is a positive definite kernel and a similarity function with } K(a,a)=1$$
From this formulation the abc conjecture in the variant above follows, because:
$$K(a,b) \le K(a,a)=1 \rightarrow \frac{(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \le \frac{1}{2} < 1$$
Since in the definition of the similarity function, some of the properties are easy to prove for $K$ such as 1. and 2. and from 3. follows abc, I am asking if it is possible to prove $\require{enclose}\enclose{horizontalstrike}{\text{ property 4. and / or}}$ that $K$ is positive definite function?
Thanks for your help.
Edit: I think that $4.$ can be proved by considering that if $K(x,y)=1$ and assuming there exists a prime $p$ which divides $x$ but not $y$, leads to a contradiction.
Second edit: The abc conjecture (in the version above) would follow, if one could prove that the function
$$K(a,b) := \frac{2(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} $$
is positive definite over the natural numbers.
Proof: If $K$ is a positive definite function over the natural numbers, then by the Moore-Aronszajn theroem , there exists a Hilbert space $H$ and a feature mapping $\phi: \mathbb{N} \rightarrow H$ such that:
$$K(a,b) = \left < \phi(a),\phi(b) \right >_{H}$$
But by Cauchy-Schwarz inequality we get:
$$K(a,b) = |K(a,b)| = |\left < \phi(a),\phi(b) \right >_{H}|$$ $$\le |\phi(a)||\phi(b)| = \sqrt{K(a,a)K(b,b)} = \sqrt{1 \cdot 1} = 1$$
But then we get by division of $2$ and application of the last inequality:
$$\frac{(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \le \frac{1}{2} < 1$$
hence it would follow that: $$\forall a,b \in \mathbb{N}: \frac{a+b}{\gcd(a,b)}< \operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2 $$
Modified question: Is there a way to speed up the computations to try and test if the Gram matrix for $n=1,\cdots,10000$ is positive definite?
Here is my code so far, but it is rather slow:
# inductive cholesky decomposition
def rad(n):
return prod(prime_divisors(n))
def kk(a,b):
return (2(a+b)a1*b1/gcd(a,b))1/rad(ab(a+b)/gcd(a,b)3)*2
def notPositiveDefinite(a,b):
return rad(a*b/gcd(a,b))
#kk = notPositiveDefinite
def grammat(kk,n):
return matrix([[kk(a,b) for a in range(1,n+1)] for b in range(1,n+1)],ring=QQ)
PHI = dict([])
def phi(n,useN=False):
if n in PHI.keys():
return PHI[n]
if n==1:
if not useN:
x = vector([sqrt(kk(1,1))])
else:
x = vector([sqrt(kk(1,1)+0.0)])
PHI[n] = x
return x
else:
w = CC(n-1)**(-1)*vv(n-1)
lw = list(w)
if not useN:
lw.append(sqrt(kk(n+1,n+1)-w.norm()**2))
else:
lw.append(sqrt(kk(n+1,n+1)-w.norm().n()**2).n())
x = vector(lw)
PHI[n]= x
return x
def vv(n):
return vector([kk(i,n+1) for i in range(1,n+1)])
CCDict = dict([])
def CC(n):
if n in CCDict.keys():
return matrix(CCDict[n])
mm = []
for i in range(1,n+1):
vi = list(phi(i))
if len(vi) < n:
vi.extend((n-i)*[0])
mm.append(vi)
M= matrix(mm)
CCDict[n] = mm
return M
for n in range(1,200):
print(n,all(x in RR for x in phi(n,useN=True)))
#print(grammat(kk,n).rational_form())
Comment on code: It is based on inductive Cholesky decomposition as describe below and keeps track in a dictionary if the $\phi(n)$ or $C_n$ have already been computed. It uses float approximation for greater speed and in the last step we check if each component of the vector $\phi(n)$ is a real number. If it is for all $1 \le k \le n$ then we conclude that the Gram matrix of $n$ is positive definite. If it is not, then at least the Gram matrix for $n$ is not positive definite:
inductive Cholesky decomposition to find an embedding $\phi(n)$ for each number $n$:
Equation 1): $$\phi(n+1) = \left(\begin{array}{r} {C_{n}^{-1} v_n} \\ \sqrt{k(n+1,n+1)-{\left| {C_{n}^{-1} v_n} \right|}^{2} } \end{array}\right)$$
Equation 2):
$$v_n = \left(\begin{array}{r} k(1,n+1) \\ k(2,n+1) \\ \cdots\\ k(n,n+1)\\ \end{array}\right)$$
Equation 3): $$C_n = \left(\begin{array}{r} \phi(1)^T \\ \phi(2)^T \\ \cdots\\ \phi(n)^T\\ \end{array}\right)$$
Equation 4): $$\phi(1) := \sqrt{k(1,1)} e_1$$
If we assume that the function $k$ is positive definite, then this inductive method, should give us the vectors $\phi(a),\phi(b)$ such that under the normal inner product we have:
$$k(a,b) = \left < \phi(a), \phi(b) \right > $$
