# A reinterpretation of the $abc$ - conjecture in terms of metric spaces?

I hope it is appropriate to ask this question here:

One formulation of the abc-conjecture is

$$c < \text{rad}(abc)^2$$

where $$\gcd(a,b)=1$$ and $$c=a+b$$. This is equivalent to ($$a,b$$ being arbitrary natural numbers):

$$\frac{a+b}{\gcd(a,b)} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$

Let $$d_1(a,b) = 1- \frac{\gcd(a,b)^2}{ab}$$ which is a proven metric on natural numbers. Let $$d_2(a,b) = 1- 2 \frac{\gcd(a,b)}{a+b}$$, which I suspect to be a metric on natural numbers, but I have not proved it yet. Let $$d(a,b) = d_1(a,b)+d_2(a,b)-d_1(a,b)d_2(a,b) = 1-2\frac{\gcd(a,b)^3}{ab(a+b)}$$

Then we get the equivalent formulation of the inequality above:

$$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$

which is equivalent to :

$$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{1}{1-d_1(a,b)}\cdot\frac{2}{1-d_2(a,b)} )^2$$

My question is if one can prove that $$d_2$$ and $$d$$ are distances on the natural numbers (without zero)?

Result: By the answer of @GregMartin, $$d_2$$ is a metric. By the other answer $$d$$ is also a metric.

Edit: By "symmetry" in $$d_1$$ and $$d_2$$, this interpretation also suggests that the following inequality is true , which might be trivial to prove or very difficult or might be wrong and may be of use or not in number theory:

$$\frac{1}{1-d_1(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$

which is equivalent to

$$\frac{ab}{\gcd(a,b)^2} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$

(This is not easy to prove, as the $$abc$$ conjecture $$c=a+b < ab < \text{rad}(abc)^2$$ would follow for all $$a,b$$ such that $$a+b < ab$$.)

Second edit: Maybe the proof that $$d_2,d$$ are distances can be done with some sort of metric transformation, for example maybe with a Schoenberg transform (See 3.1, page 8 in https://arxiv.org/pdf/1004.0089.pdf) The idea, that this might be proved with a Schoenberg transform comes from the fact that:

$$d_1(a,b) = 1-\exp(-\hat{d}(a,b))$$ so $$d_1$$ is a Schoenberg transform of $$\hat{d}(a,b) = \log( \frac{ab}{\gcd(a,b)^2}) = \log( \frac{\text{lcm}(a,b)}{\gcd(a,b)})$$ which is proved to be a metric (see Encyclopedia of Distances, page 198, 10.3 )

Third edit: Here is some Sage Code to test the triangle inequality for triples (a,b,c) up to 100:

def d1(a,b):
return 1-gcd(a,b)**2/(a*b)

def d2(a,b):
return 1-2*gcd(a,b)/(a+b)

def d(a,b):
return d1(a,b)+d2(a,b)-d1(a,b)*d2(a,b)

X = range(1,101)
for a in X:
for b in X:
for c in X:
if d2(a,c) > d2(a,b)+d2(b,c):
print "d2",a,b,c
if d(a,c) > d(a,b)+d(b,c):
print "d",a,b,c


so far with no counterexample.

• Your speculated "by symmetry" inequality is not valid: take $a=2^j$ and $b=3^k$. – Greg Martin Sep 30 at 1:23
• You may write something like $\ \mbox{rad}^2(x)\$ rather than $\ \mbox{rad}(x)^2$ – Wlod AA Sep 30 at 3:53
• @WlodAA The question clearly says one formulation . – orgesleka Sep 30 at 4:10
• Thus, is it or is it not? Yes or No? – Wlod AA Sep 30 at 4:17
• @GregMartin: Can there be done anything to include $a=2^j$ and $b=3^k$ in terms of the constant $K_{\epsilon}$ or $\epsilon$ in $1/(1-d_1)< K_{\epsilon} \text{rad}(2/(1-d))^{1+\epsilon}$? – orgesleka Sep 30 at 14:18

$$d_2$$ is indeed a metric. Abbreviating $$\gcd(m,n)$$ to $$(m,n)$$, we need to show that \begin{align*} 1-\frac{2(a,c)}{a+c} &\le 1-\frac{2(a,b)}{a+b} + 1-\frac{2(b,c)}{b+c} \end{align*} or equivalently \begin{align*} \frac{2(a,b)}{a+b} + \frac{2(b,c)}{b+c} &\le 1 + \frac{2(a,c)}{a+c}. \end{align*} Furthermore, we may assume that $$\gcd(a,b,c)=1$$, since we can divide everything in sight by that factor.

Note that if $$a=(a,b)\alpha$$ and $$b=(a,b)\beta$$ with $$(\alpha,\beta)=1$$, then $$\frac{2(a,b)}{a+b} = \frac2{\alpha+\beta}$$. The only unordered pairs $$\{\alpha,\beta\}$$ for which this is at least $$\frac12$$ are $$\{1,1\}$$, $$\{1,2\}$$, and $$\{1,3\}$$. Further, if neither $$\frac{2(a,b)}{a+b}$$ nor $$\frac{2(b,c)}{b+c}$$ is at least $$\frac12$$, then the inequality is automatically valid because of the $$1$$ on the right-hand side.

This leaves only a few cases to check. The case $$\{\alpha,\beta\} = \{1,1\}$$ (that is, $$a=b$$) is trivial. The case $$\{\alpha,\beta\} = \{1,2\}$$ (that is, $$b=2a$$) can be checked: we have $$(a,c)=\gcd(a,2a,c)=1$$, and so the inequality in question is \begin{align*} \frac23 + \frac{2(2,c)}{2a+c} &\le 1 + \frac2{a+c}, \end{align*} or equivalently $$\frac{(2,c)}{2a+c} \le \frac16 + \frac1{a+c};$$ there are only finitely many ordered pairs $$(a,c)$$ for which the left-hand side exceeds $$\frac16$$, and they can be checked by hand.

The proof for the case $$\{\alpha,\beta\} = \{1,3\}$$ (that is, $$b=3a$$) can be checked in the same way, as can the cases $$a=2b$$ and $$a=3b$$.

Not an answer but an observation.

Set $$r_2(a,b,c)=d_2(a,c)/(d_2(a,b)+d_2(b,c))$$ (when defined), and similarly for $$r(a,b,c)$$. Then Greg Martin's proof shows that the values of $$r_2$$ should be discrete, and indeed experimentally the values are in decreasing order

$$(1,9/10,6/7,5/6,9/11,...)$$

The same experiment done for $$d$$ gives

$$(1,27/40,40/63,28/45,...)$$

Thus, apart from trivial cases such as $$a=b$$ one should have the stronger triangle inequality $$d(a,c)\le0.675(d(a,b)+d(b,c))$$.

• Thank you for this observation – orgesleka Sep 30 at 10:25

$$d$$ is also a metric. Proof:

First let us call a metric on natural numbers $$d$$ such that $$d(a,b)<1$$ and $$d(a,b)$$ is a rational number for all $$a,b$$ a "rational metric". Second let $$d_1,d_2$$ be two rational metrics such that if we set $$d=d_1+d_2-d_1 d_2$$ then for all $$a \neq c, a \neq b$$ we have $$d(a,b)+d(a,c)>1$$. If this is the case for $$d_1,d_2$$ we will call $$d_1$$ and $$d_2$$ "paired". If $$d_1,d_2$$ are such paired rational metrics, then $$d=d_1+d_2-d_1 d_2$$ is a metric. Proof:

1) $$d(a,b) = 0$$ iff $$0 \le d_1(a,b)(1-d_2(a,b)) = -d_2(a,b) \le 0$$ hence since $$1-d_2(a,b)>0$$ we must have $$d_1(a,b) = 0$$ hence $$a=b$$. If on the other hand $$a=b$$ then plugging this in $$d$$ and observing that $$d_1(a,b)=d_2(a,b)=0$$ gives us $$d(a,b)=0$$.

2) $$d(a,b) = d(b,a)$$ since $$d_i(a,b) = d_i(b,a)$$ for $$i = 1,2$$.

3) Triangle inequality: If $$a=c$$ or $$a=b$$ the triangle inequality is fullfilled and becomes an equality because of 1) : $$d(b,c) \le d(a,b)+d(a,c)$$ First observe that $$d(x,y) < 1$$ for all $$x,y$$. Let therefore $$a\neq c, a\neq b$$. Since $$d_1,d_2$$ are paired rational metrics we have: $$d(b,c) < 1 < d(a,c)+d(a,b)$$ and the triangle inequality is proved.

This proves also that $$d$$ is a rational metric (if $$d_1,d_2$$ are paired rational metrics.)

What remains to show is that $$d_2(a,b) = 1-\frac{2 \gcd(a,b)}{a+b}$$, $$d_1(a,b) = 1-\frac{\gcd(a,b)^2}{ab}$$ are paired (rational) metrics, hence it remains to show that $$d(a,b) = 1- \frac{2 \gcd(a,b)^3}{ab(a+b)}$$ satisfies:

$$d(a,c)+d(a,b)>1, \text{ whenever } a\neq c, a \neq b$$

The last inequality is equivalent to, after some algebra:

$$\frac{abc(a+b)(a+c)}{2} - \gcd(a,b)^3c(a+c) - \gcd(a,c)^3b(a+b)>0$$

Let $$U=\gcd(a,b,c)$$. Then there exist natural numbers $$R,S,T,A,B,C$$ such that:

$$RU = \gcd(a,b), SU = \gcd(a,c), TU = \gcd(b,c), a = RSUA, b = RTUB, c = STUC$$

Plugging this in the last inequality and after some algebra, we find:

$$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0$$

We can pair each of the positive summand with a negative summand to give for example:

$$(A^3*B*C*R^2*S^2*T-2*A*C*R^2)=(A^2*B*S^2*T - 2)*A*C*R^2$$

The condition $$a \neq b$$ translates to $$SA \neq TB$$ and similarily $$a \neq c$$ translates to $$RA \neq TC$$. Suppose that $$A^2*B*S^2*T - 2 \le 0$$. The case $$A^2*B*S^2*T=1$$ contradicts $$SA \neq TB$$. Hence we can only have at most $$A^2*B*S^2*T=2$$ which leads to $$A=S=1$$, $$BT=2$$ and plugging this in the definition of $$a,b$$ we get $$b=2a$$ and $$d(a,b)=\frac{2}{3}$$.

Now we must show that the other pairings give the desired result:

$$( A^2*B^2*C*R^2*S*T^2-2*B**2*S*T)=(A^2*C*R^2*T - 2)*B^2*S*T$$ A similar argument to the above leads to: If $$A^2*C*R^2*T = 2$$, then $$A=R=1$$, $$CT=2$$ which leads to (with $$S=A=1$$) $$a=RSUA=U,b=RTUB=2U=2a,c=STUC=2U=2a$$ and it follows that $$d(a,c)=\frac{2}{3}$$, so $$d(a,b)+d(a,c)=\frac{4}{3}>1$$, and this case is done.

If $$A^2*C*R^2*T > 2$$ and $$A^2*B*S^2*T=2$$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0$$ is true.

If $$A^2*C*R^2*T > 2$$ and $$A^2*B*S^2*T>2$$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0$$ is true. This shows, that $$d_1,d_2$$ are paired metrics and completes the proof.

This question has already very good answers. I justed wanted to highlight that it is possible to shorten the proofs, using the following:

If $$X_a = \{ a/k | 1 \le k \le a \}$$ then $$X_a \cap X_b = \gcd(a,b)$$, which is straightforward to prove. Then $$d_1(a,b) = 1-\gcd(a,b)^2/(ab) = 1-|X_a \cap X_b|^2 / (|X_a||X_b|)$$ is the squared cosine metric (see Encyclopedia of Distances) and $$d_2(a,b) = 1-2\gcd(a,b)/(a+b) = 1-2|X_a \cap X_b| / (|X_a|+|X_b|)$$ is the Sorensen Metric (Encyclopedia of Distances). Hence $$d_1,d_2$$ are metrics of the form $$d_i = 1- s_i$$ where $$s_i$$ is a similarity. But then $$s=s_1 \cdot s_2$$ is also a similarity and $$d=d_1 +d_2 -d_1 d_2 = 1-s=1-s_1 s_2$$ is a metric.