# Large radical of an integer and three AB conjectures

In this Note, We propose a new definition called "large radical of an integer". Using this definition, three very useful $$AB$$ conjecture are given.

1. Large counter examples of the ABC conjecture

$$A=a_1^{x_1}a_2^{x_2}...a_n^{x_n}$$, $$B=b_1^{y_1}b_2^{y_2}...b_m^{y_m}$$, $$C=c_1^{z_1}c_2^{z_2}...c_k^{z_k}$$.

The ABC conjecture states that:

ABC Conjecture. For every positive real number $$\varepsilon$$, there exist only finitely many triples $$(A, B)$$ of coprime positive integers, such that: $$A + B > \operatorname{rad}(abc)^{1+\varepsilon}$$.

If the $$ABC$$ conjecture is true, it should true with $$\varepsilon \approx 0$$.

In this table, there are $$14482065$$ case $$A+B > rad(AB(A+B))$$ in ranges $$1 \le A < B <10^{18}$$.

Maybe we can prove that in the most counter example of $$A+B < rad(AB(A+B)$$ when has at least one condition $$A \ge rad(A)^3$$ or $$B \ge rad(B)^3$$ or $$A+B \ge rad(A+B)^3$$ there are maximum $$\approx 508544975$$ numbers such that the condition $$X \ge rad(X)^3$$ (update because computation checked again, can you help to improving). So in range $$1 \le A < B <10^{18}$$ there are $$14482065$$ counter examples, this is LARGE COUNTER EXAMPLES.

II. The first AB conjecture

Given a positive integer $$P>1$$, let its prime factorization be written $$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_k^{a_k}$$

Define the functions $$h(P)$$, $$d(P)$$ and $$rad'(P)$$ by $$h(1)=1$$ and $$h(P)=min(a_1, a_2,..,a_k)$$

Let $$g=\gcd(a_1, a_2,..., a_k)$$

$$d(P)=\frac{h(P)}{g}=min(\frac{a_1}{g},\frac{a_2}{g},...,\frac{a_k}{g})$$ $$rad'(P)=(p_1p_2...p_k)^{d(P)}$$

Some examples:

1) Let $$P=2^5.5^7.11^8$$ then $$rad(P)=2.5.11$$ and $$rad'(P)=2^5.5^5.11^5$$

2) $$P=17^8$$ then $$rad(P)=17$$ and $$rad'(P)=17$$

There are some simple properties of $$rad'(P)$$

1) $$rad'(P)=rad(P)^{d(P)}$$

2) $$rad(P) \le rad'(P) \le P$$

3) $$rad'(P^n)=rad'(P) \le P$$

4) In general case $$rad'(AB) \ne rad'(A)rad'(B)$$

Conjecture1: For every positive real number $$\varepsilon >0$$, the inequality $$A+B > (rad'(A).rad'(B).rad'(A+B))^{1+\varepsilon}$$ has only finitely relatively prime integers $$A$$ and $$B$$.

Remarks: By the definition, the first AB conjecture is weaker than the ABC conjecture. The proof of first AB is simpler than the proof of ABC conjecture. The true possibility of the first AB conjecture is higher than the ABC conjecture because $$rad'(A) \ge rad(A)$$. But the first AB conjecture is as useful as the ABC conjecture. Because useful ABC conjecture based on two peroperties $$rad(A^n)=rad(A)$$ and $$rad(A) \le A$$. The first AB conjecture also have two properties $$rad'(A^n)=rad'(A)$$ and $$rad'(A) \le A$$.

Let $$\varepsilon=0$$ here are some examples $$rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$$

Other two conjecture in here

COMPUTER CHECKED

My question: Could You help me full fill the table 2 above? OR Hopefully, the conjectures would be interested by a professor in the field and be researched further.

• With $1 \le A < B \le X=10^5, \varepsilon=0$, my laptop spent 178 hour to check the first AB conjecture. I don't know how long to check $X$ up to $10^6$, $10^7$, $10^8$, $10^9$....$10^{18}$ with my laptop. I think need a year to check $1 \le A < B \le X=10^6$. So I need everyone help. – Đào Thanh Oai Sep 5 '19 at 3:10
• I don't think it is the role of Mathoverflow to help you doing computer calculations. – abx Sep 6 '19 at 10:10
• @abx Hopefully, the conjectures would be interested by a professor in the field and be researched further. I think this conjecture is very deep and useful. So first of all, there are one people check the conjecture before they research them. There are many answer in mathoverflow were given by calculations. – Đào Thanh Oai Sep 6 '19 at 10:19
• I am sorry, because I edited many times. I stop. – Đào Thanh Oai Sep 7 '19 at 10:58

To answer your query, I think that you would find helpful the ABC@Home project and its output file.

I also have found a counterexample to the second (third?) AB conjecture: consider the triple $$A+B=C$$ with

\begin{aligned} A&=5^{12}*17^2*31^2*1699, \\ B&=23^{14}*29, \\ C&=2^{19}*3^2*11*13^{10}*{47}. \end{aligned}

The sum of the cubes of the (large) radicals is equal to:

$$(5×17×31×1699)^3 + (23×29)^3 + (2×3×11×13×47)^3 =8.972\ldots × 10^{19},$$

which is less than

$$A+B=3.363\ldots × 10^{20}.$$

To find this interesting triple, I used eye-skim on the list of abc triples by quality (here q=1.4578). The triple was discovered by Abderrahmane Nitaj.

• Thank You very much, I will use the database on your link to research more. – Đào Thanh Oai Sep 8 '19 at 6:49