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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

By Fundamental theorem of arithmetic

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

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 mustshould true with $\varepsilon=0$$\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 > rad(A)^2$$A \ge rad(A)^3$ or $B> rad(B)^2$,$B \ge rad(B)^3$ or $A+B \ge rad(A+B)^3$ there are maximum $\approx 508544975$ numbers such that the condition $X \ge rad(X)^3$ (pleaseupdate 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

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 $A > rad(A)^2$ or $B> rad(B)^2$, there are maximum $\approx 508544975$ numbers (please help). 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

deleted 174 characters in body
Source Link

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

In this table, there are $14482065$ case $A+B > rad(AB(A+B))$ in ranges $1 \le A < B <10^8$$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 exponents of $A$ or $B$ or $(A+B)$ greater than $2$ (the exponents of $A$ or $B$, $A+B$ greater than $2$ refer at least one $x_i$ orcondition $y_j$$A > rad(A)^2$ or $z_k$ $>2$). But in the range $1 \le A < B <10^8$$B> rad(B)^2$, there are maximum $\approx 508544975$ numbers have an exponents greater than $2$(please help). So in range $1 \le A < B <10^8$$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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

In this table, there are $14482065$ case $A+B > rad(AB(A+B))$ in ranges $1 \le A < B <10^8$.

Maybe we can prove that in the most counter example of $A+B < rad(AB(A+B)$ has at least one exponents of $A$ or $B$ or $(A+B)$ greater than $2$ (the exponents of $A$ or $B$, $A+B$ greater than $2$ refer at least one $x_i$ or $y_j$ or $z_k$ $>2$). But in the range $1 \le A < B <10^8$, there are maximum $\approx 508544975$ numbers have an exponents greater than $2$. So in range $1 \le A < B <10^8$ 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

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 $A > rad(A)^2$ or $B> rad(B)^2$, there are maximum $\approx 508544975$ numbers (please help). 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

added 2 characters in body
Source Link

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

In this table, there are $14482065$ case $A+B > rad(AB(A+B))$ in ranges $1 \le A < B <10^8$.

Maybe we can prove that in the most counter example of $A+B < rad(AB(A+B)$ has at least one exponents of $A$ or $B$ or $(A+B)$ greater than $2$ (the exponents of $A$ or $B$, $A+B$ greater than $2$ refer at least one $x_i$ or $y_j$ or $z_k$ $>2$). But in the range $1 \le A < B <10^8$, there are maximum $\approx 1037611$$\approx 508544975$ numbers have an exponents greater than $2$. So in range $1 \le A < B <10^8$ 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

In this table, there are $14482065$ case $A+B > rad(AB(A+B))$ in ranges $1 \le A < B <10^8$.

Maybe we can prove that in the most counter example of $A+B < rad(AB(A+B)$ has at least one exponents of $A$ or $B$ or $(A+B)$ greater than $2$ (the exponents of $A$ or $B$, $A+B$ greater than $2$ refer at least one $x_i$ or $y_j$ or $z_k$ $>2$). But in the range $1 \le A < B <10^8$, there are maximum $\approx 1037611$ numbers have an exponents greater than $2$. So in range $1 \le A < B <10^8$ 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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

By Fundamental theorem of arithmetic

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

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 must true with $\varepsilon=0$.

In this table, there are $14482065$ case $A+B > rad(AB(A+B))$ in ranges $1 \le A < B <10^8$.

Maybe we can prove that in the most counter example of $A+B < rad(AB(A+B)$ has at least one exponents of $A$ or $B$ or $(A+B)$ greater than $2$ (the exponents of $A$ or $B$, $A+B$ greater than $2$ refer at least one $x_i$ or $y_j$ or $z_k$ $>2$). But in the range $1 \le A < B <10^8$, there are maximum $\approx 508544975$ numbers have an exponents greater than $2$. So in range $1 \le A < B <10^8$ 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 $rad(AB(A+B)) < A+B< rad'(A).rad'(B).rad'(A+B)$

enter image description here

Other two conjecture in here

COMPUTER CHECKED

enter image description here

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.

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