ABC - Conjecture

"In this age in which mathematicians are supposed
to bring their research into the classroom, even at
the most elementary level, it is rare that we can turn
the tables and use our elementary teaching to help
in our research. However, in giving a proof of Fermat’s
Last Theorem, it turns out that we can use
tools from calculus and linear algebra only."

In this note, I give the proof that the abc conjecture is false because, in the case c > rad(abc), for 0 <\epsilon < 1 we can not find the constant K(\epsilon) so that c < K(\epsilon).rad^{ 1+\epsion}(abc) for c very large. A counter-example is given.

We show that the size of sets A having the property that with some
non-zero integer n, a1a2 + n is a perfect power for any distinct a1; a2 2 A, cannot be
bounded by an absolute constant. We give a much more precise statement as well,
showing that such a set A can be relatively large. We further prove that under the abc-
conjecture a bound for the size of A depending on n can already be given. Extending a
result of Bugeaud and Dujella, we also derive an explicit upper bound for the size of A
when the shifted products a1a2 + n are k-th powers with some xed k  2. The latter
result plays an important role in some of our proofs, too.

In this paper, we consider the $abc$ conjecture in the case $c=a+1$. Firstly, we give the proof of the first conjecture that $c<rad*2(ac)$. It is the key of the proof of the $abc$ conjecture. Secondly, the proof of the $abc$ conjecture is given for $\epsilon \geq 1$, then for $\epsilon \in ]0,1[$ for the two cases: $ c\leq rad(ac)$ and $c> rad(ac)$. We choose the constant $K(\epsilon)$ as $K(\epsilon)=e^{ \left(\frac{1}{\epsilon*2} \right)}$. A numerical example is presented.

In this paper, we consider the abc conjecture. Assuming that c<rad2(abc) is true, we give the proof of the abc conjecture for \epsilon is positive,  then for the case \epsilon \in ]0,1[, we consider that the abc conjecture is false, from the proof, we arrive in a contradiction.

In this paper, we consider the abc conjecture. Assuming that c<rad*2(abc) is true, we give the proof of the abc conjecture for \epsilon≥ 1, then for the case \epsilon ∈]0, 1[, we consider that the abc conjecture is false, from the proof, we arrive in a contradiction.

In this paper, we assume that Beal conjecture is true, we give a complete proof of the ABC conjecture. We consider that Beal conjecture is false $\Longrightarrow$ we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true $\Longrightarrow$ Beal conjecture is true. But, if the Beal conjecture is true, then we deduce that the ABC conjecture is true

In this book, I present my collection of 23 papers written, with different approaches to try to resolve the $abc$ conjecture and others conjectures related to it like $c<rad^2(abc)$.

This monograph can give an idea about the advancement of the comprehension of the conjectures related to the problem cited above.

In this paper, using the recent result that c < rad(abc)^2 , we will give the proof of the abc conjecture for \epsilon ≥ 1, then for all \epsilon ∈]0, 1[. We choose the constant K(\epsilon) as K(\epsilon) = e^(1/\epsilon^2). Some numerical examples are presented.

In this paper, from a, b, c positive integers relatively prime with c = a + b, we consider a bounded of c depending of a, b, then we do a choice of K(\epsilon) and finally we obtain that the ABC conjecture is true. Four numerical examples confirm our proof.

In this paper, we consider the abc conjecture. As the conjecture c&lt;rad^2(abc) is less open, we give firstly the proof of a modified conjecture that is c&lt;2rad^2(abc). The factor 2 is important for the proof of the new conjecture that represents the key of the proof of the main conjecture. Secondly, the proof of the abc conjecture is given for \epsilon \geq 1, then for \epsilon \in ]0,1[. We choose the constant K(\epsion) as K(\epsilon)=2e^{\frac{1}{\epsilon^2} } for $\epsilon \geq 1 and K(\epsilon)=e^{\frac{1}{\epsilon^2}} for \epsilon \in ]0,1[. Some numerical examples are presented.

In this paper, we consider the abc conjecture. In the first part, we give the proof of the conjecture c<rad^{1.63}(abc) that constitutes the key to resolve the abc conjecture. The proof of the abc conjecture is given in the second part of the paper, supposing that the abc conjecture is false, we arrive in a contradiction.

In this paper about the $abc$ conjecture, we propose a new conjecture about an upper bound for $c$ as $c<R.exp(\frac{3\sqrt[3]{2}}{2}Log^{2/3}R)$. Assuming the last condition holds, we give the proof of the $abc$ conjecture by proposing the expression of the constant $K(\ep)$, then we approve that $\forall \epsilon>0$, for $a,b,c$ positive integers relatively prime with $c=a+b$, we have $c< K(\epsilon).rad^{1+\epsilon}(abc)$. Some numerical examples are given.

In this paper, we consider the abc conjecture in the case c = a + 1. Firstly, we give the proof of the first conjecture that c < rad 2 (ac) using polynomial functions. It is the key to the proof of the abc conjecture. Secondly, the proof of the abc conjecture is given for\epsilon ≥ 1, then for \epsilon ∈]0, 1[ for the two cases: c ≤ rad(ac) and c > rad(ac). We choose the constant K(\epsilon) as K(\epsilon) = e^{ 1/\epsilon* 2}. A numerical example is presented.

In this paper, we consider the abc conjecture. Firstly, we give a proof of a the first conjecture that c<rad*2(abc). It is the key of the proof of the abc conjecture. Secondly, a proof of the abc is given for \epsilon  \geq 1, then for \ep \in ]0,1[ for the two cases:  c\leq rad(abc) and c> rad(abc). We choose the constant K(\ep) as K(\ep)= 6*{1+\ep}e*{\ds \left(\frac{1}{\ep*2}-\ep \right)}. Five numerical examples are presented.