(texts below from Wikipedia English )
In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions.[1] In order to directly prove a conditional statement of the form "If p, then q", it suffices to consider the situations in which the statement p is true. Logical deduction is employed to reason from assumptions to conclusion.
The type of logic employed is almost invariably first-order logic, employing the quantifiers for all and there exists. Common proof rules used are modus ponens and universal instantiation.[2]
In contrast, an indirect proof may begin with certain hypothetical scenarios and then proceed to eliminate the uncertainties in each of these scenarios until an inescapable conclusion is forced.
Proof methods that are not direct include proof by contradiction, including proof by infinite descent.
Direct proof methods include proof by exhaustion and proof by induction.
Examples:
1/ Direct proof which tackles a specific problem ( FLT conjecture.) and proposes a solution dedicated to the problem.
The proof is comprehensible by one person and understandable by a normal student in mathematics.
- Dirichlet (in 1825) solved the equation x^5+y^5=z^5 for n=5 only.
- Sophie Germain (in 1823) generalized the result of Dirichlet for prime p if 2p+1 is prime..
2/ Indirect proof which tackles a general problem and the solution of a specific problem becomes a consequence .
Sometimes the proof is hard to understand by one person, and not comprehensible by a normal student in mathematics.
- Sophie Germain (in 1823) proposed a solution of FLT for prime p if 2p+1 is prime..
- Dirichlet's (1825) solution becomes a consequence of Sophie Germain's solution in spite of the date of his previous work (1825) .
Andrew Wiles solved a high level problem in modular forms about elliptic curves and the consequence is a solution for FLT.
I acknowledge Wiles to be the best in his subject, his domain. I accept his advanced work and the consequence about FLT. I think he opens a space for mathematicians to search proofs for FLT comprehensible by a normal student in mathematics and may be to find new concepts or ideas. It should imply a direct proof of FLT.
Here we present a Direct proof of FLT x^n +y^n = z^n based on Induction on z not on n.
Your comment are welcome.
https://drive.google.com/drive/folders/0BxtwW9tCskdULXVQTWg2ZGFKZUk?usp=sharing
Induction Proof:
The induction proof is on z not on n.
The induction proof starts from z =1 and z =2, z =3 … until z=p.
Observe that for z=1, the equation x^n + y^n = 1^n has no solutions in integers x and y whatever n > 2. There is no equality between the sum (x^n+y^n) and 1^n.
A. Basis step / anchoring
Let’s assume z=2 with 1 2^n; 2^n+2^n > 2^n.
The basis case with z = 2 does not need any proof, because you can calculate that regardless of what you choose for x, y, or n. It is true that (x^n + y^n) ≠ 2^n and never equal to 2^n .
B. Inductive step:
Now assume the Fermat’s conjecture is true until z=(p-1) and we will prove it for z=p.
For more details , please , read the PDF file from the link:
https://drive.google.com/drive/folders/0BxtwW9tCskdULXVQTWg2ZGFKZUk?usp=sharing
Your comment are welcome.
Yousif Yaqoob Yousif
Thank you. My proof of FLT is comprehensible for a normal student in mathematics.
Regards
This link maybe useful:
https://hal.archives-ouvertes.fr/file/index/docid/37462/filename/Fermat4.pdf
Dear Mohamed Azzedine
I appreciate your keen efforts to use mathematical induction in solving Diophantine equations. Unfortunately, this technique doesn't work in this domain.
Your article shows an attempt by using this technique to prove the (stone wall: LFT). I will show your mistake soon.
Regards
T.B.C
Dear Issam Kaddoura
As you know I appreciate to work with you and I like your valuable comment on mathematical subjects.
Waiting your review on my proof of FLT.
Best regards
Dear Mohamed Azzedine
I read your article about the proof of LFT using the mathematical induction on z.
Where z is an integer value coprime to x and y in the equation
xn + yn = zn
You have used the following ( Wrong fact)
If we have two polynomials in P are equal and have that same degree, then
the corresponding coefficients are equal.
Polynomials can be considered as a vector space if they are considered as functions of variable x. (x should be variable )
If we fix x, then the above relation is wrong.
Counterexample:
Consider
f(x)= x³ - 10x² +48x -60 and
g(x)= -x³ + 20x² -100x +180,
f(x) and g(x) have different coefficients.
But, f(P)=g(P) for P=4,5,6
f(4)=(4)³-10(4)²+48(4)-60= 36
g(4)=-(4)³+20(4)²-100(4)+180= 36
f(5)=(5)³-10(5)²+48(5)-60= 55
g(5)=-(5)³+20(5)²-100(5)+180= 55
f(6)=(6)³-10(6)²+48(6)-60= 84
g(6)=-(6)³+20(6)²-100(6)+180= 84
Best regards
Issam Kaddoura
I do not agree with your comment.
In my proof , the two polynomials has p as variable and it is not fix, it varies with z from 1 to infinity. Here is the text extracted from my proof
***********************************
p^n – (p-j)^n=n [ j*p^(n-1) – j^2*(n-1)/2 p^(n-2)………….(+/-j^n)/n ]
is a polynomial of the variable p with degree (n-1).
(p-i)^n =(p-i)*(p-i)^(n-1)=(p-i) [p^(n-1) –(n-1) p^(n-2) i +(n-1)(n-2)/2 p^(n-3) i^2 ....+/-i^(n-1)]
is also a polynomial of p with degree (n-1).
p^n – (p-j)^n=n*j [ p^(n-1) – j*(n-1)/2 p^(n-2)………….(+/-j^(n-1))/n ]
p^n- (p- j)^n=(p-i)^n for all values of p, leads to the equality of the two polynomials:
***************************************
Polynomials can be considered as a vector space with the basis {1, x, x^2, x^3.......x^n} because they are considered as functions of variable p which varies as z from 1 to infinity. This is why the induction is on z not on n.
Best regards
Mohamed Azzedine
I know what a vector space is!!! Your mathematical induction is on z, and your polynomials are considered over discrete integer values, so it is not vector space. The counterexample is obvious. The complete results are not correct. Induction doesn't work. Please don't defend the wrong ideas. For the readers, Counterexample:
Consider
f(x)= x³ - 10x² +48x -60 and
g(x)= -x³ + 20x² -100x +180,
f(4)=(4)³-10(4)²+48(4)-60= 36
g(4)=-(4)³+20(4)²-100(4)+180= 36
f(5)=(5)³-10(5)²+48(5)-60= 55
g(5)=-(5)³+20(5)²-100(5)+180= 55
f(6)=(6)³-10(6)²+48(6)-60= 84
g(6)=-(6)³+20(6)²-100(6)+180= 84
The polynomials f(x) ≠ g(x), but f(P) = g(P) for P = 4, 5, 6.
Regards
In Your counter example you start with two different polynomials and you compute the values for differents values of x.
I think You have to start with f(x)=g(x) for all values of x. The variable x is real which implies rational and integer numbers and then compare the equality of similar coefficients. f'x)=g(x) for all x implies equalities between coefficients and the reverse implies the equality of two polynomials.
Regards
You have repeated the same information that I told you !!
You used integer discrete values for P.
In conclusion, the polynomials over discrete values are not vector space.
And the article proof is not accepted.
I wish you good luck in a new attempt.
Hello dear colleagues,
I've published some time ago in this July my new preprint in ResearchGate.net with my simple proof of the Fermat's last theorem:
https://wwоw.researchgate.net/publication/342624684_APPLICATION_OF_THE_FERMAT%27S_THEOREM_FOR_PRACTICAL_PROBLEMS_SOLUTION_IN_BIOLOGY_Article_is_under_consideration_for_IJB
If you are interested you could write me to my email id [email protected]
Best regards,
Sergey Klykov,PhD
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