Dear Mohammed,

I agree with the comments of Wulf and Peter; I would like to shed light on the following point:

To prove that a statement P(n), n ∈ N, is correct, you can apply the mathematical induction ( or the strong mathematical induction) on n.

But if P (x, y, z, n) a statement that includes n, x, y, and z, all are variables. We can't apply the mentioned principle of induction to prove that P is true.

P.S.

Recall that,the principle of mathematical induction:

Let S = { n ∈ N, P(n) is true }

If (a) 1 ∈S.

(b) If k ∈S, then k+1 ∈S.

Then S = N the set of all natural numbers.

( That is P(n) is true for all n ).

Best regards

Hello,

Thank you (all) for your comments.

First

The approach above is used for the well known equation x^2+y^2=z^2 and gives the first couple of solutions.

n is fixed to n=2 and z is fixed to z=5, x and y are variable x2.

I agree with the last comment of Peter Breuer. The approach used is right and correct may be it must be well presented.

Regards

@Wulf Rehder

A valid induction step would have to show that for any combination a, b, and all n, a^n + b^n ≠ (p-1)^n implies that for any combination of u, v, and all n, u^n + v^n ≠ p^n. I am using u and v as precaution because they may be different from the set of a, b.

I am sorry, the induction step from z=3 to z=4 is clearly demonstrated in the PDF file. This is why I mentionned it at the beginning of the proof,

because it is a critical point.

The same method is used to imply x^n +y^n /=p^n from x^n+y^n/=(p-1)^n. May be it is not well presented. I wiil rewrite it and send it.

There is not nothing "oddities" but it is new and creative.

Regards

@Wulf Rehder

A valid induction step would have to show that for any combination a, b, and all n, a^n + b^n ≠ (p-1)^n implies that for any combination of u, v, and all n, u^n + v^n ≠ p^n. I am using u and v as precaution because they may be different from the set of a, b.

This step was not proven in the document.

Please find the proof of the inductive step from z=(p-1) to z=p in the attached file; Pminus1impliesP.pdf.

Thank you very much for your valuable comment.

Regards

Dear Mohamed,

Once more, induction is not accepted for a general proof to LFT.

I agree that, in your tables, LFT is obvious for the values

(x,y,z)∈{1,2,3,4} and n > 2.

Consider the following example:

(x,y,z)=(8,9,10)

your calculations considered cases as:

(1)ⁿ+(2)ⁿ< (10)ⁿ,.......,(1)ⁿ+(9)ⁿ< (10)ⁿ,......,(8)ⁿ+(9)ⁿ< (10)ⁿ

Also inequalities with greater than.

But this not acceptable.

Induction should be done for each fixed triple of numbers (x,y,z) with n varies. Also, all other cases where (x,y,z) and n vary simultaneously.

How you can tackle the following:

(8)²+(9)²> (10)²

(8)³+(9)³> (10)³

(8)⁴+(9)⁴> (10)⁴

BUT

(8)⁵+(9)⁵< (10)⁵ ,

(8)k+(9)k ≤ (10)k where k ≥5.

In general if

(a)^{s}+(b)^{s}> (c)^{s}, it is excpected to find m such that

(a)^{m} + (b)^{m} ≤ (c)^{m}

CLAIM: Induction as applied in the listed numbers can't guarantee to avoid equality when (less than) switch into (greater than) after finite number of steps.

( To confirm the previous claim is equivalent to LFT itself)

Best regards

@wulf Rehder

Thank you very much for your patience and your rigorous analysis.

I understand your reservation but the variables u and v different from x and y are already taken into account in my proof.

Refer to the last table in PDF file, there are two columns one for (p-1) and another for p. The column p contains all the new sums different from

x^n +y^n coming from the case (p-1).The new sums are not equal to p^n. Therefore (c) for all pairs x,y, and n , the sum x^n+y^n is not equal to p^n. We can claim that x^n+y^n /= p^n and FLT has no integers solution x,y,p for any value of n>2.

I am trying to tackle this big problem (FLT) . Wile' s proof is not understandable by high school student and even by some members of our community. It is a good and important work but very long demonstration ( 300 pages) using advanced tools of mathematics.

I would like to thank all of you for your comment and help in order to offer to our community a simple and elementary proof of FLT.

Best regards

@Wulf,

Simple cases for x,y,z are 2,3, 4 are trivial for any n.

In general, the companion tables don't offer at least the proof of the well-known case n=3.

Could you please show me how you accept the previous proof for the case n = 3?

How I can accept that, it is true for n = p-1?

[ (cf. my very first answer, also @Peter Breuer’s two earlier notes), but intended to try induction on z (sometimes called c in the posts above)]

Which is fixed, which is varying? to apply the induction?

What is induction on Z? It has no meaning at all.

For fixed n =3, and fixed Z=10 there are values (x,y) such that

x < y < z and

(8)³+(9)³> (10)³

(1)³+(9)³< (10)³

Which is fixed, which is varying? to apply the induction?

Each case has an independent interest, and Induction can never tackle

the LFT.

If so I appreciate if you can show the proof of the simple case where n =3

using induction.

@Wulf,

First of all, I would like to say that your comments, as well as peters, are logical, helpful and constructive.

No doubt, many famous problems were proved by using the mathematical induction.

My objection didn't mean that your ideas or Mohamed proof are wrong. But I wonder if we can use the mathematical induction in this situation.

The discussion will improve our point of view and make things are clear

and helps the dialogue to reach the best results in this concern.

It is not clear to me how one can use mathematical induction in this case.

In brief, If we consider LFT is true for n=3.

How we employ the mathematical induction to prove that LFT for n=4?

Which is fixed and which is variable in x^4 + y^4 = z^4 to apply the induction?

According to me: companion tables are not enough and can't be accepted as a proof.

The discussions for this particular case may convince me that math. induction works in such type of Diophantine equations.

Best regards

@Peter Brewer

I agree with your answer to Issam' s question.In initial PDF file and in PDF Pminus1impliesP we provided the proof. The case z=3 implies z=4

Here is the answer of @Rehder about that z=3 implies z=4 in the PDF file Pminus1impliesP "Regardless, your “claim” that the Fermat equation has no solution for z=3 or =4 is correct."

@Wulf Rehder

Thank you very much for the links about multidimentional induction.It is very clear and applicable to FLT. Remember I mentioned it in my previous answer.

"It is generalization of induction method to the two dimensional space (dimension 2).

x^n +y^n =z^n is equivalent to Norm(x,y) = Norm(z,0). The induction method is applied to the norm as in the one dimensional space.

The method could be extended to higher dimension >2."

We have made a lot of progress in our discussion. Induction method can be applied to FLT.

Thank you

Best regards

@Breuer

Where is a reduction to z=3?

From z=4^n the possible solutions (6) are: 1+1=4^n; 1+2^n=4^n; 1+3^n=4^n; 2^n+2^n=4^n; 2^n+3^n=4^n; 3^n+3^n=4^n

The reduction to z=3 is inside the list of possible solutions: 1+1=3^n; 1+2^n=3^n; 2^n+2^n=3^n

They are ruled out since they are not divide by 3 (3 does not divide 1 and 2)

From 2^n+2^n=4^n parity are right but for 2^n+2^n=3^n the parity is not right.

Let me take the reverse problem. From the case z=3 can we deduce the case z=4

The case x^n + y^n=3^n has no integers solutions for n>2. Then we get the following inequalities:

InfG3:

1/ 1+1 < 3^n

2/ 1+2^n < 3^n

3/ 2^n +2^n < 3^n

SupG3

4/ 1+3^n > 3^n

5/ 3^n+2^n > 3^n

6/ 3^n+3^n > 3^n

From the case z=2 and z=3 we can deduce the following inequlities:

1/ From 1+3^n > 3^n we infer 1 +3^n < 4^n

2/ From 3^n + 2^n > 3^n we infer 3^n + 2^n < 4^n

3/ From 3^n+3^n > 3^n we infer 3^n+3^n < 4^n

1/ From 1+2^n > 2^n we infer 1 +2^n < 3^n < 4^n

2/ From 2^n+2^n > 2^n we infer 2^n+2^n < 3^n 4^n

8/ 4^n +2^n > 4^n

9/ 4^n+3^n > 4^n

10/ 4^n+4^n > 4^n

It is clear that the case z=3 implies z=4.

We can claim that x^n+y^n=4^n has no integers solutions for n>2.

Using the same reasoning, the case z=(p-1) implies z=p and we can claim we get a simple proof of FLT by Induction method.

Best regards

@ Mohamed,

In the tables, you restrict your statements to values for

(x,y,z)∈{1,2,3,4} and n > 2.

As

[1/ 1+1 < 3^n

2/ 1+2^n < 3^n

3/ 2^n +2^n < 3^n

SupG3

4/ 1+3^n > 3^n

5/ 3^n+2^n > 3^n

6/ 3^n+3^n > 3^n]

Yes for the simple cases in the listed tables:

2^n +3^n < 4^n

2^n +2^n < 4^n

are true for all n > 2.

But for a little bit larger values all such results turn illusion.

Now consider the case: (you called it inf G10 )

(8)ⁿ+(9)ⁿ virsus (10)ⁿ .

There are an obvious contradiction.

(8)³+(9)³> (10)³

(8)⁴+(9)⁴> (10)⁴ following tables for inf G10

BUT

(8)⁵+(9)⁵< (10)⁵ .

produces contradiction it stops and not in inf G10.

It follows the values in supG10.

Conclusion:

Using the only 1,2, 3 and 4 create such illusion.

2n +3n < 4n is true for all n > 2.

But

(8)ⁿ+(9)ⁿ < (10)ⁿ is NOT true for all n.

Best regards

@Issam

I am soory I do not restrict to {1,2,3,4}. This set is only for x^n+y^n=4^n because x and y must be less than z in x^n+y^n=z^n.

In the previous answer I demonstrate that the case z=3 implies z=4. (you asked early this point)

The same reasonning is applied to the general case, z=(p-1) implies z=p.

I think it is now clear that the Induction method applies to FLT and allow to get a simple proof of FLT.

I am sorry to repeat myself.

Now I come to 8^n +9^n < 10^n. I know that it is not true for all n, but it is not equal to 10^n . It is sometimes less than 10^n and somstimes greater than 10^n.

This why in the original file PDF I mentioned the evolution of some value of (x^n+y^n) when n varies. Please return to this PDF file.

There is no illusion. We can say there is no equality between (x^n+y^n ) and z^n (10^n in your example) which is the most important point in FLT.

Best regards

@ Mohamed,

You agree with my example.

In this case, your partitions infG and SupG in the tables are not hold

How one can proceed.

Dear Mohamed,

Can you show in clear steps:

If LFT is true for n=3, then LFT is true for n =4.

This will help to go further.

Best regards

@issam

@Breuer

Here is the clear steps of the induction method. The variable of induction is z not n. z varies from 1 to p.

Basic step

Observe that x^n + y^n = 1^n has no solutions in integers x , y and z=1 for any value of n > 2.

It is obvious that the equation x^n + y^n =2^n has no solution in integers x, y and z=2 for any value of n >2.

(1+1< 2^n )

Induction step

If FLT is true for n=3, then FLT is true for n =4.

From the case z=3 can we deduce the case z=4 The case x^n + y^n=3^n has no integers solutions for n>2. Then we get the following inequalities which is equivalent to x^n+y^n/=3^n: InfG3: 1/ 1+1 < 3^n 2/ 1+2^n < 3^n 3/ 2^n +2^n < 3^n SupG3 4/ 1+3^n > 3^n 5/ 3^n+2^n > 3^n 6/ 3^n+3^n > 3^n From the case z=3 we can deduce the following inequlities: 1/ From 1+3^n > 3^n we infer 1 +3^n < 4^n 2/ From 3^n + 2^n > 3^n we infer 3^n + 2^n < 4^n 3/ From 3^n+3^n > 3^n we infer 3^n+3^n < 4^n 1/ From 1+2^n > 2^n we infer 1 +2^n < 3^n < 4^n 2/ From 2^n+2^n > 2^n we infer 2^n+2^n < 3^n < 4^n As you see we get the 6 inequalities of InG4 from the previous cases z=3 InfG4: 1/ 1+1 < 4^n 2/ 1+2^n < 4^n 3/ 2^n +2^n < 4^n 4/ 1+3^n < 4^n 5/ 2^n+3^n < 4^n 6/ 3^n+3^n < 4^n The inequlities of SupG4 are obvious. SupG4 7/ 1+4^n > 4^n 8/ 4^n +2^n > 4^n 9/ 4^n+3^n > 4^n 10/ 4^n+4^n > 4^n Conclusion

It is clear that the case z=3 implies z=4. We can claim that x^n+y^n=4^n has no integers solutions for n > 2. Using the same reasoning (technique), If FLT is true for the case z=(p-1) then it is true for z=p . I did this technique in the file Pminus1impliesP.

I can repeat it again. Let we go step by step. I would like your agreement for the proof above . the case z=3 implies z=4. Best regards

mostly big conjectures are proved but aft er some time past somebody is saying that proof has mis t akes....

@Breuer

First , do you agree with the demonstration in my previous answer about Induction method ? (if FLT is true for z=3 then FLT is true for z=4).

(A), what about x or y more than 3

From x^n+y^n =3^n , if x>3 or y>3 the sum (x^3+y^3) is greater than 3^n. No equality, no solution

(B) why do you think that x^n+y^n < 3^n .....

The case x^n+y^n < 3^n is examinated in SupG3. Remember we have two types of inequalities InfG3 and SupG3.

When I say 'we infer' It is a way to avoid to repeat the demonstration for each inequality.I agree with your comment.I have got the same idea and all the inference used in the proof are right. It is better to avoid words like strange,...

Yes my induction proof is designed to work for" any z and any x and any y and any n" except n=2

Remember for each equation I use matrix form for the integer values of x and y. I cannot forget any integer value.

z varie for 1 to p and n from 2 to N. There is no restrictions nor conditions.

I remain ready to answer to any question about this proof. Thank for your valuable comment .

Best regards

@Breuer

It seems we have a problem of vocabulary, may be my english is not accurate but the meaning is right.

If you read carefully my original paper and my previous answer about a question from @Issam (if FLT is true for z=3 then FLT is true for z=4)." )

I have followed the 3 steps of Induction (basis step, induction step, conclusion).I would like your agreement on this point.

Please one objection at a time.

I have already answered for the general case (all z) in the file Pminus1impliesP. The case z=(p-1) implies z=p. I will rewrite it and send it with more details.

I must convince you by answering to your questions.( all your questions.) because I know the proof works.If there is a flaw I will correct it or acknowledge my error.

Best regards

Dear Mohamed,

Peter's and my objections are serious and true.

I summarize your method as

( always n > 2)( without loss of generality x < y < z ) .

If x^n +y^n = p^n ..(1) is not solvable in integers for all x and y,

then x^n +y^n = (p+1)^n ..(2) is not solvable in integers for all x and y,

And using the same argument using (2) you deduced that

x^n +y^n =( p+2)^n..(3) is not solvable in integers for all x and y,

and so on, then you reach the general case:

x^n +y^n = z^n is not solvable in integers for all x and y,

and for all z too. Hence LFT is true.

To show you work, you started with z=3 and you proved that

x^n +y^n = 3^n is not solvable, where x < y 2 and for all x < y < 4 .

( The proof is direct).

The serious issue starts with larger values for Z.

Consider the following example Z = 10:

Following the tables, we obtain 8⁴+9⁴ >10⁴

Now for Z = 10 +1= 11, the table values didn't work

and you can see

8⁴+10⁴ < 11⁴.

9⁴+10⁴ >11⁴.

There is a dramatic behavior for the tables even we add 1 to z.

Conclusion: Induction didn't work.

Adding 1 to z change all expected results.

Best regards

If I summarize the proof:

I stated from z=1 until z=4 going through z=2, z=3 with n>2 and x

@Issam

Consider the following example Z = 10:

Following the tables, we obtain 8⁴+9⁴ >10⁴

Now for Z = 10 +1= 11, the table values didn't work

and you can see

8⁴+10⁴ < 11⁴.

9⁴+10⁴ >11⁴.

There is a dramatic behavior for the tables even we add 1 to z.

There is only one table which grows with z. All the inequalities are overlaped or interlinked. The inequalities of InfG10 are included in InfG11.

SupG10 moves to InfG11 when z moves to z+1.

Here is the content of the table for z=9, z=10 and z=11. You will see the evolution but nothing is dramatic.The sum x^n+y^n is sometimes < z^n and sometimes > z^n but never equal to z^n when n varies.

z=9

InfG9

1+1

@Rehder

Thank you very much for your positive contribution. I will be in touch with the team of "Extensions of FLT"

@Breuer

I answered to @Issam question about z=10 and z=11. In the same answer I mentionned that I will examine the general case with

z=11111111111111111 or z=9999999999999999999

Please give time to the time.

Best regards

@Mohammed,

Following your reply to my post, you said,

7^n+8^n10^3

We can't trust tables.

Best wishes

@ Followers,

If one interested in some extensions of the LFT:

{On an infinite number of solutions to the Diophantine equation x^n + y^p = z^q over the square integer matrices }

Also, I have an accepted paper, to appear soon in ( Linear and Multilinear Algebra} concerning the same issue.

Best regards

@ Peter,

The case q = 1 is obvious in all domains, even the ring Z,

but the mentioned results hold for all n, p and q.

Best regards

FLT seems to be the bone of contention. Why is a proof to this theorem so important?

@ Wulf,

Thank you for reading the article.

(*)[ thus C^n + C^p – C^q = 0. Nothing new here.]

I don't agree with you; it is new and original.

What is the 2x2 matrix with characteristic polynomial

xn +xp +xq =0 ? try to generate any one!

(*) [ U^n + V^p + W^q = 0. (But I agree, that is cheating.) ]

I think you didn't get the point.

Kindly give an example that shows any cheating!!

(*)[ i.e. when 6k+1 and/or 6k+5 are prime? ]

Dear Wulf, all primes have the form 6k+1 or 6k+5 in addition to 2 and 3.

You can substitute in the mentioned general form any prime number with any number of digits.

(*) For your question in (4) about x^2 + x^3 –x^5 = 0.

My answer is yes.

Soon, I will send you my accepted paper that shows solutions for FLT and Beal's conjecture in any dimension nxn matrix.

Best regards

Dear Muneeb,

I think you are a doctor and not mathematician.

In my life, I noticed that almost all successful doctors are curious to know and are very close to arts and all branches of science.

In fact, the last Fermat's theorem represents a long story of a struggle for more than 3 hundreds of years. In brief, it is a story of challenge and beauty of the human creation to overcome the impossible.

Just google Fermat's Last Theorem to read thousands of articles about that legend.

Best regards

@ Wulf,

(*)Could you please read my conference paper once more.

I didn't claim that I have a proof for FLT.

Who said there is a proof to FLT included?

I show some simple solutions in other domains.

Regardless you like them or not.

(*)Concerning all other points, what do you want to say?

In brief, you don't like them. Ok. So what?

The main point, is there anything wrong? Any plagiarism?

(*)Cayley-Hamilton is an elementary topic in Linear Algebra.

Yes, it is. Thanks for the information.

(*)[ If you have a proof that is accepted by a reputable refereed mathematical journal, I will give you $1,000.]

Are you sure about this offer?

If so, do you think the following journal:

"Linear And Multilinear Algebra" is reputable refereed Journal?

@ Peter,

What do you want to say?

Your answers are not clear to me.

Thank you

@ Peter,

We have a nice idea with a new approach.

But take care if all x, y, and z are nilpotent then all terms x^n, y^p, and z^q reduce them all zeros.

Best

@Issam

About the Table.

It is a symmetric matrix (p,p) which contains the elements (x^n+y^n) when x and y varie.

The diagonal contains the sum (x^n+x^n) when x=y. The elements (x^n+x^n) are sometimes < z^n and sometimes > z^n. There are grouped together in “NotEqualGroup” (NEqG).

When we change from z=p to z=p+1 the elements of NEqGp move to InfGpplus1 and others elements coming from SupGp move to NEqGpplus1.

Example:

p=5

4^n+4^n /= 5^n

n=3 64+64 > 125

n=4 256+256 < 625

p=5+1=6

5^n+5^n /= 6^n comes from SupGp (1+5^n>5^n; 2^n+5^n>5^n........5^n+5^n>5^n)

n=3 125+125 >216

n=4 625+625 < 1296

but

4^n+4^n /= 5^n moves to 4^n+4^n < 6^n

n=3 64+64 < 216

n=4 256+256 < 1296

Your post

Following your reply to my post, you said,

7^n+8^n

@ Mohamed,

For any fixed n >2 and x< y z^n (2)

and

x^n +y^ = z^n (3)

You considered Case(1) and (2) but case (3) is excluded.

Why? Mathematical induction as shown in your tables will not guarantee that equality will not appear!!

Best regards

@Wulf,

I think you want to post this answer in the thread of Muneeb question:

“Why is a proof of this theorem [FLT] so important?”

Isn't it?

Best regards

Dear Wulf,

Yes, it seems good anywhere.

Best

@Issam

You considered Case(1) and (2) but case (3) is excluded.

Why? Mathematical induction as shown in your tables will not guarantee that equality will not appear!!

I am sorry I do not exclude the case (3). It excludes itself because the sum (x^n+y^n) is sometimes > z^n and sometimes < z^n with successive n. There is no integer between n and n+1, then the equality is impossible to appear. The same sum (x^n+y^n) will be < z^n or > z^n for the next value of z+1.

Example:

4^n+4^n /= 5^n belongs to NEqG of z=5

n=3 64+64 > 125

n=4 256+256 < 625

p=5+1=6

4^n+4^n < 6^n the sum (4^n+4^n) becomes < (5+1)^n

n=3 64+64 5^n; 2^n+5^n>5^n........5^n+5^n>5^n)

n=3 125+125 >216

n=4 625+625 < 1296

Best regards

Dear Mohamed,

Reread my post, I said for fixed n>2 and x < y < z.

How will you guarantee that equality will not occur?

Few examples of greater than or less than is not a general proof.

Best regards

Hello,

Please find the new text of the Fermat's Last Theorem based on induction proof on z. Your comment are appreciated.

Best regards

@Rehder

Thank you very much. It is a pleasure to read your mathematical objection.

You will find the answer when you follow all the steps of the "Approach based on overlapping of the triangular matrixes" specially in the last paragraph of the proof just before the conclusion.

All the triangular matrixes from z=1 to z=p-1 are nested and in each matrix we checked that there is no equality between the sum (x^n+y^n) and z^n. This argument (no equality) is part of the induction proof.

The sum (x^n+y^n) is positive, increasing and discontinuous , it never cross z^n proved by induction . The crossing can happen in the real field not in natural numbers.

Thank you again

Best regards

@Breuer

You are right the sentence

1/ "All the triangular matrixes from z=1 to z=p-1 are nested".is not comprehensible. I mean "All triangular matrixes from the case z=1 to the case z=p-1".

2/ "in each matrix we checked that there is no equality between the sum (x^n+y^n) and z^n" Matrices do not contain sums, they contain entries.

From the anchoring z=2 to z=p-1 , I use matrix with entries and elements, each element is the sum (x^n+y^n). In all the document I introduce the matrix (2,2) , matrix(3,3) and so on until matrix (p,p).

Each table can be seen as a triangular matrix. In all the document we use in bold character "All the element of the symmetrical matrix (k,k) with the form (x^n+y^n) are not equal to z^n" .

It means that "no equality" is the main argument of the induction proof.

3/ "The sum (x^n+y^n) is positive, increasing and discontinuous , it never cross z^n proved by induction ."

Please read the last paragraph of the proof, just before the conclusion . It contains the proof that the elements of the column (p-1), the sum (x^n + (p-1)^n) switches from less than p^n to greater than p^n for consecutive values of x. It means that this sum cannot be equal to p^k with k

@Rehder

1/ "you haven’t grasped the concept of induction"

I am sorry this is not true. I introduce the case z=3, z=4 and z=5 because there is a disruption in the behavior of the sum x^n +y^n at x=y=4

I write it in bold in order to introduce the change in the behavior of the sum x^n + (p-1)^n at each column (p-1) of the symmetrical matrix (p-1, p-1).

Later in the document, I write ,in italic character, a paragraph about the introduction of the three cases z=3, z=4 and z=5. You can check it.

I know that I do not need these cases for the concept of induction. Anchoring is enough.

2/ Mainly Issam: Even if we leave aside the logical deficiency of (1), there is an essential case missing in your tables, to repeat Issam once more:

I am sorry there is no logical deficency, there is a quick reading or may be my writing is not clear. But the meaning is right.

The case x^n+y^n=z^n does not exist in natural numbers when n>2.

I show it in the anchoring case and in the following case z=3, z=4 and z=5.

In all the document I repeat at each case "

"All the element of the symmetrical matrix (k,k) with the form (x^n+y^n) are not equal to z^n" .

It means that the argument "no equality" is the main point of my induction proof.

Please read the last paragraph of the proof, just before the conclusion . It contains the proof that the elements of the column (p-1), the sum (x^n + (p-1)^n) switches from less than p^n to greater than p^n for consecutive values of x. It means that this sum cannot be equal to p^k with k

Hello,

@Rehder

@Breuer

Please find the short proof (3 pages) followed by your two examples revisited. The case (3) is not valid in integers values as shown in different examples.

In RG reviewers and researchers look for mathematical truth. We need each other and we learn from each other. I am sorry if I bother someone.

I would like to thank all followers for their comments.

Best regards

@Breuer

"All you would be proving [if we accept your explanation of what you have done] would be that there is no solution to x^n+y^n=(y+1)^n in integers. That is not FLT."

I am sorry, what we proved is (x^n+y^n) not equal to z^n for x,y,z integers and n>2, using proof by induction on z.

I can say today we have make progress (half way). You agree with the expression x^n+y^n=(y+1)^n.

The expreeion x^n+y^n=(y+1)^n does not exist in my proof. I guess how you build it. To be more precise it is correct if you have written it as

x^n + (p-1)^n = p^n. We examined on the column (p-1) in the matrix ((p-1), (p-1)).

We can repeat the same reasoning for others columns (p-2), (p-3)……until 1. It means the variable y is varying from (p-1) to 1.

x^n +(p-2)^n /=p^n

x^n +(p-3)^n /= p^n and so on

At the end the set of (x^n +(p-1)^n), (x^n+(p-2)^n) and so on

is never equal to p^n for a fixed z=p

We can repeat the same process with another value of z until infinity.

This is FLT.

At the final stage we can claim that x^n+y^n is never equal to z^n with integers x, y, z and n>2.

Proof by induction on z not on n works for FLT.

Thank you very much for your valuable comment. I hope to convince you.

I remain confident to answer to your question.

Best regards

Hello,

A picture is better than a thousand of speeches.

May I suggest you to choose any value of integer z=p . We assume that the triple of integers (x,y,p) verify FLT. x^n +y^n is never equal to p^n for all integers x,y and n>2.

Upon reception I will show you that this assumption implies x^n+y^n is never equal to (p+1)^n for all integers x,y and n>2.

After this pragmatic attempt we'll go to the theorical proof.

About the symmetrical matrixes they are well defined in my paper.

M (1,1) with A11=2

M(2,2) with A11=2, A12=A21=1+2^n, A22=2^n +2^n

M(p,p) with Aij=(xi)^n +(yj)^n

Best regards

Dear Professor

Every body knows that M(p,p) means a matrix with p rows and p columns

When I write Aij=(xi)^n +(yj)^n [and not (xi)^n + (yi)^n] is for our context equivalent to x^n+y^n.

I am polite and I'll stay polite.

In RG we are not interested in feeling and opinion . Only mathematical fact has a value and meaning , they allow us to make progress.

I have my approach to prove FLT you do not like it. OK.

Professor: remember me experience, science , phylosophy and wisdom .

Best regards

Hello every body

Please find attached a simple proof of FLT based on induction on z and not on n

Could someone review it

Thank you

Noori Abdul-nabi Nasir

The value for Q are powers of n because we try to solve x^n+y^n=z^n

According to the table above in my paper we compare (p-1)^n + k^n to p^n

Please see the last cloumn (p-1) which contains the set (p-1)^n +1, (p-1)^n+2^n

(p-1)^n + 3^n and so on.... to (p-1)^n+(p-1)^n

Sorry RG display Noori

I disagree wiyh you , but it is obvious thrugh the table and the context .

Igael Azoulay

" But this proof is not consistent because Q may be a set of anything, not only of integer powers of n. I spent also some time to understand what you do mean by the first table which brings nothing. "

The table is an array with two variables x and y. 1

Peter Breuer

OK for comment

Here is the new version.

Regards