## New approach to FLT Proof for prime powers of n

 Quote by ramsey2879 Yes adjacent binomial factors of the expansion (c-b)^(n-1) sum to zero mod n. One is 1 and the other is -1 mod n. they are not all 1 mod n. But the odd powers of b are negative so by simply adding or subtracting a n*c*b*I part will no get you to the desired expression. I forgot to make amends for the negative powers of b. But you claim that "c-b" is coprime with I. I don't see where you get that. Please explain. Then you say that "If we apply (1*) in (3) we see that (c - b)^2 part is square AND (c – b)^2 does not divide a^n since n is prime. So we got a contradiction. The second (2*) case when applied would give the contradiction too. It follows that if (c – b) = n*m when we apply this in (3) we get that n^2 divides a^n. This does not hold true since n is prime so the only remaining possibility would be that n^(n-2) is hidden in the c*b*l but we know that l is coprime to (c – b) and so are c and b, so this could not be the case.. We see now that in order that the solution exists (c - b) has to be coprime to n and since (c - b) is coprime to l," I don't follow any of this since a^2 divides a^7 and 7 is prime. So (c-b)^2 could divide a^n
Yes but what you get when you divide a^7 with a^2 is not coprime to a. So when you set (c-b) = n you can move out of the brackets in (#) another (c-b) part. Since when we prove l is coprime to (c-b) what is left in the brackets is coprime to (c-b) so because of that and triangle inequality, (c-b)^2 divides one whole part of a on the n power, and this never has a solution.I will take now some days off. In summer I will try to prove something else, when I have more time.I am tired of this FLT problem. Have you seen Ramsey the other one? That an odd perfect number does not exist? I have a good intuition sometimes and I was always good in abstract logic, But in Math formulation I am not. I am preety sure the last proof goes this way. There could be some mistakes. It would be the best for me if I could connect with someone who is good at math and when I have an idea, he proves it or not within few minutes.

I know nothing about modularity sorry. All I know is that the equation (3) satisfy the condition. When (c-b) =n it follows that (c-b)^2 divides c^n - b^n OR when (a+b) = n (a+b)^2 divides a^n + b^n alternating binomial factors form (c-b)^(n-1) when transformed to (2) they give rise to equation (3) no doubt about that. Now I will take a break for few days. I have enough of FLT. Ramsey are there any other Math puzzles in Number theory? The ones you dont need complicated mathematical tools to understand them? I mean I would like to start something, which is formulated very simple...Is there any list?

Thanks and we hear in few days.

I claimed that (c-b) is coprime to l. This I havent proved, it arise from the experiences.Thats thats the thing I am asking MATH EXPERTS several times. Is it so hard to prove that (c-b) is coprime to l? I mean when this is not working simply leave it.
 And Ramsey, please move the questions to my private mail, I have sent it to you days ago. But till Wednesday, I am occupied with my professional work . Greetings
 I have found the missing link and assistant on Uni found another special case which I proved. I will put this elementary proof into more readable form in few weeks. Thank you for all your patience.
 Its possible to show (c - b) = (a1)^n when we apply in equation c = d + b, it follows directly, when we rearrange equations. Its possible to show with the same trick (c - a) = (b1)^n but its impossible to show, that (a + b) = (c1)^n, so I am closing this thread. This exact way is not a Fermats proof. Thread closed.

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