Can u prove this simple but very important fact?

Question

Show that in the equation of FLT
a^(n) + b^(n) = c^(n)
we have to always consider "c" as odd natural number.

We have to always cosider "c" as odd for finding natural coprime solution of "c" & "b" for a given "n" & "a".

Answer 1

suppose c is even. then this means that a is even and b is even, or a is odd and b is odd. that is, a and b have the same parity. we could not have a even, b even and c even because otherwise, a, b and c would not be coprime. so a is odd and b is odd, if c is even. I am unable to do this right now (I have forgotten my math, sorry), but I believe you could dispense with a proof which, loosely speaking, allows you to conclude that c must be odd on the basis of the fact that c^n = 0 (mod 2) is too restrictive (i.e. not all nth powers are even (except for those containing 2^n as a factor), or in other words, most nth powers are odd). sorry, am not much help.

incidentally though (if I may mention it here), a proof for FLT has already been published (and verified) in 1994. the original proof comprised a whole edition of the Annals of Mathematics for that year and is composed of two papers, to wit:

Modular Elliptic Curves and Fermat's Last Theorem

AND

(again, it is easy to forget article titles, so please just search on the WWW web for it... :-))

the proof is subtle and surprisingly involved, and there is actually a page on the Internet dedicated to presenting the mathematics behind the proof to the general mathematical audience. try googling "The Mathematics of Fermat's Last Theorem".

happy reading!

Answer 2

PROOF:

If the numbers A, B and C are a solution to Fermat's equation for the power of 6, namely A6+B6=C6, then the numbers A2, B2 and C2 are a solution to the power of 3, contradictory to the proof that FLT is true for this power. Generally, if we had proven FLT to any power k, than the theorem is valid to all the multiples of k. The reason for this is that if the numbers A, B and C are a solution for the power mk, then the numbers Am, Bm and Cm are a solution to the power of k, contradictory to the proof that FLT is true for this power.

It follows from this proof that it is enough to prove FLT just for powers that are primes greater than 2, since each other power is a multiple of these primes (except 4, but for this power the theorem has already been proven by Fermat). It is a big step toward reduction of the scope of the problem, but it is still leaving us with the need to prove FLT for an infinite set of powers.

id Fermat prove this theorem?



No he did not. Fermat claimed to have found a proof of the theorem at
an early stage in his career. Much later he spent time and effort
proving the cases n = 4 and n = 5 . Had he had a proof to his theorem,
there would have been no need for him to study specific cases.

Fermat may have had one of the following ``proofs'' in mind when he
wrote his famous comment.

* Fermat discovered and applied the method of infinite descent,
which, in particular can be used to prove FLT for n = 4 . This
method can actually be used to prove a stronger statement than FLT
for n = 4 , viz, x^4 + y^4 = z^2 has no non-trivial integer
solutions. It is possible and even likely that he had an incorrect
proof of FLT using this method when he wrote the famous
``theorem''.
* He had a wrong proof in mind. The following proof, proposed first
by Lame' was thought to be correct, until Liouville pointed out
the flaw, and by Kummer which latter became and expert in the
field. It is based on the incorrect assumption that prime
decomposition is unique in all domains.

The incorrect proof goes something like this:

We only need to consider prime exponents (this is true). So
consider x^p + y^p = z^p . Let r be a primitive p -th root of
unity (complex number)

Then the equation is the same as:

(x + y)(x + ry)(x + r^2y)...(x + r^(p - 1)y) = z^p

Now consider the ring of the form:

a_1 + a_2 r + a_3 r^2 + ... + a_(p - 1) r^(p - 1)

where each a_i is an integer

Now if this ring is a unique factorization ring (UFR), then it is
true that each of the above factors is relatively prime.

From this it can be proven that each factor is a p th power from
which FLT follows. This is usually done by considering two cases,
the first where p divides none of x , y , z ; the second where p
divides some of x , y , z . For the first case, if x + yr = u*t^p
, where u is a unit in Z[r] and t is in Z[r] , it follows that x =
y (mod p) . Writing the original equation as x^p + (-z)^p = (-y)^p
, it follows in a similar fashion that x = -z (mod p) . Thus 2*x^p
= x^p + y^p = z^p = -x^p (mod p) which implies 3*x^p = 0 (modp)
and from there p divides one of x or 3|x . But p > 3 and p does
not divides x ; contradiction. The second case is harder.

The problem is that the above ring is not an UFR in general.



Another argument for the belief that Fermat had no proof -and,
furthermore, that he knew that he had no proof- is that the only place
he ever mentioned the result was in that marginal comment in Bachet's
Diophantus. If he really thought he had a proof, he would have
announced the result publicly, or challenged some English
mathematician to prove it. It is likely that he found the flaw in his
own proof before he had a chance to announce the result, and never
bothered to erase the marginal comment because it never occurred to
him that anyone would see it there.

Some other famous mathematicians have speculated on this question.
Andre Weil, writes:

Only on one ill-fated occasion did Fermat ever mention a curve of
higher genus x^n + y^n = z^n , and then hardly remains any doubt
that this was due to some misapprehension on his part [ ... ] for a
brief moment perhaps [ ... ] he must have deluded himself into
thinking he had the principle of a general proof.

Answer 3

PROOF:

If the numbers A, B and C are a solution to Fermat's equation for the power of 6, namely A6+B6=C6, then the numbers A2, B2 and C2 are a solution to the power of 3, contradictory to the proof that FLT is true for this power. Generally, if we had proven FLT to any power k, than the theorem is valid to all the multiples of k. The reason for this is that if the numbers A, B and C are a solution for the power mk, then the numbers Am, Bm and Cm are a solution to the power of k, contradictory to the proof that FLT is true for this power.

It follows from this proof that it is enough to prove FLT just for powers that are primes greater than 2, since each other power is a multiple of these primes (except 4, but for this power the theorem has already been proven by Fermat). It is a big step toward reduction of the scope of the problem, but it is still leaving us with the need to prove FLT for an infinite set of powers.

id Fermat prove this theorem?



No he did not. Fermat claimed to have found a proof of the theorem at
an early stage in his career. Much later he spent time and effort
proving the cases n = 4 and n = 5 . Had he had a proof to his theorem,
there would have been no need for him to study specific cases.

Fermat may have had one of the following ``proofs'' in mind when he
wrote his famous comment.

* Fermat discovered and applied the method of infinite descent,
which, in particular can be used to prove FLT for n = 4 . This
method can actually be used to prove a stronger statement than FLT
for n = 4 , viz, x^4 + y^4 = z^2 has no non-trivial integer
solutions. It is possible and even likely that he had an incorrect
proof of FLT using this method when he wrote the famous
``theorem''.
* He had a wrong proof in mind. The following proof, proposed first
by Lame' was thought to be correct, until Liouville pointed out
the flaw, and by Kummer which latter became and expert in the
field. It is based on the incorrect assumption that prime
decomposition is unique in all domains.

The incorrect proof goes something like this:

We only need to consider prime exponents (this is true). So
consider x^p + y^p = z^p . Let r be a primitive p -th root of
unity (complex number)

Then the equation is the same as:

(x + y)(x + ry)(x + r^2y)...(x + r^(p - 1)y) = z^p

Now consider the ring of the form:

a_1 + a_2 r + a_3 r^2 + ... + a_(p - 1) r^(p - 1)

where each a_i is an integer

Now if this ring is a unique factorization ring (UFR), then it is
true that each of the above factors is relatively prime.

From this it can be proven that each factor is a p th power from
which FLT follows. This is usually done by considering two cases,
the first where p divides none of x , y , z ; the second where p
divides some of x , y , z . For the first case, if x + yr = u*t^p
, where u is a unit in Z[r] and t is in Z[r] , it follows that x =
y (mod p) . Writing the original equation as x^p + (-z)^p = (-y)^p
, it follows in a similar fashion that x = -z (mod p) . Thus 2*x^p
= x^p + y^p = z^p = -x^p (mod p) which implies 3*x^p = 0 (modp)
and from there p divides one of x or 3|x . But p > 3 and p does
not divides x ; contradiction. The second case is harder.

The problem is that the above ring is not an UFR in general.



Another argument for the belief that Fermat had no proof -and,
furthermore, that he knew that he had no proof- is that the only place
he ever mentioned the result was in that marginal comment in Bachet's
Diophantus. If he really thought he had a proof, he would have
announced the result publicly, or challenged some English
mathematician to prove it. It is likely that he found the flaw in his
own proof before he had a chance to announce the result, and never
bothered to erase the marginal comment because it never occurred to
him that anyone would see it there.

Some other famous mathematicians have speculated on this question.
Andre Weil, writes:

Only on one ill-fated occasion did Fermat ever mention a curve of
higher genus x^n + y^n = z^n , and then hardly remains any doubt
that this was due to some misapprehension on his part [ ... ] for a
brief moment perhaps [ ... ] he must have deluded himself into
thinking he had the principle of a general proof.

Answer 4

What a nonsense .... FLT is for all positive integers , not jkust c odd.