Gangerolf
Last edited 1/6/2010
Thursday, October 15, 2009

Fermat's Last Theorem

Fermat's last theorem has fascinated pedestrians and professionals alike over several centuries. The theorem itself is little more than a curiosity, its attraction and fame are based on the fact that until recently it resisted all attempts at finding a rigorous proof, and in the process spawned new approaches to number theory. Finally, Wiles and Taylor provided a proof  in 1994. It is long and complicated, if Fermat had a proof, this was not it.

If!  - Most, if not all, writers express doubts that he had a proof. The argument goes that the famous annotation in his book was written around 1630 and he was striving mightily thereafter to find proof for special cases. If he had a general proof, why bother with special cases? It doesn't make sense.

It doesn't, unless the time line is wrong. We do not know the date of the annotation; it could have been written much later. Suppose he struggled with the issue in the 1630s and gave up, to devote his time to more rewarding quests. In 1653 Pascal went public with his observations on the marvelous  triangle that bears his name. In that period Fermat corresponded with Pascal on the theory of chance. He may have seen  that Pascal's triangle offered the key to solve his old problem. He wrote the jubilant annotation, and, problem solved, he lost interest. The foundation of probability calculus was far more interesting.

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The Equation

aⁿ + bⁿ =c ⁿ

has no non-zero integer solution when n > 2.If there is no integer solution when when a, b, and c are mutual primes and n is an odd prime, no other integer solution is possible. The special case, n is a power of 2, was proven by Fermat.

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Given Pascal's Triangle;

the n'th line provides the coefficients for expanding the expression

(b+x)ⁿ

When n is prime, n divides all coefficients greater than 1.
Set c = b+x; x being integer, x < a
then:

aⁿ  + bⁿ= (b+x)n = bⁿ + nb^n-1x + … … + nbx^n-1 + xⁿ

and:

aⁿ = nb^n-1x + … … + nbx^n-1 +xⁿ

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Show that a and n are coprime:

If n divides a, it must also divide x and it cannot divide b,

aⁿ =  nb^n-1x + (n(n-1)/2)b^n-2x² … + nbx^n-1 + xⁿ

If n divides a, divide both sides by n³ to get

aⁿ /n³ =  nxb^n-1/n³ + {(n(n-1)/2)b^n-2x² … + nb x^n-1 +  xⁿ}/n³

Then if n divides a, an integer left side is equated to a non-reducible rational right, n and a must be  coprime; Fermat's Little Theorem is applicable.

Show that x does not divide a:

(x does also not divide b. If it did, b and c would not be coprime.)

aⁿ  =  nb^n-1x + … … + nbxn-1 + xⁿ

then aⁿ /x is integer.
Although x divides aⁿ ; unless x is prime it does not necessarily divide a.
If x divides a, divide by x². Then

aⁿ /x² =  nb^n-1 /x + … … + nbx^n-3+x^n-2

If x divides a, an integer on the left side is equated with  a fraction on the right.

Based on our premises, then

aⁿ  ≡  (xⁿ)Mod (n)

We know, from Fermat's "Little Theorem" that

aⁿ  ≡ a Mod (n)

We have  x < a. Divide by a:

a^n-1  ≡ 1 Mod (n)

and

  a^n-1  ≡ (xⁿ)/a Mod (n)

leading to a contradiction: If a = xⁿ, then x must divide a, contrary to the demonstration above,

 x = 1 is not a valid result.

Fermat's "Little Theorem"

If p is a prime number and a a natural number, then

   a^P ≡ a (mod p)                                                        (1)

Furthermore, if p ≠ a (p does not divide a ), then there exists some smallest exponent d such that

a^d ≡ 0(mod p)                                                   (2)

and d  divides p-1 . Hence,

a^p-1 ≡ 0(mod p)                                                   (3)

                                                                                                   (Wolfram)                                                                                                                                                                      

 Pierre de Fermat first stated the theorem in a letter dated October 18, 1640 to his friend and confidant Frénicle de Bessy as the following[1] : p divides a^p-1 − 1 whenever p is prime and a is coprime to p                                                                                                      (Wikipedia)

The converse is not always true. Some exponents, although they are composite numbers,  may satisfy the congruence. They are called pseudoprimes.

The special case: a = 2 is nicely illustrated in Pascal's triangle, The sum of the coefficients in the n'th line is 2ⁿ. If n is prime we have

2n ≡ 2 (mod n)

The smallest pseudoprime to base 2 is 341. It does not divide all the  coefficients  >1 on its line in Pascal's triangle.

Comment:
Ever since I got bit by the bug I have had several "mirabilem" proofs that all have fizzled, No doubt the attempt above will suffer the same fate. I shall be in good company; it was fun (and educational) trying,

These were tools at Fermat's disposal. He did not have modular arithmetic but the arguments can be made without its help.

The case n = 2  is special because the polynomials are reduced to simply 2bx + x².

a²  = 2bx + x².

It has nontrivial integer solutions if x  = 1. a  is then odd; any odd value for a will yield a solution,