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Brief synopsis of Fermat's Last Theorem

Introduction:
Fermat's Last Theorem has long been a magnet to the amateur and professional mathematician alike because of its seeming simplicity; yet, extraordinary difficulty. Although there is a proof by Andrew Wiles, I think it is understandable that the problem still would incite curiosity. I would also assume that a simpler solution would also be of interest.

1. Statement of the Problem: Fermat's Last Theorem

Given x, y, z, relatively prime, n odd prime

No solution exists for the equation x n + y n = z n

2. Definitions:

f = (z - y), g = (z - x), h = (x + y);

Now using the substitution for y with n = 3 gives

x 3 = 3fy 2 + 3f 2 y + f 3

from which it is obvious by inspection that x 3 is divisible by f only once, except for possibly one other factor of 3. In general, that also requires that those factors of f be raised to the nth power except for the n factor. So now, using a,b,c for the other factors

x = af or naf, y = bg or nbg, z = ch or nch

x + y = h n or n {n - 1}h n, z - x = g n or n {n - 1} g n, and z - y = f n or n {n - 1} f n.

now in general, (x + y - z) n = n(z - x)(z - y)(x + y) Q + x n + y n - z n

where Q represents all those other terms that are hard to write out for the general case.

For n = 3 it is one. And for n = 5

Q = z 2 - (x + y) z + x 2 + xy + y 2

I also use q where Q = q n or n {n - 1} q n

Finally, I use a term p, where nfgp = (f + g) n - (f n + g n)

From which it is obvious that p is divisible by (f + g)


now by Fermat's Little Theorem, since x n + y n = z n,

x n - (z - y) = 0 (mod n), therefore x n - f n = 0 (mod n), x - f = 0 (mod n)

and also y - g = 0 (mod n) and z - h = 0 (mod n)

And therefore that (a - 1) = 0 (mod n) with similar relations for b and c


3. General Proof:

Now from before remembering that x n + y n = z n

(x + y - z) n = n (z - x)(z - y)(x + y) Q which gives

(x + y - z) = nfghq

To use the above it is easier to consider individual cases where one of x,y or z is divisible by n or none of them are. Note that if z is divisible by n, then (x + y) and therefore (f + g) must be divisible by n, as explained in the definitions section.

a. Case where z is divisible by n:

x + y - z = x - f n ; y + x - z = y - g n; x + y - z = n {n - 1} h n - z = nfghq

Adding them gives

x + y -z - f n - g n + n {n - 1}hn = 3nfghq

Which is just

-f n - g n + n {n - 1} h n = 2nfghq

From which I get

f n + 2nfghq + g n = n {n - 1} h n and subtracting f n + nfgp + g n = (f + g) n:

nfg (p - 2hq) = (f + g) n - n {n - 1} h n

Neither f or g or q can be divisible by n, because that would mean that x or y would have to share a factor of n with z which violates the condition of relative primeness. Therefore, the above requires that h be further divisible by n.


The same comes up with x or y divisible by n since you get

g n + 2nfghq - h n = n {n - 1} f n subtracting

g n + nghp - h n = (g - h) n

gives

ngh (p - 2fq) = (g - h) n - n {n - 1} f n

which requires that f, g, h or q be divisible by n because (g - h) is divisible by n which again requires that f be further divisible by n for the reasons stated before.


4. Proof for Case x, y, z not divisible by n
lemma(1): Extension of Fermat's Little Theorem:

Given a - b divisible by n, a n - b n must be divisible by n 2

Proof: a - b divisible by n implies that a equals some jn + r and b equals some kn + r

Then,

a n - b n = (jn) n +...+ n (jn) r {n - 1} + r n - [(kn) n +...+ n (kn) r {n - 1} + r n]

by inspection r n can be subtracted off and the other terms are multipled by n 2


So by lemma(1) if neither x, y nor z are divisible by n and q is then

(x + y) n - (x n + y n) must be divisible by n 2.

But since f has the same modulus with respect to n as x and g has the same as y, it also requires that (f + g) n - (f n + g n) be divisible by n 2.

and from before

(f + g) n - (f n + g n) = nfgp

and p is seen to be required to be divisible by n

But like before I can write

f n + 2nfghq + g n = h n and again subtract f n + nfgp + g n = (f + g) n

which gives:

nfg (p - 2hq) = (f + g) n - h n

and since both sides are divisible by n 2, q is then forced to be further divisible by n.


5. Case for (x + y - z) divisible by n 2 or higher powers of n

Now then, these cases force me to rewrite my expression for (x + y - z). Considering the situation with divisibility by n 2, I have

(x + y - z) = (n 2) fghq


But still for my other relations at least two aren't divisible by n. Considering the possibility x, y not divisible by n

x + y - z = af - f n = (n 2) fghq

This requires that af - f n be divisible by n 2. But if I rewrite using moduli

f = kn + r a = mn + 1 giving (kn + r)(mn + 1) - (kn + r) n

Expanding (kn + r) n all terms are divisible by n 2 except the last r n term

Now (kn + r)(mn + 1) = km (n 2) + (rm + k)n + r , r < n

which requires that (rm + k) be divisible by n and that r n - r be divisible by n 2


But using s = n - r gives n n - ... + n (n) s {n-1} - s n - (n - s)

Which it can be seen requires that s n - s + n must be divisible by n 2


So then by observation s n - s can't then be divisible by n 2

But kn + r can be written as (k + 1)n + (r - n) = (k + 1) n - s

Expanding [(k + 1) n - s] n all terms are divisible by n 2 except the last s n term

and like before [(k + 1) n - s](mn + 1) = (k + 1) m (n 2) - [sm - (k + 1)] n - s , s < n

and as before [sm - (k + 1)] must be divisible by n and s n - s must be divisible by n 2.


So I try, f = k (n 2 )+ r n, a = m (n 2) + 1

Now if y is also not divisible by n and I have z divisible by n, then I can also write

bg - g n = (n 2) fghq

And the same argument applies so that g and b must be in the same form as f and a.

For instance, g = j (n 2) - r n, from which f + g = (j + k)(n 2)

But then (f + g) and therefore p would have to be divisible by (n 2) and I would have

f n + 2(n 2) fghq + g n = n {2n - 1} h n and subtracting

f n + nfgp + g n = (f + g) n giving

nfg (p - 2nhq) = (f + g) n - n {2n - 1} h n

which would require both sides to be divisible by n{2n - 1}.

Since p is divisible by n 2 this still requires that h be further divisible by n, and a similar argument applies for x or y divisible by n 2 like before.

Also, if neither x, y nor z is divisible by n, then f, g and h would all have to be of the above format and a similar expression would require that q be further divisible by n.

In general, whatever power t of n I use, p will be divisible by n to that factor and the term on the other side will always have to be further divisible by n.

nfg [p - 2(n {t - 1}) hq] = (f + g) n - n{tn - 1} h n


Then it can be seen that n has to always continuously be raised to a higher power and since there are an infinite number of n's there is no integer solution, and Fermat's Last Theorem is proven.


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