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.

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)

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:

(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)

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.