*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})

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}}h^{n} = 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.