 Contents Mathematics Fermat Physics Wordlist Notes Chemistry Quick Facts Reference LINKS

# 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.

Contents Mathematics Fermat Physics Wordlist Notes Chemistry Quick Facts Reference LINKS

Visit Sciencepages 2 for more mathematics

Home