Fermat's Last Theorem (FLT for short)
The equation x^n + y^n = z^n has no solution for non-zero integers x, y, and z if n is an integer greater than 2.
Proof
We shall denote the greatest commom divisor of the numbers x and y by (x, y).
Suppose that d^n + e^n = f^n
with any integer n greater than 2, then we may always assume that
0 < d < e < f (1)
d + e > f (2)
(d, e) = (e, f) = (f, d) = 1 (3)
From (3) and
it follows that there only exist three following cases:
Case 1: f is an even number and d, e are odd numbers.
Case 2: d is an even number and f, e are odd numbers.
Case 3: e is an even number and f, d are odd numbers.
Obviously, with the above suppositions we have
d^2 + e^2 > f^2
e^2 + f^2 > d^2
f^2 + d^2 > e^2
Thus, the numbers a = d^2, b = e^2, c = f^2 can represent as three sides of a triangle ABC from a geometrical point of view.
In the triangle ABC whose sides are represented by a, b, c we have
c^2 = a^2 + b^2 - 2abcosC (4)
a^2 = b^2 + c^2 - 2bccosA (5)
b^2 = c^2 + a^2 - 2cacosB (6)
4S^2 = (absinC)^2 = (bcsinA)^2 = (casinB)^2 (7)
S^2 = p(p-a)(p-b)(p-c) (8)
in which S is the area and 2p is the perimeter of the triangle ABC.
We shall first consider case 1.
(4) can be written as
abcosC = (a^2 + b^2 - c^2)/2
In this case, we have p, (p-c), abcosC are odd numbers and (p-a), (p-b), (a^2 + b^2 – c^2), S2 are even numbers.
If n > 4, from
it follows that a^2 + b^2 – c^2 > 0 then 0 < cosC < 1 and we can put
cosC = u/v in which (u, v) = 1 and 0 < u < v
Then abcosC = abu/v (9)
Since abcosC is an integer and (u, v) = 1, ab must be divisible by v, i.e. ab has the form
ab = kv where k is a positive integer (10)
Since abcosC is an odd number, from (9) and (10) we have ku is an odd number. Hence, k, u are odd numbers.
Otherwise, we have kv is an odd number. Hence, k, v are odd numbers.
(7) can be written as
4S^2 = (ab)^2(sinC)^2
= (ab)^2[1 - (cosC)^2]
= (ab)^2[1 - (u/v)^2]
= (ab/v)^2(v^2 - u^2)
= k^2(v^2 - u^2)
It follows that v^2 - u^2 = 4S^2/k^2 (11)
Since (4, k^2) = 1, S^2 must be divisible by k^2. Therefore, we can put
S^2 = mk^2 where m is a positive integer
And (11) becomes
v^2 - u^2 = 4m (12)
Notice that S^2 is even then m is, too.
By putting m = 2^i*t in which t is an odd number and i is a positive integer then (12) can be rewritten as
v^2 - u^2 = 42^i*t (13)
Solving (13) with respect to v, u we obtain
v = 2^i + t , u = 2^i - t if 2^i > t (14)
or v = t + 2^i , u = t - 2^i if t > 2^i (15)
And v = 2^i*t + 1 , u = 2^i*t - 1 (16)
It is evident that the above conditions of v, u are satisfied.
With the solution (16), we have
(p-a)(p-b) = [(b+c-a)(c+a-b)]/4
= [2ab(1-cosC)]/4
= [kv(1-u/v)]/2
= k(v-u)/2
= k(2)/2
= k (17)
This is unreasonable because the right hand side of equation (17) is an odd number while its left hand one is an even number. From here it follows that cosC must equal 1 and then C = 0. This means that the triangle ABC does not exist.
Similiary for the solution (14), we have the fact that the triangle ABC does not exist.
With the solution (15), it follows that the triangle ABC exists.
Consider the following propositions:
(H): d^n + e^n = f^n where d, e, f, n are positive integers and n greater than 2.
(K): the triangle ABC exists.
It is clear that from (H) it follows that (K) exists and non-(K) exists. Therefore, (H) is a false proposition. Consequently, FLT holds true in this case.
Similiary, if n =3, from
it follows that a^2 + b^2 – c^2 < 0 then -1 < cosA < 0 and we can put
cosC = - u/v in which (u, v) = 1 and 0 < u < v
In case the solutions (16) and (14), it follows that cosC must equal -1 and then C = 180 degrees. This means that the triangle ABC does not exist.
If n = 4, from
it follows that a^2 + b^2 – c^2 = 0 then cosC = 0, C = 90 degrees. That’s the reason why Fermat gave another method.
Similiary, FLT also holds true in the other cases when we begin with the equality (5) in case 2 and the one (6) in case 3.
FLT is now proved completely.
