Fermat's great theorem short solution

a^n + b^n = c^n

We can suppose that numbers a,b,c are pairly comprime. (if numbers a and b have common divisor than с has it either and gcd(a,b,c)>1)

\frac {a^n}{cn} + \frac {b^n}{cn} = 1, n = 2k + 1
Suppose it has solutions a, b, c in positive integers, then exists \triangle ABC such that
AC = \frac {b^k}{ck}, BC = \frac {a^k}{ck} and AB = 1.

The strategy is to show that in the triangle ABC altitude CD is such that
AD = \frac {b^{2k + 1}}{c^{2k + 1}}, BD = \frac {a^{2k + 1}}{c^{2k + 1}} and get new quality from law of cosines which will lead to a contradiction.
(see attachment 1.)

Figure 1.
let there be a line l and points A_1, D_1 lying on l such that A_1D_1 = \frac {b^{2k + 1}}{c^{2k + 1}} (see attachment 2) We'll build a line e perpendicular l through point D_1.
C_1 is such point on e that A_1C_1 = \frac {b^k}{ck}
Let P_1 be such point on A_1C_1 that A_1P_1 = \frac {b^{k + 1}}{c^{k + 1}}, B_1 is such point that B_1 lies on line l and B_1P_1 \bot A_1C_1. It is easy to note that \triangle P_1A_1B_1 \sim \triangle D_1A_1C_1.
Hence, \frac {A_1B_1}{A_1C_1} = \frac {A_1P_1}{A_1D_1}, A_1B_1 = \frac {b^{k + 1} b^k c^{2k + 1}}{c^{k + 1} c^k b^{2k + 1}} = 1

Figure 2.
By analogy, we'll build another configuration (see att.3) where
B_2C_2 = \frac {a^k}{ck}, B_2D_2 = \frac {a^{2k + 1}}{c^{2k + 1}}, B_2M_2 = \frac {a^{k + 1}}{c^{k + 1}}. From the similarity of triangles \triangle A_2M_2B_2 \sim \triangle C_2D_2B_2 follows equality A_2B_2 = 1.

From figures 1 and 2 it follows that this constructions coincide (AB = 1)... and the claim is prooved.

From the cosines law it follows:
BC^2 = AB^2 + AC^2 - 2(AB\cdot AC\cdot \cos \angle CAB) = AB^2 + AC^2 - 2(AB\cdot AD).
Finally it follows that
AD = \frac {AB^2 + AC^2 - BC^2}{2AB}
\frac {b^{2k + 1}}{c^{2k + 1}} = \frac {1 + \frac {b^{2k}}{c{2k}} - \frac {a^{2k}}{c{2k}}}{2}
b^{2k + 1} = c\cdot \frac {c^{2k} + b^{2k} - a^{2k}}{2}
b^{2k + 1} = c\cdot M , it is easy to note that M is a positive integer.

There is a contradiction since b and c are comprime positive integer numbers. So the equation doesn't have any solution if n is an odd number. The statement for even n follows from the inequality
a^4 + b^4 \neq c^4 which can be prooved in an easy way.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1055728