example of Fermat’s last theorem
Fermat stated that for any the Diophantine equation has no solution in positive integers. For this follows from the following
Theorem 1.
has no solution in positive integers.
Proof.
Suppose we had a positive such that holds. We may assume . Then must be odd, and have opposite parity. Since is a primitive Pythagorean triple, we have
(1) |
where , are coprime and have opposite parity. Since is a primitive Pythagorean triple, we have coprime , of opposite parity satisfying
(2) |
From it follows that , which implies . Since is a square, each of is a square.
Setting , , leads to
(3) |
where . Thus, equation 3 gives a solution where . Applying the above steps repeatedly would produce an infinite sequence of positive integers, each of which was the sum of two fourth powers. But there cannot be infinitely many positive integers smaller than a given one; in particular this contradicts to the fact that there must exist a smallest for which (1) is solvable. So there are no solutions in positive integers for this equation. ∎
A consequence of the above theorem is that the area of a right triangle with integer sides is not a square; equivalently, a right triangle with rational sides has an area which is not the square of a rational.
Title | example of Fermat’s last theorem |
---|---|
Canonical name | ExampleOfFermatsLastTheorem |
Date of creation | 2013-03-22 14:09:51 |
Last modified on | 2013-03-22 14:09:51 |
Owner | Thomas Heye (1234) |
Last modified by | Thomas Heye (1234) |
Numerical id | 9 |
Author | Thomas Heye (1234) |
Entry type | Example |
Classification | msc 11F80 |
Classification | msc 14H52 |
Classification | msc 11D41 |
Related topic | InfiniteDescent |
Related topic | X4Y4z2HasNoSolutionsInPositiveIntegers |