PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (192)
Orphanage
Unclass'd (1)
Unproven (490)
Corrections (5)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Low Entry average rating: Very high
[parent] example of Fermat's last theorem (Example)

Fermat stated that for any $ n > 2$ the Diophantine equation $ x^n+y^n=z^n$ has no solution in positive integers. For $ n=4$ this follows from the following

Theorem 1   $ x^4+y^4 =z^2$ has no solution in positive integers.
Proof. Suppose we had a positive $ z$ such that $ x^4+y^4=z^2$ holds. We may assume $ \gcd(x,y,z)=1$. Then $ z$ must be odd, and $ x,y$ have opposite parity. Since $ (x^2)^2 +(y^2)^2 =z^2$ is a primitive Pythagorean triple, we have
$\displaystyle x^2=2pq, y^2 =q^2-p^2, z=p^2+q^2$ (1)

where $ p,q \in {\mathbb{N}}$, $ p<q$ are coprime and have opposite parity. Since $ y^2+p^2=q^2$ is a primitive Pythagorean triple, we have coprime $ s,r \in {\mathbb{N}}$, $ s<r$ of opposite parity satisfying
$\displaystyle q=r^2+s^2, y=r^2-s^2, p=2rs.$ (2)

From $ \gcd(r^2, s^2)=1$ it follows that $ \gcd(r^2, r^2+s^2)=1=\gcd(s^2, r^2+s^2)$, which implies $ \gcd(rs, r^2+s^2)=1$. Since $ \left(\frac{x}{2}\right)^2 = \frac{pq}{2} = rs(r^2+s^2)$ is a square, each of $ r,s,r^2+s^2$ is a square.

Setting $ Z^2 =q$, $ X^2 =r$, $ Y^2=s$ $ q=r^2+s^2$ leads to

$\displaystyle Z^2=X^4+Y^4$ (3)

where $ Z^2=q<q^2+p^2=z <z^2$. Thus, equation 3 gives a solution where $ Z< z$. Applying the above steps repeatedly would produce an infinite sequence $ z > Z > z_2 > \ldots$ 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 $ z$ for which (1) is solvable. So there are no solutions in positive integers for this equation. $ \qedsymbol$

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.



"example of Fermat's last theorem" is owned by Thomas Heye. [ full author list (2) ]
(view preamble)

View style:

See Also: infinite descent, $x^4-y^4=z^2$ has no solutions in positive integers


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: rational, sides, right triangle, area, consequence, solvable, sum, sequence, infinite, equation, square, implies, coprime, primitive Pythagorean triple, odd, integers, positive, solution, Diophantine equation
There are 3 references to this entry.

This is version 6 of example of Fermat's last theorem, born on 2004-02-16, modified 2008-06-25.
Object id is 5588, canonical name is ExampleOfFermatsLastTheorem.
Accessed 2179 times total.

Classification:
AMS MSC: 11D41 (Number theory :: Diophantine equations :: Higher degree equations; Fermat's equation)
 14H52 (Algebraic geometry :: Curves :: Elliptic curves)
 11F80 (Number theory :: Discontinuous groups and automorphic forms :: Galois representations)
Pending Errata and Addenda
None.
[ View all 4 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)