Results 1 to 5 of 5

Math Help - Application of Fermat's last theorem

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    3

    Application of Fermat's last theorem

    So here's my issue.
    I'm not trying to write a formal proof, I'm just trying to find all cases for this problem.

    Question: Assuming Fermat's Last Theorem (FLT), show that x^n+y^n=1 (n integer >2), contains no points with rational coordinates except those points where the curve crosses the axis.

    I have graphed the equation for an odd n and an even n. They both have a distinct shape so I think I need two main cases where n is even or odd with subcases for x,y being <0 and >0. It's easy to show that there are rational coordinates at the intersection of the axes.

    I then solved the equation for y. I thought it would be possible to show that if x was rational (positive or negative) that y=(1-x^n)^(1/n) makes y irrational. While I've read most nth roots are irrational, this doesn't seem conclusive enough to say y will always be irrational.

    Any help or direction to conquering this problem?

    Thanks in advance

    Edit:
    It can also be noted that for
    n->odd : -∞ <x < ∞ -∞ < y < ∞
    n->even -1< x < 1 -1 < y < 1
    Last edited by yamahabob; March 31st 2011 at 11:00 AM. Reason: more info
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by yamahabob View Post
    So here's my issue.
    I'm not trying to write a formal proof, I'm just trying to find all cases for this problem.

    Question: Assuming Fermat's Last Theorem (FLT), show that x^n+y^n=1 (n integer >2), contains no points with rational coordinates except those points where the curve crosses the axis.

    I have graphed the equation for an odd n and an even n. They both have a distinct shape so I think I need two main cases where n is even or odd with subcases for x,y being <0 and >0. It's easy to show that there are rational coordinates at the intersection of the axes.

    I then solved the equation for y. I thought it would be possible to show that if x was rational (positive or negative) that y=(1-x^n)^(1/n) makes y irrational. While I've read most nth roots are irrational, this doesn't seem conclusive enough to say y will always be irrational.

    Any help or direction to conquering this problem?

    Thanks in advance

    Edit:
    It can also be noted that for
    n->odd : -∞ <x < ∞ -∞ < y < ∞
    n->even -1< x < 1 -1 < y < 1


    Suppose \displaystyle{\left(\frac{a}{b}\,,\,\frac{c}{d}\ri  ght)\mbox{ is a solution to }x^n+y^n=1\,,\,\,2<n\in\mathbb{N}\,,\,\,a,b,c,d\in  \mathbb{Z} , then

    \displaystyle{\left(\frac{a}{b}\right)^n+\left(\fr  ac{c}{d}\right)^n=1\Longrightarrow (ad)^n+(cb)^n=(bd)^n , and this

    contradicts FLT unless the solution is trivial, so...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    3
    Quote Originally Posted by tonio View Post
    Suppose \displaystyle{\left(\frac{a}{b}\,,\,\frac{c}{d}\ri  ght)\mbox{ is a solution to }x^n+y^n=1\,,\,\,2<n\in\mathbb{N}\,,\,\,a,b,c,d\in  \mathbb{Z} , then

    \displaystyle{\left(\frac{a}{b}\right)^n+\left(\fr  ac{c}{d}\right)^n=1\Longrightarrow (ad)^n+(cb)^n=(bd)^n , and this

    contradicts FLT unless the solution is trivial, so...

    Tonio
    While I follow this example, I am confused about one thing.
    Consider the solutions
    \displaystyle{\left(\frac{-a}{b}\,,\,\frac{c}{d}\right)\mbox
    OR
    \displaystyle{\left(\frac{a}{b}\,,\,\frac{-c}{d}\right)\mbox
    OR
    \displaystyle{\left(\frac{-a}{b}\,,\,\frac{-c}{d}\right)\mbox
    I'm saying what if x<0 or y<0 or both <0? FLT only works for positive integers correct? I'm also assuming n is odd in this case.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by yamahabob View Post
    While I follow this example, I am confused about one thing.
    Consider the solutions
    \displaystyle{\left(\frac{-a}{b}\,,\,\frac{c}{d}\right)\mbox
    OR
    \displaystyle{\left(\frac{a}{b}\,,\,\frac{-c}{d}\right)\mbox
    OR
    \displaystyle{\left(\frac{-a}{b}\,,\,\frac{-c}{d}\right)\mbox
    I'm saying what if x<0 or y<0 or both <0? FLT only works for positive integers correct? I'm also assuming n is odd in this case.

    Who cares what the sign is if, by FLT, only the trivial solutions exist?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    3
    Quote Originally Posted by tonio View Post
    Who cares what the sign is if, by FLT, only the trivial solutions exist?

    Tonio
    Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation $a^n + b^n = c^n$.

    That makes me think I would need to do something else to show FLT works for negative values.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 10th 2011, 08:51 AM
  2. Replies: 1
    Last Post: March 9th 2009, 12:51 PM
  3. Fermat's little theorem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 23rd 2008, 06:37 PM
  4. Fermat's little theorem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: November 8th 2007, 07:28 AM
  5. fermat's little theorem application
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: May 1st 2007, 05:06 PM

Search Tags


/mathhelpforum @mathhelpforum