# Application of Fermat's last theorem

• Mar 31st 2011, 07:39 AM
yamahabob
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?

Edit:
It can also be noted that for
n->odd : -∞ <x < ∞ -∞ < y < ∞
n->even -1< x < 1 -1 < y < 1
• Mar 31st 2011, 12:37 PM
tonio
Quote:

Originally Posted by yamahabob
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?

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 , 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
• Mar 31st 2011, 01:07 PM
yamahabob
Quote:

Originally Posted by tonio
Suppose $\displaystyle{\left(\frac{a}{b}\,,\,\frac{c}{d}\ri ght)\mbox{ is a solution to }x^n+y^n=1\,,\,\,2 , 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

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.
• Mar 31st 2011, 01:11 PM
tonio
Quote:

Originally Posted by yamahabob
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
• Mar 31st 2011, 01:24 PM
yamahabob
Quote:

Originally Posted by tonio
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.