# Fermat's Theorem

• Feb 16th 2013, 04:44 PM
Gibo
Fermat's Theorem
Hi,

I have difficulties with below problem:

Let $p$ and $\theta$ be primes with $\theta>p$ such that $p$ is not a factor of $\theta - 1$. As $\theta$ is a prime, we know that for any $x\epsilon Z$ we can write $x^{\theta-1}$ in the form $x^{\theta-1} = k\theta+1$ for some integer $k$. Furthermore, since the greatest common divisor of $p$ and $\theta-1$ is $1$, we can write $1=ap+b(\theta-1)$ for some integers $a$ and $b$.
- Show that every integer $x$ can be written in the form $x=y^p+j\theta$ for some integer $y$ and some integer $j$.
- Prove the First Case of Fermat's Last Theorem for the exponents $13,17$ and $19$. (That is, show that if there is a nonzero integer solution to $x^p+y^p=z^p$ for $p=13$ then $13$ is a factor of $xyz$. and so on.)
Any help would be highly appreciated.
Thank you.
• Feb 16th 2013, 05:06 PM
HallsofIvy
Re: Fermat's Theorem
• Feb 16th 2013, 05:12 PM
HallsofIvy
Re: Fermat's Theorem
Quote:

Originally Posted by Gibo
Hi,

I have difficulties with below problem:

Let $p$ and $\theta$ be primes with $\theta>p$ such that $p$ is not a factor of $\theta - 1$. As $\theta$ is a prime, we know that for any $x\epsilon Z$ we can write $x^{\theta-1}$ in the form $x^{\theta-1} = k\theta+1$ for some integer $k$.

Are your sure we know this? If, for example, $\theta= 3$ and $x= 6$, this says that $6^{3-1}= 6^2= 36= 3k+ 1$. And I don't believe this is true.

Quote:

Furthermore, since the greatest common divisor of $p$ and $\theta-1$ is $1$, we can write $1=ap+b(\theta-1)$ for some integers $a$ and $b$.
- Show that every integer $x$ can be written in the form $x=y^p+j\theta$ for some integer $y$ and some integer $j$.
- Prove the First Case of Fermat's Last Theorem for the exponents $13,17$ and $19$. (That is, show that if there is a nonzero integer solution to $x^p+y^p=z^p$ for $p=13$ then $13$ is a factor of $xyz$. and so on.)
Any help would be highly appreciated.
Thank you.
• Feb 17th 2013, 12:41 AM
Gibo
Re: Fermat's Theorem
Thank you very much for pointing that, I found it really helpful.