# Fermat Last Theorem for 13,17,19

I am asked to prove the Case I of Fermatâ€™s Last Theorem for the exponents 13, 17 and 19. FLT states that the equation $x^n+y^n=z^n$ has no integer solutions for and $x,y,z$ different from zero. Case I states that in oder for the theorem to hold we need that none of the three values $x,y,z$ is divisible by $p$ (prime ie 13,17,19 ).

i thought about using Sophie Germain theorem: If $x^n + y^n = z^n$ and $n$ is a prime $\ge 3$ and $2n+1$ is a prime, then $n$ must divide $xyz$. but this doesn't work for neither 13, 17, 19

So i tried to assume that $x,y,z$ are relatively prime to each other and assume that 13 doesn't divide $xyz$.

Since 13 is odd, we can set $z'$ to $-z$ and get: $x^{13} + y^{13} + (z')^{13} = 0$.

I can rewrite this into $-x^{13} = (y+z)(y^{13-1} - y^{13-2}z + \ldots - yz^{13-2} + z^{13-1})$ but even from there I couldn't get any result. Is there any method to do this?

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Write your formulas inside  to use TeX. – Sigur Feb 4 '13 at 19:35
There is a uniform proof for all regular primes (which 13, 17, 19 all are) by Kummer which you can find in Marcus number fields. – user58512 Feb 4 '13 at 19:40
I know that Kummer proved it holds for all regular primes ( but the proof wasn't discussed during the course due to his mathematical topics beyond course ). As in this case I am asked to show the first case I think I have to find a link with Sophie's Germain method. – Lola Feb 4 '13 at 19:57
there is some notes about it here fermatslasttheorem.blogspot.co.uk/2005/08/sophies-proof.html but as you said they are not sophie germain primes, I don't know what adaptation needs to be made. – user58512 Feb 4 '13 at 20:05
I had this idea. Maybe they want me to PROVE they are regular primes. hence I can use Lame' method ( which has been showed in the course) and show that they have unique factorization in $\mathbb{Z}[\sqrt{-p}]$.and after I proved they are regular primes I can tell that Kummer theorem holds. – Lola Feb 4 '13 at 20:09