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 2019-10-20, 13:20 #1 Alberico Lepore   May 2017 ITALY 52×19 Posts Proof of Fermat's last theorem h^n+k^n=N^n with N even Proof of Fermat's last theorem h^n+k^n=N^n with N even solve (N^n/2)^2-[((N^n/2)-h^n)]^2=h^n*k^n ,h h=e^(-(i π)/n)* (k^n - N^n)^(1/n) h is immaginary paper: https://www.academia.edu/40675631/Pr..._n_with_N_even What do you think?
2019-10-20, 15:19   #2
Alberico Lepore

May 2017
ITALY

1DB16 Posts

Quote:
 Originally Posted by Alberico Lepore Proof of Fermat's last theorem h^n+k^n=N^n with N even solve (N^n/2)^2-[((N^n/2)-h^n)]^2=h^n*k^n ,h h=e^(-(i π)/n)* (k^n - N^n)^(1/n) h is immaginary paper: https://www.academia.edu/40675631/Pr..._n_with_N_even What do you think?
n>2
n<=2

2019-10-20, 16:38   #3
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2·19·241 Posts

Quote:
 Originally Posted by Alberico Lepore Proof of Fermat's last theorem h^n+k^n=N^n with N even solve (N^n/2)^2-[((N^n/2)-h^n)]^2=h^n*k^n ,h h=e^(-(i π)/n)* (k^n - N^n)^(1/n) h is immaginary paper: https://www.academia.edu/40675631/Pr..._n_with_N_even What do you think?
Take N=5, k=4, n=2. You attempt to demonstrate that h=3 "is immaginary".
So, either you just proved that 3 is an "imaginary" number - or your proof is nonsense.

And of course, the second is correct.

2019-10-20, 17:20   #4
Alberico Lepore

May 2017
ITALY

52·19 Posts

Quote:
 Originally Posted by Batalov Take N=5, k=4, n=2. You attempt to demonstrate that h=3 "is immaginary". So, either you just proved that 3 is an "imaginary" number - or your proof is nonsense. And of course, the second is correct.
N even

n<=2 -> does not admit imaginary solutions

https://www.wolframalpha.com/input/?...n+%2Cn%3D2%2Ch

https://www.wolframalpha.com/input/?...n+%2Cn%3D3%2Ch

where am I wrong?

 2019-10-20, 17:21 #5 LaurV Romulan Interpreter     Jun 2011 Thailand 22×7×11×29 Posts Haha, Serge, I can't believe you clicked that link
2019-10-20, 17:43   #6
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

915810 Posts

Quote:
 Originally Posted by LaurV Haha, Serge, I can't believe you clicked that link
I am a seasoned Monte-Carlo sampler. TL;DR version of biased-probability MC: when the search space is large, visit partitions with best research yield most of the time, but visit even hopeless partitions (albeit very rarely).

 2019-10-20, 18:27 #7 Alberico Lepore   May 2017 ITALY 52·19 Posts If it can help someone this h^t^t+k^t^t=(N^t/2-n/2)^t+(N^t/2+n/2)^t=N^t^t
2019-10-20, 21:06   #8
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2·19·241 Posts

Quote:
 Originally Posted by Alberico Lepore If it can help someone
It can't.
Ex falso quodlibet.
if you say something meaningless - it is in many senses worse than to say nothing at all.

I will make just one attempt to help you. Think about this: how many values does an n-th degree root have? I'll give you a hint: it is not one.

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