20191020, 13:20  #1 
May 2017
ITALY
5^{2}×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? 
20191020, 15:19  #2  
May 2017
ITALY
1DB_{16} Posts 
Quote:
n<=2 

20191020, 16:38  #3  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·19·241 Posts 
Quote:
So, either you just proved that 3 is an "imaginary" number  or your proof is nonsense. And of course, the second is correct. 

20191020, 17:20  #4  
May 2017
ITALY
5^{2}·19 Posts 
Quote:
n<=2 > does not admit imaginary solutions https://www.wolframalpha.com/input/?...n+%2Cn%3D2%2Ch n>=3 > admit imaginary solutions https://www.wolframalpha.com/input/?...n+%2Cn%3D3%2Ch where am I wrong? 

20191020, 17:21  #5 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×7×11×29 Posts 
Haha, Serge, I can't believe you clicked that link

20191020, 17:43  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9158_{10} Posts 
I am a seasoned MonteCarlo sampler. TL;DR version of biasedprobability 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).

20191020, 18:27  #7 
May 2017
ITALY
5^{2}·19 Posts 
If it can help someone this
h^t^t+k^t^t=(N^t/2n/2)^t+(N^t/2+n/2)^t=N^t^t 
20191020, 21:06  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·19·241 Posts 
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 nth degree root have? I'll give you a hint: it is not one. 
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