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Old 2019-10-20, 13:20   #1
Alberico Lepore
 
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Default 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?
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Old 2019-10-20, 15:19   #2
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2019-10-20, 16:38   #3
Batalov
 
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Quote:
Originally Posted by Alberico Lepore View Post
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.
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Old 2019-10-20, 17:20   #4
Alberico Lepore
 
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Originally Posted by Batalov View Post
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

n>=3 -> admit imaginary solutions

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

where am I wrong?
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Old 2019-10-20, 17:21   #5
LaurV
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Haha, Serge, I can't believe you clicked that link
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Old 2019-10-20, 17:43   #6
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Originally Posted by LaurV View Post
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).
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Old 2019-10-20, 18:27   #7
Alberico Lepore
 
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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
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Old 2019-10-20, 21:06   #8
Batalov
 
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Originally Posted by Alberico Lepore View Post
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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