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Can someone prove this for me (reward) (1 Viewer)

Will

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Can someone prove for me the following:
show there are no non-zero integers a, b, c, n with n > 2 such that a^n + b^n = c^n

I will give $50 to anyone who can prove it for me. It's urgent, thanks if you can!

(tehehe for anyone who got the allusion)
 

Lazarus

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What's the fun in that, Bon? I can't give away all the secrets. :p
 

Lazarus

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It's been proved for n = 1, n = 2, n = 3, n = 4 etc and I think all the way up to n = 167 or something, but not for all natural numbers n.
 

Lazarus

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Then I stand corrected. I think I was confusing it with Goldbach's conjecture.
 

Will

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ok good stuff peoples
anyone wanna prove riemann's hypothesis? i'll give you $150 for that
 

Will

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the riemann hypothesis can be neither proved or disproved
there's a 7 million dollar prize for solving it though :D
 

Will

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lol, funny that
you know sir, my interest in fermat's last theorem stemmed from visiting your website months ago
 

KeypadSDM

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Originally posted by Will
the riemann hypothesis can be neither proved or disproved
there's a 7 million dollar prize for solving it though :D
Nah, it's only a 1 million prize. 7 million for 7 different prizes, making 1 million each.
 

turtle_2468

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It was US though... which was a lot better a year ago than now, but still.
And btw I think Goldbach is now worth nothing, faber & faber offered 2 million on it for a while but then that expired last year...

also note that the poincare conjecture (one of the 7 problems) has a proof which seems correct, I think drbuchanan pointed it out before as well..
 

ND

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Actually he had posted in this thread. I don't know what happened to his posts...

The allusion is to drbuchanan. I think this whole thread is a mockery of him.
Check the date, this thread was posted long, long ago.
 
Last edited:

Grey Council

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o

who made this alive? And Buchanan deletes his posts, for some reason that escapes me. :|
 

+Po1ntDeXt3r+

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Originally posted by drbuchanan

Prove that if n is a positive integer and
d(n)=sum of the positive divisors of n and
h(n)=1/1+1/2+1/3+...+1/n then
d(n)<=h(n)+e<sup>h(n)</sup>ln(h(n)).
Stick tat as Question 9 on a 4 u exam.. dun tell the students there is no answer.. and go claim the prize "for 1 million dollars"... evil NSW Board of studies examiners ...
 

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