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Old 2006-06-06, 02:27   #1
devarajkandadai
 
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May 2004

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Default Fermat's Theorem-tip of the iceberg?

For a glimpse of what lies beneath the tip you may visit
www.crorepatibaniye.com/failurefunctions
and read the third conjecture.Tks to those taking the trouble.
A.K.Devaraj
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Old 2006-06-07, 21:29   #2
akruppa
 
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You can rewrite as \large \left(a^b^n\right)^{b^{k \Phi(\Phi(m))}} + c = (m-c)^{b^{k Phi(Phi(m))}} + c

For odd b, the right hand side is obviously ≡ 0 (mod m) as the lone c cancel. For even b, the term will be ≡ 2c (mod m).

Alex
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Old 2006-06-16, 08:50   #3
devarajkandadai
 
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Quote:
Originally Posted by akruppa
You can rewrite as \large \left(a^b^n\right)^{b^{k \Phi(\Phi(m))}} + c = (m-c)^{b^{k Phi(Phi(m))}} + c

For odd b, the right hand side is obviously ≡ 0 (mod m) as the lone c cancel. For even b, the term will be ≡ 2c (mod m).

Alex
Thank you for your kind comments.My point,however, was to emhasise
the layered structure and to point out that Fermat's theorem is only the tip.In other words, we shd not be surprissed if


a^(b^d^n + k*phi(phi(phi(m)))) + c is also congruent to 0 (mod m)

subject to a^(b^d^n) + c being = m.

A.K.Devaraj
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