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 2006-06-06, 02:27 #1 devarajkandadai     May 2004 31610 Posts 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
 2006-06-07, 21:29 #2 akruppa     "Nancy" Aug 2002 Alexandria 2,467 Posts 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
2006-06-16, 08:50   #3

May 2004

22·79 Posts

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
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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