Carmichael's totient function conjecture: Difference between revisions

Revision as of 18:23, 17 September 2008

Revision as of 18:24, 17 September 2008

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Carmichael proved that any counterexample to his conjecture (that is, a value ''n'' such that φ(''n'') is different from the totients of all other numbers) must be at least 10³⁷, and [[Victor Klee]] extended this result to 10⁴⁰⁰. Carmichael's conjecture has since been verified computationally for all ''n'' less than or equal to 10^10⁷ by Schlafly and Wagon. The current lower bound for a counterexample to the Conjecture is 10^10¹⁰, which was determined by Kevin Ford in 1998. Ford went on to prove that if there exists a counterexample to the Conjecture, then a positive fraction (that is infinitely many) of the integers are likewise counterexamples.

Carmichael's Conjecture is one of the first among open problems with very high lower bounds and which are relatively easy to determine. The computational technique basically depends on some key results of Klee which makes it possible to show that the smallest counterexample must be divisible by squares of the primes dividing it. Klee's results imply that 8 and Fermat primes (primes of the form 2^''k''</sup)+1) excluding 3 do not divide the smallest counterexample. Consequently, proving the conjecture is equivalent to proving that the conjecture holds for all integers congruent to 4 (''mod'' 8)

Although the conjecture is widely believed, [[Carl Pomerance]] gave a sufficient condition for an integer ''n'' to be a counterexample to the conjecture. According to this condition, ''n'' is a counterexample if for every prime ''p'' such that ''p''&nbsp;&minus;&nbsp;1 divides &phi;(''n''), ''p''² divides ''n''. However Pomerance showed that the existence of such an integer is highly improbable. Essentially, one can show that if the first ''k'' primes ''p'' congruent to 1 (''mod q'') (where ''q'' is a prime) are all less than ''q''^''k''+1</sup>, then such an integer will be divisible by every prime and thus cannot exist. In any case, proving that Pomerance's counterexample does not exist is far from proving Carmichael's Conjecture. However if it exists then infinitely many counterexamples exist as asserted by Ford.