Carmichael's totient function conjecture: Difference between revisions

Revision as of 18:00, 17 September 2008

Revision as of 18:01, 17 September 2008

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The conjecture states that this phenomenon of repeated values holds for every ''n''. That is, for every ''n'' there is at least one other integer ''m'' ≠ ''n'' such that φ(''m'') = φ(''n'').

[[Robert Daniel Carmichael|Robert Carmichael]] first stated this conjecture 1907, but as a theorem rather than as a conjecture. However, his proof was faulty and in 1922 he retracted his claim and stated the conjecture as an open problem.

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. However if it exists then infinitely many counterexamples exist as asserted by Ford.

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.

Another way of stating Carmichael's conjecture is that, if