2
$\begingroup$

Hi all. There is a conjecture by Erdős that states "There doesn't exist an integer covering system with all moduli odd AND distinct." The link is http://en.wikipedia.org/wiki/Covering_system

I think that, if a covering exists with all moduli distinct, not only the lcm of the moduli must be even, it must also be abundant. In this sense, the odd perfect number conjecture and this conjecture are related in my opinion.

Is my conjecture true or any counterexample exists? Has the same conjecture been made before? Any help would be appreciated. Thanks in advance.

$\endgroup$
5
$\begingroup$

Each modulus $m$ covers a proportion $1/m$ of the integers. To cover all the integers, we need $\sum 1/m\ge1$, where the sum is over all the moduli. Multiply by the lcm, $L$, to get $\sum(L/m)\ge L$. But all the terms in the sum are distinct, proper divisors of $L$, so $L$ is abundant (unless we have equality, but it's known that a distinct cover can't be exact, so in fact all the inequalities are strict).

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ One should add that there are infinitely many odd abundant numbers; the smallest is 945. So this observation does not prove Erdos's conjecture. $\endgroup$ – David E Speyer Sep 6 '11 at 13:33
  • $\begingroup$ In fact, for covering systems with the moduli square-free and odd one requires at least 22 distinct prime divisors by a result of Song Guo and Zhi-Wei Sun. But 3*5*7*11*13 has only five prime divisors and is already abundant. $\endgroup$ – Pace Nielsen Sep 6 '11 at 15:40
  • $\begingroup$ @ Gerry Myerson: Thanks a lot, your observation clarified on why the lcm of moduli must be abundant. @ David Speyer: Point taken. @ pace Nielson: True, but does there exist a non-trivial covering system with 3*5*7*11*13 as lcm? $\endgroup$ – nb1 Sep 7 '11 at 4:48
  • $\begingroup$ Did you read what Pace wrote? "...for covering systems with the moduli square-free and odd one requires at least 22 distinct prime divisors..." $\endgroup$ – Gerry Myerson Sep 7 '11 at 6:04
  • $\begingroup$ Oh, true, mea culpa. $\endgroup$ – nb1 Sep 7 '11 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.