Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am reading this paper by Cohen and Selfridge that deals with covering systems. Its link is

http://www.ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/S0025-5718-1975-0376583-0.pdf

Here, they demonstrate the existence of a number that is not the sum or difference of two primes powers; using covering systems.

First, they construct the covering systems and obtain some congruences that demonstrate that the given number is not the sum or difference of a power of 2 and a prime. This I understand without any difficulty.

Then, for proving that the number such obtained is not the sum of a power of two and a POWER of prime,they introduce some extra congruences. The basic idea is that if one could show the reqd. number to be multiple of a product of two or more primes, then it won't be a prime power.

That is fine, but take a look at the primes in extra congruences. From where are these extra primes selected? e.g. on page 2 of the paper, we have:

$M + 2^8 \equiv 0\pmod{5^2 \times 11}$, $M + 2^{34} \equiv 0\pmod{97^2 \times 389}$, $M + 2^6 \equiv 0 \pmod{17^2\times 137}$, $M + 2^{18} \equiv 0\pmod{241^2\times 1447}$ $M+ 2^2 \equiv 0 \pmod{13^2 \times 53}$ $M + 2^{10} \equiv 0 \pmod{257\times 673}$

In the above congruences, the primes 5,97,17,241,13,257 all come from the covering (1,2), (0,4), (6,8), (10,12), (2,48), (10,16), (18,24).

But where do the primes 11, 389, 137, 1447, 53, 673 etc. come? The same question I have regarding the extra congruences for the other covering given in the paper.

The authors have given no reasoning for the same in the paper.

I checked the prime factors of $2^{48}-1$ & $2^{180}-1$ as well. The additional primes in both cases are different from those.

Hence the question: what is the rationale behind selecting the primes in the extra congruences?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.