A question on divisibility of a product of primes - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:20:47Z http://mathoverflow.net/feeds/question/81816 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81816/a-question-on-divisibility-of-a-product-of-primes A question on divisibility of a product of primes Ralph 2011-11-24T16:02:01Z 2011-11-24T18:45:10Z <p>I'm not sure if this question has the appropriate level for MO. If not, feel free to vote for closing. </p> <p>Let $m \ge 2$. Are there odd primes $p_1 \le ... \le p_m$ and non-negative integers $n_1,...,n_m$ such that $$\prod_{i=1}^m (p_i^{n_i}-1) \quad \text{ divides }\quad (\prod_{i=1}^m p_i^{n_i})-1 \quad\quad ?$$</p> <p><strong>Background:</strong> If there aren't such primes, that would answer a conjecture in another question affirmatively (if one requires $|A|$ to be odd): </p> <p><a href="http://mathoverflow.net/questions/81799/a-conjecture-on-finite-commutative-rings" rel="nofollow">http://mathoverflow.net/questions/81799/a-conjecture-on-finite-commutative-rings</a></p> <p>So far, I could show that for $m=2$ the described constellation isn't possible. By writing $$\prod_{i=1}^m a_i \quad \text{divides}\quad \prod_{i=1}^m (a_i + 1) -1 \quad ?$$ , I think the left hand side and the right hand side are too close together such that $\prod_{i=1}^m a_i$ can divide the difference. But I may be wrong.</p> <p><strong>Edit:</strong> Thank you all very much for the comments. </p> <p>My statement about $m=2$ wasn't quite correct, since $(3 -1) \cdot (3-1) \mid 3 \cdot 3 -1$ is a (the only) solution. </p>